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Quadratic Equations: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 139 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. Two products already set against zero, and one factor that is bare . 12 points. Question 1 of 10.

Both equations below arrive as a product on one side and $0$ on the other, so no rearranging is needed. One of them has a number in front of the $x$ inside a factor, and the other has a factor that is nothing but $x$ .
1. Part A.
  Solve $(2x - 7)(x + 6) = 0$ , reporting both solutions.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Solve $x(3x + 12) = 0$ , reporting every solution.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Both parts relied on the same property, and that property names the number $0$ specifically. Explain what is true of $0$ that is true of no other number, and say what can and cannot be concluded from an equation whose product equals $12$ instead.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

$x = \tfrac{7}{2}$ or $x = -6$ .
- $x = 3.5$ is the same as $\tfrac{7}{2}$ ; what is not the same is $x = 7$ , which has not divided out the $2$ in front of $x$
Part B

$x = 0$ or $x = -4$ .
- Both values are needed; reporting only $x = -4$ has dropped the root that the bare factor $x$ carries
Part C

Zero is the only number a product can reach with no factor equal to it: two nonzero numbers always multiply to something nonzero. With $12$ on the other side nothing follows about either factor, since $12$ splits in endlessly many ways, so the equation must be expanded and re-standardized first.
Worked solution

Part A

A product is $0$ only when one of its factors is $0$ , so each factor gives its own linear equation.

$2x - 7 = 0 \quad \text{or} \quad x + 6 = 0$

The first gives $2x = 7$ , so $x = \tfrac{7}{2}$ ; the second gives $x = -6$ . A number in front of $x$ inside a factor still produces exactly one root, and dividing it out is what makes that root a fraction.

Part B

The two factors here are $x$ and $3x + 12$ , so set each to zero.

$x = 0 \quad \text{or} \quad 3x + 12 = 0$

The first factor hands over the root $x = 0$ directly, and the second gives $3x = -12$ , so $x = -4$ . The bare factor $x$ is a factor like any other, and dividing both sides by it instead would have thrown its root away.

Part C

What is special about $0$ . If a product $AB = 0$ and $A \ne 0$ , then multiplying both sides by $\tfrac{1}{A}$ leaves $B = 0$ :

$\tfrac{1}{A} \cdot (A \cdot B) = \tfrac{1}{A} \cdot 0 \quad\Longrightarrow\quad B = 0.$

So one factor is forced to vanish. No other number behaves this way.

Why $12$ settles nothing. From $AB = 12$ neither factor is pinned down, because $12$ has endlessly many factorizations, $3 \times 4$ and $2 \times 6$ and $\tfrac{1}{2} \times 24$ among them. An equation with $12$ on the right has to be expanded and gathered to standard form, with $0$ alone on one side, before anything can be split.

In one line

$(2x - 7)(x + 6) = 0$ gives $x = \tfrac{7}{2}$ or $x = -6$ , and $x(3x + 12) = 0$ gives $x = 0$ or $x = -4$ . The property works only against $0$ , because $0$ is the one number two nonzero factors can never produce; against $12$ nothing follows, since $12$ splits in endlessly many ways, so such an equation must first be expanded and gathered to standard form.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Sets each factor equal to zero separately rather than working on the product as a whole. . Worth 2 points.

Divides out the number in front of $x$ in the first factor and reports both roots, each with the sign that makes its own factor vanish. . Worth 2 points.

Part B 3 points

Treats the bare $x$ as a factor in its own right and keeps the root it produces. . Worth 2 points.

Solves the second factor correctly and reports both solutions. . Worth 1 point.

Part C 5 points

Identifies the property of zero that forces a factor to vanish, rather than restating the rule as something to be remembered. . Worth 3 points. needs an explanation, not just an answer

Says what a nonzero right-hand side does not license, and names the step that has to come first instead. . Worth 2 points.
2. One equation, and two ways in that need different things first . 15 points. Question 2 of 10.

The equation $3(x + 2)^2 - 75 = 0$ can be approached from two directions. One works on the squared group where it stands; the other flattens the equation out first. This question runs both and then puts them side by side.
1. Part A.
  Isolate the squared group and solve $3(x + 2)^2 - 75 = 0$ , reporting both solutions.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Expand the same equation into standard form $ax^2 + bx + c = 0$ , then substitute one of the values you found in part A and confirm the left side comes out to $0$ .
  
  Carry your own answer forward Substitute whichever value you reported in part A, even if it was not the expected one, and say honestly whether the left side reaches $0$ . The credit here is for expanding correctly and for carrying out the check, not for a particular verdict.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Compare the two routes on this equation. Say what each one needs to be in place before it can begin, and identify which feature of this particular equation makes one route shorter than the other. Then say what would have to change about the equation for that advantage to disappear.
  
  Carry your own answer forward Compare the two routes as you actually carried them out in parts A and B, whatever they produced. The credit here is for naming what each route requires before it can start and for identifying the feature that decides between them.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer
Part A

$x = 3$ or $x = -7$ .
- Both are needed; reporting only $x = 3$ has dropped the branch where the square root is taken as $-5$
Part B

$3x^2 + 12x - 63 = 0$ , and substituting $x = 3$ gives $27 + 36 - 63 = 0$ .
- $x^2 + 4x - 21 = 0$ is the same equation divided through by $3$ , which has the same solutions; the question asks for the expanded form, so say which you are giving
Part C

The square-root route needs the squared group alone on one side; the standard-form route needs everything gathered and a factor pair found. The first is shorter here only because $x$ appears once, inside a single square. A loose $x$ outside the square leaves nothing to isolate, so that route could not start.
Worked solution

Part A

Get the squared group alone before taking any root: add $75$ to both sides, then divide by $3$ .

$(x + 2)^2 = 25 \quad\Longrightarrow\quad x + 2 = \pm 5$

The $\pm$ covers the whole group, so $x + 2$ is either $5$ or $-5$ , giving $x = 3$ or $x = -7$ . Both check: $3(5)^2 - 75 = 0$ and $3(-5)^2 - 75 = 0$ .

Part B

Expand the square first, then distribute the $3$ and collect the constants.

$3\left(x^2 + 4x + 4\right) - 75 = 3x^2 + 12x + 12 - 75 = 3x^2 + 12x - 63 = 0$

Substituting $x = 3$ gives $3(9) + 12(3) - 63 = 27 + 36 - 63 = 0$ , so that value satisfies the standard form as well.

Part C

What each route needs first. Taking square roots needs the squared group by itself on one side, which part A arranged in two steps. Expanding to standard form needs every term gathered with $0$ alone on the other side, after which a factor pair has to be found:

$3x^2 + 12x - 63 = 3\left(x^2 + 4x - 21\right) = 3(x + 7)(x - 3),$

which returns the same two values, $-7$ and $3$ .

