Chapter test · nothing is marked until you submit

The Language of Algebra: 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, 114 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 brackets opened, and one sign with further to travel than it looks . 9 points. Question 1 of 10.

Two expressions are printed below.

$P = 3(2w + 5), \qquad Q = 4(w - 2).$
1. Part A.
  Write $P$ and $Q$ again with no brackets, each in simplest form.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Write $P - Q$ in simplest form.
  
  Carry your own answer forward Work from the two bracket-free forms you produced in part A, whatever they were. The credit here is for subtracting a whole expression rather than only its first term, and for gathering what is left.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  The constant term of $P - Q$ is larger than the constant term of $P$ on its own, even though something was taken away. Explain what makes that happen, and state what the constant would have come to had the subtraction reached only the first term of $Q$ .
  
  Carry your own answer forward Account for the constant your own part B produced, even if it was not the expected one. The credit is for the account of what a leading minus sign does to each term it reaches, not for a particular number.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer
Part A

$P = 6w + 15$ and $Q = 4w - 8$ .
- $6w + 15$ may be written $15 + 6w$ ; what is not the same is $6w + 5$ , which has left the outside factor off the second term
Part B

$P - Q = 2w + 23$ .
- $2w + 23$ may be written $23 + 2w$ ; $2w + 7$ is what comes out if the minus sign never reaches the second term
Part C

Subtracting $Q$ subtracts its $-8$ , and taking away a negative adds, so the constant runs $15 + 8 = 23$ . Had the minus sign stopped at the $4w$ , the $-8$ would have been carried down unchanged and the constant would have been $15 - 8 = 7$ .
Worked solution

Part A

The outside factor reaches both terms inside, carrying each term's sign with it.

$P = 3(2w) + 3(5) = 6w + 15, \qquad Q = 4(w) + 4(-2) = 4w - 8.$

Neither result gathers any further, because a $w$ term and a constant are unlike.

Part B

Subtracting a whole expression means adding its opposite, and the opposite reverses the sign of every term, so the $-8$ arrives as $+8$ .

$P - Q = (6w + 15) - (4w - 8) = 6w + 15 - 4w + 8 = 2w + 23.$

Part C

Subtracting an expression is adding its opposite, and the opposite of $4w - 8$ reverses both signs:

$-(4w - 8) = -4w + 8.$

So what joins the constant $15$ is $+8$ , and the difference carries the constant $15 + 8 = 23$ , larger than either of the constants it came from. Nothing has gone wrong: the quantity removed was itself $8$ short of $4w$ , and returning that shortfall is what pushes the constant up.

Had the minus sign stopped at the first term, the line would have read $6w + 15 - 4w - 8$ , giving the constant $15 - 8 = 7$ .

In one line

$P = 6w + 15$ and $Q = 4w - 8$ , so $P - Q = 2w + 23$ . The constant grows because subtracting the whole of $Q$ subtracts its $-8$ , and taking away a negative adds: the constant runs $15 + 8 = 23$ . Had the minus sign reached only the $4w$ , the constant would have been $15 - 8 = 7$ .

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

Multiplies the outside factor against both terms of each bracket, keeping the sign of the second term. . Worth 2 points.

Leaves each answer as two terms, recognising that a variable term and a constant do not combine. . Worth 1 point.

Part B 3 points

Reverses the sign of every term