What makes the difference here. The equation contains exactly one appearance of $x$ , inside a single square, so isolating that square is possible at all. Add a loose $x$ outside the square, as in $3(x + 2)^2 + 8x = 75$ , and $x$ appears in two places at once; nothing can be isolated, and only the expand-and-factor route remains.

In one line

Isolating the square gives $(x + 2)^2 = 25$ and the solutions $x = 3$ and $x = -7$ ; expanding gives the same equation as $3x^2 + 12x - 63 = 0$ , which factors as $3(x + 7)(x - 3)$ and returns the same pair. Taking square roots is shorter here only because $x$ appears once, inside a single square, so that square can be isolated. Put a loose $x$ outside the square and nothing can be isolated, leaving only the expand-and-factor route.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Isolates the squared group completely, undoing both the $-75$ and the factor of $3$ , before taking any square root. . Worth 2 points.

Keeps both signs at the root and lets the $\pm$ apply to the whole group, then finishes each branch. . Worth 2 points.

Reports both solutions as the answer. . Worth 1 point.

Part B 5 points

Expands the squared binomial into three terms rather than squaring each piece separately. . Worth 2 points.

Distributes the leading number across all three terms and combines the constants into standard form. . Worth 2 points.

Substitutes a value from part A and reports what the left side comes out to. . Worth 1 point.

Part C 5 points

States a precondition for each route separately, rather than one shared description of solving. . Worth 2 points. needs an explanation, not just an answer

Locates the advantage in $x$ appearing only once, inside the square, rather than in one route being generally faster. . Worth 2 points. needs an explanation, not just an answer

Describes a change to the equation that removes the advantage, and says which route survives it. . Worth 1 point.
3. A wrong pair of roots, and the single misreading behind it . 13 points. Question 3 of 10.

For the equation $x^2 - 7x - 18 = 0$ , the pair $x = -9$ and $x = 2$ is offered as the answer. It is wrong, but it is not random: one specific misreading produces exactly that pair and no other.
1. Part A.
  Factor $x^2 - 7x - 18$ and state the two solutions of $x^2 - 7x - 18 = 0$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Identify the single misreading that turns your part A work into the offered pair $-9$ and $2$ , and show that it does produce exactly that pair.
  
  Carry your own answer forward Work from whichever factorization you produced in part A, even if it was not the expected one, and account for the offered pair against it honestly. The credit here is for naming one specific misreading and showing it reproduces the pair, not for reaching a fixed wording.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
3. Part C.
  This misreading always reports the opposite of each true solution, so it is usually detectable. Determine the condition on a monic trinomial $x^2 + bx + c$ under which the misreading would report the correct pair anyway, and justify that your condition is the only one.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$x^2 - 7x - 18 = (x - 9)(x + 2)$ , so the solutions are $x = 9$ and $x = -2$ .
- $(x + 2)(x - 9)$ is the same product written the other way round; the solutions are the same pair either way
Part B

The factors are read as if the numbers written inside them were already the solutions, with no zero-product step. Those numbers are $-9$ and $+2$ , which is the offered pair exactly. The step that is missing is solving each factor: $x - 9 = 0$ gives $9$ , not $-9$ .

Part C

It happens exactly when $b = 0$ . Writing the factors as $(x + p)(x + q)$ , the misreading reports $\{p, q\}$ while the truth is $\{-p, -q\}$ ; those two sets agree exactly when $q = -p$ , and $b = p + q$ is then $0$ . Both directions hold, so $b = 0$ is the whole condition, as with $x^2 - 9 = (x + 3)(x - 3)$ .
Worked solution

Part A

The constant $-18$ is negative, so the two numbers have opposite signs, and their sum is $-7$ , so the one with the larger size is negative. That points to $-9$ and $2$ .

$x^2 - 7x - 18 = (x - 9)(x + 2) = 0$

Setting each factor to zero gives $x - 9 = 0$ , so $x = 9$ , and $x + 2 = 0$ , so $x = -2$ .

Part B

The factored form is $(x - 9)(x + 2)$ , and the numbers written inside its two factors are $-9$ and $+2$ . Reporting those as the solutions, without ever setting a factor to zero, gives the offered pair on the nose:

$\text{numbers inside the factors: } -9, \ +2 \qquad \text{true solutions: } 9, \ -2.$

So the misreading is skipping the zero-product step, which is the step that flips each sign, because $x + p = 0$ solves to $x = -p$ . Every offered value is the exact opposite of a true one, which is the signature of this particular slip.

Part C

Write the factorization as $(x + p)(x + q)$ . The misreading reports the set $\{p, q\}$ ; the true solutions are $\{-p, -q\}$ . The question is when those two sets are the same.

$\{p, q\} = \{-p, -q\} \quad\text{exactly when}\quad q = -p, \quad\text{that is, when}\quad p + q = b = 0.$

Both directions matter and both hold. If $b = 0$ then $q = -p$ , so the numbers are $p$ and $-p$ and the true solutions are $-p$ and $p$ : the same two values. If the sets agree, then either $p = -p$ and $q = -q$ , forcing both to be $0$ , or $p = -q$ ; in every case $p + q = 0$ , so $b = 0$ . An example is $x^2 - 9 = (x + 3)(x - 3)$ , where the misreading reports $3$ and $-3$ , which is the correct pair.

In one line

$x^2 - 7x - 18 = (x - 9)(x + 2)$ , so the solutions are $x = 9$ and $x = -2$ . The offered pair is produced by reporting the numbers written inside the factors as if they were the solutions, skipping the zero-product step that flips each sign. Writing the factors as $(x + p)(x + q)$ , the misreading gives $\{p, q\}$ against the true $\{-p, -q\}$ , and those agree exactly when $q = -p$ , that is, exactly when $b = 0$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Finds a pair whose product is the constant term and whose sum is the middle coefficient, testing both conditions and not just one. . Worth 2 points.

Converts the factored form into the two solutions, each with the sign that makes its own factor vanish. . Worth 2 points.

Part B 4 points

Names one specific misreading rather than describing the answer as generally wrong or as a sign error somewhere. . Worth 2 points. needs an explanation, not just an answer

Demonstrates that the named misreading reproduces the offered pair exactly, and identifies the step it skips. . Worth 2 points.

Part C 5 points

Reaches a condition stated on the coefficients of the trinomial, not on a particular example. . Worth 2 points. needs an explanation, not just an answer

Argues both directions, that the condition forces agreement and that agreement forces the condition, rather than checking one example. . Worth 3 points. needs an explanation, not just an answer
4. A leading coefficient that reaches into both factors . 14 points. Question 4 of 10.

The trinomial $8x^2 - 2x - 15$ begins with a number in front of the squared term, so that number has to be shared out between the two factors rather than sitting quietly outside them. Its three coefficients share no common factor, so there is nothing to pull out first.
1. Part A.
  Factor $8x^2 - 2x - 15$ , and confirm your answer by expanding it.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Solve $8x^2 - 2x - 15 = 0$ , reporting both solutions.
  
  Carry your own answer forward Solve from whichever factorization you produced in part A, even if it was not the expected one. The credit here is for setting each factor to zero and for dividing out the number in front of $x$ , not for landing on a particular pair.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Both solutions came out as fractions, which did not happen for any monic trinomial you have factored. State precisely what decides whether the factor $px + r$ produces a whole-number solution or a fractional one, then apply your test to each of your two factors from part A.
  
  Carry your own answer forward Apply your test to whichever factors you wrote in part A, and report honestly what it says about each of them. The credit here is for the general test and for using it on your own factors, not for a particular verdict.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
The answer
Part A

$8x^2 - 2x - 15 = (4x + 5)(2x - 3)$ , and expanding gives $8x^2 - 12x + 10x - 15$ , the original.
- $(2x - 3)(4x + 5)$ is the same product written the other way round; $(8x + 5)(x - 3)$ is not, since it expands to $8x^2 - 19x - 15$
Part B

$x = -\tfrac{5}{4}$ or $x = \tfrac{3}{2}$ .
- $x = -1.25$ and $x = 1.5$ are the same two numbers; $x = -5$ and $x = 3$ are not, since neither has been divided by the number in front of $x$
Part C

The factor $px + r$ gives the solution $-\tfrac{r}{p}$ , so it is a whole number exactly when $p$ divides $r$ evenly, and a fraction otherwise. In $4x + 5$ , four does not divide five, and in $2x - 3$ , two does not divide three, so both solutions here are fractions. A monic trinomial has $p = 1$ throughout, and $1$ divides everything.
Worked solution

Part A

Aim the search: the middle term splits into two numbers with sum $-2$ and product $a \cdot c = 8 \times (-15) = -120$ . The product is negative, so they have opposite signs, and $-12$ and $10$ work.

The two middle strips are $-12x$ and $10x$ , which is what says to build $(4x + 5)(2x - 3)$ . Expand to confirm:

$(4x + 5)(2x - 3) = 8x^2 - 12x + 10x - 15 = 8x^2 - 2x - 15$

Part B

Use the factorization and set each factor to zero.

$4x + 5 = 0 \quad \text{or} \quad 2x - 3 = 0$

The first gives $4x = -5$ , so $x = -\tfrac{5}{4}$ ; the second gives $2x = 3$ , so $x = \tfrac{3}{2}$ . Checking the first: $8\left(\tfrac{25}{16}\right) - 2\left(-\tfrac{5}{4}\right) - 15 = \tfrac{25}{2} + \tfrac{5}{2} - 15 = 0$ .

Part C

Setting $px + r = 0$ gives $px = -r$ , so the solution the factor produces is

$x = -\frac{r}{p}.$

That value is a whole number exactly when $p$ divides $r$ evenly, and a fraction otherwise. Applying the test: in $4x + 5$ , four does not divide five, so $-\tfrac{5}{4}$ is a fraction; in $2x - 3$ , two does not divide three, so $\tfrac{3}{2}$ is a fraction. It also explains why the monic case never produced one: there every $p$ is $1$ , and $1$ divides every integer.

In one line

$8x^2 - 2x - 15 = (4x + 5)(2x - 3)$ , so the solutions are $x = -\tfrac{5}{4}$ and $x = \tfrac{3}{2}$ . A factor $px + r$ produces the solution $-\tfrac{r}{p}$ , which is a whole number exactly when $p$ divides $r$ evenly; four does not divide five and two does not divide three, so both solutions here are fractions, while a monic trinomial has $p = 1$ throughout and never produces one.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Aims the search with a pair whose product is the leading coefficient times the constant and whose sum is the middle coefficient. . Worth 2 points.

Places the numbers so that the two cross products, not a plain sum of two constants, come out to the middle term. . Worth 2 points.

Expands the proposed factorization and compares all three coefficients with the original. . Worth 1 point.

Part B 4 points

Sets each factor equal to zero separately rather than working on the product as a whole. . Worth 2 points.

Divides each constant by the number in front of $x$ in its own factor and flips the sign, reporting both solutions. . Worth 2 points.

Part C 5 points

States the test in terms of the two numbers in a factor, as a divisibility condition, rather than as an observation about these particular answers. . Worth 3 points. needs an explanation, not just an answer

Applies the test to each factor separately and connects the result back to why a monic trinomial behaves differently. . Worth 2 points.
5. A patio, and a second design that has to be tested rather than assumed . 14 points. Question 5 of 10.

A rectangular patio is being laid. Its length is $7$ metres greater than its width, and the finished patio is to cover exactly $60$ square metres.
1. Part A.
  Name the unknown, write both dimensions in terms of it, and turn the area requirement into an equation in standard form.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Solve your equation, report both of its solutions before deciding anything, then give the patio's width and length.
  
  Carry your own answer forward Solve whichever equation you produced in part A and read its solutions against the patio honestly, even if that equation was not the expected one. The credit here is for solving correctly, for testing each solution against what the letter stands for, and for answering in the terms the question asked.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A second design is proposed: a patio $4$ metres wide, with the length still exceeding the width by $7$ metres. Decide whether that design is admissible, by testing it against every requirement the job states rather than against your answer to part B, and say exactly which requirement decides the verdict.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

With $w$ the width in metres, the length is $w + 7$ metres, and $w(w + 7) = 60$ becomes $w^2 + 7w - 60 = 0$ .
Part B

The solutions are $w = -12$ and $w = 5$ . Only $w = 5$ is a width, so the patio is $5$ metres wide and $12$ metres long.
- $5$ metres by $12$ metres states the same patio; reporting only $w = 5$ leaves the length unstated, which the question asks for
Part C

It is not admissible. A width of $4$ metres gives a length of $11$ metres, so it meets the difference requirement, but its area is $4 \times 11 = 44$ square metres rather than the required $60$ . The area requirement is what decides it, and testing against the job keeps the verdict independent of part B.
Worked solution

Part A

Let $w$ be the width in metres. The length exceeds it by $7$ , so the length is $w + 7$ , and the area is the two multiplied together. Expand and gather onto one side.

$w(w + 7) = 60 \quad\Longrightarrow\quad w^2 + 7w - 60 = 0$

One letter carries both dimensions, so no second unknown is needed.

Part B

Two numbers with product $-60$ and sum $7$ are $12$ and $-5$ , so the trinomial factors and each factor gives a solution.

$w^2 + 7w - 60 = (w + 12)(w - 5) = 0 \quad\Longrightarrow\quad w = -12 \text{ or } w = 5$

A width is a length and cannot be negative, so $w = -12$ is set aside. The patio is $5$ metres wide, and its length is $5 + 7 = 12$ metres, which does give $5 \times 12 = 60$ square metres.

Part C

There are two requirements, and a design has to meet both. Test them one at a time.

$\text{difference: } 11 - 4 = 7 \ \checkmark \qquad \text{area: } 4 \times 11 = 44 \ne 60 \ \times$

The proposed design satisfies the difference requirement exactly and fails the area requirement by $16$ square metres, so it is not admissible, and the area is what decides it. Note that this verdict was reached without appealing to part B at all: had part B gone wrong, this test would still have returned the same answer, because it consults the job's requirements directly.

In one line

With $w$ the width in metres the model is $w^2 + 7w - 60 = 0$ , which factors as $(w + 12)(w - 5) = 0$ and gives $w = -12$ or $w = 5$ . A width cannot be negative, so the patio is $5$ metres wide and $12$ metres long. The proposed $4$ metre design meets the difference requirement, since $11 - 4 = 7$ , but covers only $44$ square metres rather than $60$ , so the area requirement rules it out.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Names the unknown with its unit and writes the second dimension in that same unknown, introducing no second letter. . Worth 2 points.

Turns the area requirement into an equation and gathers every term onto one side, with zero alone on the other. . Worth 2 points.

Part B 5 points

Solves the quadratic correctly and reports both solutions before any is discarded. . Worth 2 points.

Tests each solution against what the letter stands for and sets aside the impossible one with the reason named. . Worth 2 points.

States both dimensions in a sentence with units. . Worth 1 point.

Part C 5 points

Tests the proposal against each stated requirement separately, rather than rejecting it because it disagrees with an earlier answer. . Worth 3 points. needs an explanation, not just an answer

Names which requirement the proposal meets and which it fails, and reaches a verdict on admissibility. . Worth 2 points. needs an explanation, not just an answer
6. Two totals that were in the product all along . 13 points. Question 6 of 10.

The expression $(x - 8)(x + 3)$ is a product whose two solutions can be read at a glance. Written out as a trinomial, those solutions are hidden, and only two totals about them remain visible. This question puts the two views side by side.
1. Part A.
  Expand $(x - 8)(x + 3)$ into standard form.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Read the sum and the product of the solutions off your standard form. Separately, read the two solutions themselves off the original factored form, and compare their actual sum and product with the two totals you read.
  
  Carry your own answer forward Read the totals from whichever standard form you produced in part A, and compare them with the solutions the original factored form gives. The credit here is for reading both totals correctly and carrying out the comparison, not for reaching agreement.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The comparison in part B came out even. Decide whether such a comparison could ever come out uneven when the expansion has been done correctly, and justify your decision from what expanding a general $(x - r)(x - s)$ actually produces.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$(x - 8)(x + 3) = x^2 - 5x - 24$ .
- $-24 - 5x + x^2$ is the same expression out of order; standard form writes the squared term first
Part B

From the standard form, sum $5$ and product $-24$ . From the factored form the solutions are $8$ and $-3$ , and $8 + (-3) = 5$ with $8 \times (-3) = -24$ . The two agree.

Part C

It cannot. Expanding $(x - r)(x - s)$ gives $x^2 - (r + s)x + rs$ , so the middle coefficient IS the negative of the sum and the constant IS the product, by construction rather than by coincidence. The comparison therefore compares each quantity with itself, and it can only ever expose an arithmetic slip, never a genuine disagreement.
Worked solution

Part A

Multiply every term of the first factor by every term of the second, then collect the two middle terms.

$(x - 8)(x + 3) = x^2 + 3x - 8x - 24 = x^2 - 5x - 24$

Part B

From $x^2 - 5x - 24$ the coefficients hand over both totals directly, with the minus sign sitting only on the sum.

$r + s = -b = 5, \qquad rs = c = -24$

From $(x - 8)(x + 3)$ the solutions are $8$ and $-3$ , each the value that makes its own factor vanish. Their actual sum is $8 + (-3) = 5$ and their actual product is $8 \times (-3) = -24$ , matching both totals.

Part C

Expand the general product once and read what the coefficients are made of:

$(x - r)(x - s) = x^2 - sx - rx + rs = x^2 - (r + s)x + rs.$

Matching that against $x^2 + bx + c$ gives $b = -(r + s)$ and $c = rs$ . So the middle coefficient is built out of the sum, and the constant is built out of the product; the two totals are not facts discovered about the trinomial but the very things its coefficients were assembled from. A correct expansion therefore cannot disagree with them. What the check is genuinely good for is catching an arithmetic slip in the expansion or in the sign flip, which is why it is still worth running.

In one line

$(x - 8)(x + 3) = x^2 - 5x - 24$ , whose coefficients give sum $5$ and product $-24$ ; the factored form gives the solutions $8$ and $-3$ , which add to $5$ and multiply to $-24$ . The agreement is forced rather than lucky: expanding $(x - r)(x - s)$ gives $x^2 - (r + s)x + rs$ , so the middle coefficient is assembled from the sum and the constant from the product. A correct expansion can never disagree, and the check earns its place by catching arithmetic slips instead.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Produces all four products before collecting, rather than multiplying only the first terms and the last terms. . Worth 2 points.

Collects the two middle terms with their signs and writes the result with the squared term first. . Worth 1 point.

Part B 5 points

Reads the sum as the negative of the middle coefficient and the product as the constant, with the minus sign on the sum only. . Worth 2 points.

Takes the two solutions from the factored form with their signs flipped, then computes their actual sum and product. . Worth 2 points.

States whether the two readings agree. . Worth 1 point.

Part C 5 points

Expands the general product in letters and matches it term by term against the standard form, rather than arguing from the numerical case. . Worth 3 points. needs an explanation, not just an answer

Concludes that the coefficients are built from the two totals, so the agreement is forced, and says what the check can still detect. . Worth 2 points. needs an explanation, not just an answer
7. Two totals available before the search, and what they were worth . 15 points. Question 7 of 10.

The trinomial $6x^2 - 11x - 10$ has a leading coefficient of $6$ , so its two solutions need not be whole numbers. Some information about those solutions is available immediately, before any factoring is attempted at all.
1. Part A.
  Without factoring anything, state the sum and the product of the two solutions of $6x^2 - 11x - 10 = 0$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Now factor $6x^2 - 11x - 10$ , read its two solutions, and check them against the totals you stated in part A.
  
  Carry your own answer forward Check your solutions against whichever totals you stated in part A, even if those were not the expected ones, and report honestly whether they agree. The credit here is for factoring correctly and for carrying out the comparison.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Part A was finished before part B began. State one thing those two totals genuinely told you in advance about the factorization you were about to look for, and one thing they did not tell you, being precise about the difference between the two numbers the search uses and the two solutions the totals describe.
  
  Carry your own answer forward Argue from whichever totals and factorization you produced in parts A and B. The credit here is for naming something the totals settled in advance and something they did not, and for keeping the search numbers distinct from the solutions.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
The answer
Part A

Sum $\tfrac{11}{6}$ , product $-\tfrac{5}{3}$ .
- The product may be left as $-\tfrac{10}{6}$ , the same number before reducing; a sum of $-\tfrac{11}{6}$ is not the same, since $b$ is already negative
Part B

$6x^2 - 11x - 10 = (3x + 2)(2x - 5)$ , so the solutions are $-\tfrac{2}{3}$ and $\tfrac{5}{2}$ . They add to $\tfrac{11}{6}$ and multiply to $-\tfrac{5}{3}$ , matching part A.
- $(2x - 5)(3x + 2)$ is the same product written the other way round; the two solutions are the same pair either way
Part C

They told me the signs: a negative product means the two solutions have opposite signs, so the two constants in the factors do too. They did not hand over the solutions, and they are not the search numbers either: the search used $-15$ and $4$ , with product $a \cdot c$ and sum $b$ .
Worked solution

Part A

Both totals divide by the leading coefficient, and only the sum takes the extra minus sign. Here $a = 6$ , $b = -11$ , $c = -10$ .

$r + s = -\frac{b}{a} = \frac{11}{6}, \qquad rs = \frac{c}{a} = -\frac{10}{6} = -\frac{5}{3}$

The double negative on the sum is the step to watch: $b$ is already negative, so $-\tfrac{b}{a}$ comes out positive.

Part B

The middle term splits into two numbers with sum $-11$ and product $a \cdot c = 6 \times (-10) = -60$ , which are $-15$ and $4$ .

The two middle strips are $-15x$ and $4x$ , which is what says to build $(3x + 2)(2x - 5)$ . Expand to confirm:

$(3x + 2)(2x - 5) = 6x^2 - 15x + 4x - 10 = 6x^2 - 11x - 10$

Part C

What they told me. The product $-\tfrac{5}{3}$ is negative, and a product of two numbers is negative only when they have opposite signs. So one solution is positive and one is negative, which means the two constants inside the factors carry opposite signs too, and half of the candidate arrangements can be dropped before any expanding.

What they did not tell me. They did not hand over the solutions. Knowing a sum and a product is not the same as knowing the two numbers; separating a pair from its two totals is a further step that this chapter does not perform in general. They are also a different pair from the numbers the search uses:

$\text{search numbers: } -15 \text{ and } 4, \qquad \text{solutions: } -\tfrac{2}{3} \text{ and } \tfrac{5}{2}.$

The search numbers multiply to $a \cdot c = -60$ and add to $b = -11$ ; the solutions multiply to $\tfrac{c}{a}$ and add to $-\tfrac{b}{a}$ . Confusing the two is easy precisely because both are found by a sum-and-product hunt.

In one line

The two solutions of $6x^2 - 11x - 10 = 0$ have sum $\tfrac{11}{6}$ and product $-\tfrac{5}{3}$ , read straight off the coefficients. Factoring gives $(3x + 2)(2x - 5)$ and the solutions $-\tfrac{2}{3}$ and $\tfrac{5}{2}$ , which match both totals. The totals settled the signs in advance, since a negative product forces opposite signs, but they never delivered the solutions, and they describe a different pair from the search numbers $-15$ and $4$ , which multiply to $a \cdot c$ and add to $b$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Divides both totals by the leading coefficient rather than using the monic formulas unchanged. . Worth 2 points.

Puts the extra minus sign on the sum only, handling the double negative, and leaves the product's own sign alone. . Worth 2 points.

Part B 5 points

Aims the search with a pair whose product is the leading coefficient times the constant and whose sum is the middle coefficient. . Worth 2 points.

Reads each solution off its own factor, dividing by the number in front of $x$ and flipping the sign. . Worth 2 points.

Adds and multiplies the two solutions and compares each result with the matching total from part A. . Worth 1 point.

Part C 6 points

Names something the totals settled before the search, tied to what a sign of a product or a sum forces. . Worth 2 points. needs an explanation, not just an answer

States that the totals do not deliver the solutions themselves, treating the separation of a pair from its two totals as a further step. . Worth 2 points. needs an explanation, not just an answer

Keeps the two numbers used by the search distinct from the two solutions, with the right totals attached to each. . Worth 2 points.
8. A middle coefficient free to roam, and the short list it is held to . 14 points. Question 8 of 10.

Consider the trinomial $x^2 + kx + 24$ , where $k$ may be any integer at all. The squared term and the constant term never change; only the middle coefficient moves. Most values of $k$ leave the trinomial impossible to break into integer factors, and a few do not.
1. Part A.
  Find every integer $k$ for which $x^2 + kx + 24$ factors into two binomials with integer coefficients. Show the search that produces your list.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Take the smallest positive value of $k$ on your list. Factor the trinomial it produces, and give the solutions of the corresponding equation.
  
  Carry your own answer forward Use the smallest positive value on whichever list you produced in part A, even if that list was not the expected one, and factor the trinomial it gives. The credit here is for factoring and for reading the solutions off, not for a particular value of $k$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  The middle coefficient was free to be any integer, yet only finitely many of them made the trinomial factor. Justify why the list has to be finite, from what the search actually consults. Then say what happens to the list if the constant term is $0$ rather than $24$ , and what that reveals about where the finiteness came from.
  
  Carry your own answer forward Argue from the search you actually carried out in part A, whatever list it produced. The credit here is for locating the reason the list is finite and for saying what changes when the constant term is $0$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$k$ can be $25$ , $14$ , $11$ , $10$ , $-10$ , $-11$ , $-14$ or $-25$ : eight values, one for each integer pair whose product is $24$ .
- The list may be written as $k = \pm 10, \pm 11, \pm 14, \pm 25$ ; a list of only the four positive values has left out the pairs of negatives
Part B

The smallest positive value is $k = 10$ , giving $x^2 + 10x + 24 = (x + 4)(x + 6)$ , so the solutions are $x = -4$ and $x = -6$ .
- $(x + 6)(x + 4)$ is the same product written the other way round; the solutions $-4$ and $-6$ are the same pair either way
Part C

The search consults the factor pairs of the constant term, and $24$ has only finitely many, with each pair fixing one value of $k$ as its sum. With a constant of $0$ that changes completely: $x^2 + kx = x(x + k)$ factors for every integer $k$ , so the list becomes infinite. The finiteness came from the constant term, not from $k$ .
Worked solution

Part A

The two numbers in the factors must multiply to $24$ , and $k$ is whatever they add to. So list the integer pairs with product $24$ and record each sum.

$(1, 24) \to 25, \quad (2, 12) \to 14, \quad (3, 8) \to 11, \quad (4, 6) \to 10$

Each pair also has a negative twin, $(-1, -24)$ and so on, whose product is still $24$ and whose sum is the opposite number. That gives eight values of $k$ in all: $\pm 10$ , $\pm 11$ , $\pm 14$ and $\pm 25$ .

Part B

The smallest positive entry is $k = 10$ , which came from the pair $4$ and $6$ .

$x^2 + 10x + 24 = (x + 4)(x + 6) = 0$

Each factor gives its solution with the sign flipped, so $x = -4$ or $x = -6$ . Both totals agree: they add to $-10$ , which is $-k$ , and multiply to $24$ .

Part C

Why the list is finite. The middle coefficient is never searched for directly; what is searched is the pairs of integers multiplying to the constant term, and $k$ is only ever recorded as one of their sums. A nonzero integer has finitely many integer factor pairs, so finitely many sums can be recorded, and every other value of $k$ is unreachable.

What a constant of $0$ does. The product test becomes $pq = 0$ , which no longer restricts anything, because a common factor of $x$ is available whatever $k$ is:

$x^2 + kx = x(x + k).$

That factors for every integer $k$ , so the list of workable $k$ becomes all of them. The finiteness in the original problem therefore came from the constant term having only finitely many factorizations, not from any restriction on $k$ itself.

In one line

$x^2 + kx + 24$ factors over the integers exactly for $k = \pm 10$ , $\pm 11$ , $\pm 14$ and $\pm 25$ , one value per integer pair whose product is $24$ . The smallest positive is $k = 10$ , giving $(x + 4)(x + 6)$ and the solutions $-4$ and $-6$ . The list is finite because the search consults the factor pairs of the constant term and records their sums, and a nonzero integer has only finitely many. Replace the constant with $0$ and $x^2 + kx = x(x + k)$ factors for every integer $k$ , so the finiteness came from the constant term.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Lists the integer pairs whose product is the constant term, working from the product rather than guessing values of $k$ . . Worth 2 points.

Includes the negative pairs as well as the positive ones, giving both signs of each sum. . Worth 2 points.

Reports the complete list of values of $k$ , with no repeats. . Worth 1 point.

Part B 4 points

Selects the smallest positive value from the list and writes the trinomial it produces. . Worth 1 point.

Factors it into two binomials and reads each solution off its own factor with the sign flipped. . Worth 2 points.

Reports both solutions. . Worth 1 point.

Part C 5 points

Traces the finiteness to the constant term having finitely many integer factor pairs, with each pair fixing one middle coefficient. . Worth 3 points. needs an explanation, not just an answer

Works out the case of a zero constant term and reaches the verdict that every integer then works, exhibiting the factorization that makes it so. . Worth 2 points. needs an explanation, not just an answer
9. Two square tiles, and a number the tiles cannot use . 14 points. Question 9 of 10.

A workshop cuts square tiles. One tile has a side of $s$ centimetres. A second square tile has a side $6$ centimetres longer than the first, and the second tile covers exactly $196$ square centimetres.
1. Part A.
  Write an equation in $s$ that says what the situation says about the second tile.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Solve your equation, reporting both values it produces before deciding anything, then give the area of the first tile.
  
  Carry your own answer forward Solve whichever equation you wrote in part A, even if it was not the expected one, and read its values against the tiles honestly. The credit here is for keeping both signs at the root, for testing each value against what $s$ stands for, and for answering with an area rather than a side.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  One of the two values was set aside. Interpret it precisely: say what it genuinely is a solution of, what it is not a solution of, and identify the exact step at which the tiles, rather than the algebra, entered the problem.
  
  Carry your own answer forward Interpret whichever value your own work set aside, and check it against your own equation from part A. The credit here is for the distinction between satisfying the equation and answering the question, not for a particular number.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
The answer

Part A

$(s + 6)^2 = 196$ , since the second tile's side is $s + 6$ centimetres and a square's area is its side multiplied by itself.
Part B

The equation gives $s = 8$ and $s = -20$ . A side cannot be negative, so $s = 8$ centimetres and the first tile covers $64$ square centimetres.
- $s = 8$ cm with an area of $64$ square centimetres states the same result; reporting the area as $8$ has given the side instead of the area
Part C

It is a genuine solution of the equation: $(-20 + 6)^2 = (-14)^2 = 196$ , exactly as required. It is not a solution of the tiling problem, because no tile has a side of $-20$ centimetres. The tiles entered when $s$ was named a side length in centimetres, and that meaning is carried by me, not by the equation.
Worked solution

Part A

The second tile's side is $6$ centimetres more than the first, so it is $s + 6$ . A square's area is its side times itself, so the area of the second tile is that quantity squared.

$(s + 6)^2 = 196$

The whole side, not just the $s$ , is what gets squared, which is why the $6$ sits inside the bracket.

Part B

The squared group is already alone, so take the root of both sides and keep both signs.

$s + 6 = \pm 14 \quad\Longrightarrow\quad s = 8 \text{ or } s = -20$

A side length cannot be negative, so the first tile has a side of $8$ centimetres and an area of $8 \times 8 = 64$ square centimetres. Checking the situation: the second tile's side is $8 + 6 = 14$ centimetres and $14 \times 14 = 196$ square centimetres, as required.

Part C

Substituting the set-aside value shows it satisfies the equation completely:

$(-20 + 6)^2 = (-14)^2 = 196.$

So it is a genuine solution of the equation, not an arithmetic slip. What it fails is the situation: a tile with a side of $-20$ centimetres does not exist, so it is not a solution of the tiling problem.

The tiles entered at the moment $s$ was named as a side length in centimetres, back in part A. That naming carries a restriction to positive numbers which the equation never received and could not enforce, which is why the rejection is a decision made by the person reading the answer rather than a result the algebra produced.

In one line

The situation gives $(s + 6)^2 = 196$ , so $s + 6 = \pm 14$ and $s = 8$ or $s = -20$ . The first tile has a side of $8$ centimetres and an area of $64$ square centimetres. The value $-20$ is a genuine solution of the equation, since $(-14)^2 = 196$ , and is not a solution of the tiling problem, since no side is negative. The tiles entered at the moment $s$ was named a side length, a restriction the equation never carried.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Writes the second tile's side in terms of the first tile's side, introducing no second letter. . Worth 2 points.

Squares the whole side rather than the $s$ alone, and sets the result equal to the stated area. . Worth 2 points.

Part B 5 points

Takes the square root with both signs and reports both values before any is discarded. . Worth 2 points.

Tests each value against what $s$ stands for and sets aside the impossible one with the reason named. . Worth 2 points.

Answers with the first tile's area, in square centimetres, rather than with its side. . Worth 1 point.

Part C 5 points

Substitutes the set-aside value and confirms it really does satisfy the equation, rather than calling it an error. . Worth 2 points. needs an explanation, not just an answer

Locates the failure in what the letter stands for rather than in the algebra, and names the step where that meaning was imposed. . Worth 3 points. needs an explanation, not just an answer
10. Two whole-number totals, and what they settle about the solutions . 15 points. Question 10 of 10.

A monic quadratic $x^2 + bx + c$ is described only by its two solutions: they add to $6$ and multiply to $4$ . Nothing else about them is given, and in particular nothing says what kind of numbers they are.
1. Part A.
  Write the monic quadratic described, and state $b$ and $c$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Decide whether either of the two solutions is a whole number. Argue only from the two totals, without attempting to find the solutions themselves.
  
  Carry your own answer forward Argue from whichever totals your part A quadratic encodes. The credit here is for an exhaustive search over the integer pairs and for closing off the case where only one solution is an integer, not for a particular verdict.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
3. Part C.
  Both totals here were whole numbers, and so are $b$ and $c$ , yet part B found no whole-number solution. Decide whether whole-number coefficients ever guarantee whole-number solutions. Support the decision with the quadratic from part A and with a second quadratic of your own choosing that behaves differently.
  
  Carry your own answer forward Use your own part A quadratic as the first example and choose the second yourself. The credit here is for reaching a verdict on the general claim and for supporting it with one example on each side, not for choosing a particular second quadratic.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$x^2 - 6x + 4$ , with $b = -6$ and $c = 4$ .
- $x^2 - 6x + 4 = 0$ is the same thing written as an equation; $x^2 + 6x + 4$ is not, since the middle coefficient is the negative of the sum
Part B

Neither is. Every integer pair with product $4$ is $(1, 4)$ , $(2, 2)$ , $(-1, -4)$ or $(-2, -2)$ , adding to $5$ , $4$ , $-5$ and $-4$ , never $6$ . And one solution alone cannot be an integer either: if one were the integer $m$ , the other would be $6 - m$ , also an integer, and that pair has just been ruled out.

Part C

They never guarantee it. Part A's $x^2 - 6x + 4$ has whole-number coefficients and, by part B, no whole-number solution, while $x^2 - 6x + 8$ has whole-number coefficients and the whole-number solutions $2$ and $4$ . Two quadratics of the same shape land on opposite sides, so the coefficients cannot settle it.
Worked solution

Part A

A monic quadratic is built from its two solutions by the template whose middle coefficient is the negative of their sum and whose constant is their product.

$x^2 - (r + s)x + rs = x^2 - 6x + 4$

So $b = -6$ and $c = 4$ . The sign flip on the middle term is the step to watch: a sum of $+6$ produces a coefficient of $-6$ .

Part B

Suppose both solutions were integers. Then they would form an integer pair with product $4$ , and there are only four such pairs. List them with their sums:

$(1, 4) \to 5, \quad (2, 2) \to 4, \quad (-1, -4) \to -5, \quad (-2, -2) \to -4.$

None adds to $6$ , so the two solutions are not both integers. That leaves the possibility that exactly one of them is, and the sum rules it out: if one solution were an integer $m$ , the other would be $6 - m$ , which is an integer too, putting us back in the case just eliminated. So neither solution is a whole number.

Part C

The decision is no, and one example on each side settles it. Part A's quadratic has whole-number coefficients, and part B showed neither of its solutions is a whole number. Now change a single coefficient:

$x^2 - 6x + 8 = (x - 2)(x - 4),$

whose solutions $2$ and $4$ are whole numbers, and which passes both checks, since $2 + 4 = 6$ and $2 \times 4 = 8$ .

So two quadratics with whole-number coefficients, differing only in the constant term, land on opposite sides. What actually decides the matter is whether a pair of integers exists with the required product AND the required sum, which is exactly the search this chapter runs when factoring; the shape of the coefficients only tells you the search is worth attempting.

In one line

The quadratic is $x^2 - 6x + 4$ , with $b = -6$ and $c = 4$ . Neither of its solutions is a whole number: no integer pair with product $4$ adds to $6$ , and if one solution were an integer then the sum would force the other to be one too. Whole-number coefficients therefore guarantee nothing, since $x^2 - 6x + 8 = (x - 2)(x - 4)$ has the same shape and the whole-number solutions $2$ and $4$ . What decides it is whether an integer pair exists with both the required product and the required sum.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Uses a template in which the middle coefficient is the negative of the sum while the constant keeps the product's own sign. . Worth 2 points.

Reports both coefficients with their signs. . Worth 1 point.

Part B 6 points

Lists every integer pair with the required product, including the negative ones, and tests each against the required sum. . Worth 3 points. needs an explanation, not just an answer

Closes off the remaining case in which only one solution is an integer, using the sum to force the other one to be an integer as well. . Worth 3 points. needs an explanation, not just an answer

Part C 6 points

Reaches a verdict on the general claim rather than restating what happened in this one case. . Worth 2 points. needs an explanation, not just an answer

Supplies a second quadratic with whole-number coefficients that does have whole-number solutions, and verifies both of its totals. . Worth 3 points. needs an explanation, not just an answer

Names what actually decides the matter: whether an integer pair exists with both the required product and the required sum. . Worth 1 point.

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. Two products already set against zero, and one factor that is bare 12 points

Part A 4

Sets each factor equal to zero separately rather than working on the product as a whole. 2
Divides out the number in front of $x$ in the first factor and reports both roots, each with the sign that makes its own factor vanish. 2

Part B 3

Treats the bare $x$ as a factor in its own right and keeps the root it produces. 2
Solves the second factor correctly and reports both solutions. 1

Part C 5

Identifies the property of zero that forces a factor to vanish, rather than restating the rule as something to be remembered. 3
Says what a nonzero right-hand side does not license, and names the step that has to come first instead. 2

2. One equation, and two ways in that need different things first 15 points

Part A 5

Isolates the squared group completely, undoing both the $-75$ and the factor of $3$ , before taking any square root. 2
Keeps both signs at the root and lets the $\pm$ apply to the whole group, then finishes each branch. 2
Reports both solutions as the answer. 1

Part B 5

Expands the squared binomial into three terms rather than squaring each piece separately. 2
Distributes the leading number across all three terms and combines the constants into standard form. 2
Substitutes a value from part A and reports what the left side comes out to. 1

Part C 5

States a precondition for each route separately, rather than one shared description of solving. 2
Locates the advantage in $x$ appearing only once, inside the square, rather than in one route being generally faster. 2
Describes a change to the equation that removes the advantage, and says which route survives it. 1

3. A wrong pair of roots, and the single misreading behind it 13 points

Part A 4

Finds a pair whose product is the constant term and whose sum is the middle coefficient, testing both conditions and not just one. 2
Converts the factored form into the two solutions, each with the sign that makes its own factor vanish. 2

Part B 4

Names one specific misreading rather than describing the answer as generally wrong or as a sign error somewhere. 2
Demonstrates that the named misreading reproduces the offered pair exactly, and identifies the step it skips. 2

Part C 5

Reaches a condition stated on the coefficients of the trinomial, not on a particular example. 2
Argues both directions, that the condition forces agreement and that agreement forces the condition, rather than checking one example. 3

4. A leading coefficient that reaches into both factors 14 points

Part A 5

Aims the search with a pair whose product is the leading coefficient times the constant and whose sum is the middle coefficient. 2
Places the numbers so that the two cross products, not a plain sum of two constants, come out to the middle term. 2
Expands the proposed factorization and compares all three coefficients with the original. 1

Part B 4

Sets each factor equal to zero separately rather than working on the product as a whole. 2
Divides each constant by the number in front of $x$ in its own factor and flips the sign, reporting both solutions. 2

Part C 5

States the test in terms of the two numbers in a factor, as a divisibility condition, rather than as an observation about these particular answers. 3
Applies the test to each factor separately and connects the result back to why a monic trinomial behaves differently. 2

5. A patio, and a second design that has to be tested rather than assumed 14 points

Part A 4

Names the unknown with its unit and writes the second dimension in that same unknown, introducing no second letter. 2
Turns the area requirement into an equation and gathers every term onto one side, with zero alone on the other. 2

Part B 5

Solves the quadratic correctly and reports both solutions before any is discarded. 2
Tests each solution against what the letter stands for and sets aside the impossible one with the reason named. 2
States both dimensions in a sentence with units. 1

Part C 5

Tests the proposal against each stated requirement separately, rather than rejecting it because it disagrees with an earlier answer. 3
Names which requirement the proposal meets and which it fails, and reaches a verdict on admissibility. 2

6. Two totals that were in the product all along 13 points

Part A 3

Produces all four products before collecting, rather than multiplying only the first terms and the last terms. 2
Collects the two middle terms with their signs and writes the result with the squared term first. 1

Part B 5

Reads the sum as the negative of the middle coefficient and the product as the constant, with the minus sign on the sum only. 2
Takes the two solutions from the factored form with their signs flipped, then computes their actual sum and product. 2
States whether the two readings agree. 1

Part C 5

Expands the general product in letters and matches it term by term against the standard form, rather than arguing from the numerical case. 3
Concludes that the coefficients are built from the two totals, so the agreement is forced, and says what the check can still detect. 2

7. Two totals available before the search, and what they were worth 15 points

Part A 4

Divides both totals by the leading coefficient rather than using the monic formulas unchanged. 2
Puts the extra minus sign on the sum only, handling the double negative, and leaves the product's own sign alone. 2

Part B 5

Aims the search with a pair whose product is the leading coefficient times the constant and whose sum is the middle coefficient. 2
Reads each solution off its own factor, dividing by the number in front of $x$ and flipping the sign. 2
Adds and multiplies the two solutions and compares each result with the matching total from part A. 1

Part C 6

Names something the totals settled before the search, tied to what a sign of a product or a sum forces. 2
States that the totals do not deliver the solutions themselves, treating the separation of a pair from its two totals as a further step. 2
Keeps the two numbers used by the search distinct from the two solutions, with the right totals attached to each. 2

8. A middle coefficient free to roam, and the short list it is held to 14 points

Part A 5

Lists the integer pairs whose product is the constant term, working from the product rather than guessing values of $k$ . 2
Includes the negative pairs as well as the positive ones, giving both signs of each sum. 2
Reports the complete list of values of $k$ , with no repeats. 1

Part B 4

Selects the smallest positive value from the list and writes the trinomial it produces. 1
Factors it into two binomials and reads each solution off its own factor with the sign flipped. 2
Reports both solutions. 1

Part C 5

Traces the finiteness to the constant term having finitely many integer factor pairs, with each pair fixing one middle coefficient. 3
Works out the case of a zero constant term and reaches the verdict that every integer then works, exhibiting the factorization that makes it so. 2

9. Two square tiles, and a number the tiles cannot use 14 points

Part A 4

Writes the second tile's side in terms of the first tile's side, introducing no second letter. 2
Squares the whole side rather than the $s$ alone, and sets the result equal to the stated area. 2

Part B 5

Takes the square root with both signs and reports both values before any is discarded. 2
Tests each value against what $s$ stands for and sets aside the impossible one with the reason named. 2
Answers with the first tile's area, in square centimetres, rather than with its side. 1

Part C 5

Substitutes the set-aside value and confirms it really does satisfy the equation, rather than calling it an error. 2
Locates the failure in what the letter stands for rather than in the algebra, and names the step where that meaning was imposed. 3

10. Two whole-number totals, and what they settle about the solutions 15 points

Part A 3

Uses a template in which the middle coefficient is the negative of the sum while the constant keeps the product's own sign. 2
Reports both coefficients with their signs. 1

Part B 6

Lists every integer pair with the required product, including the negative ones, and tests each against the required sum. 3
Closes off the remaining case in which only one solution is an integer, using the sum to force the other one to be an integer as well. 3

Part C 6

Reaches a verdict on the general claim rather than restating what happened in this one case. 2
Supplies a second quadratic with whole-number coefficients that does have whole-number solutions, and verifies both of its totals. 3
Names what actually decides the matter: whether an integer pair exists with both the required product and the required sum. 1

Multiple choice

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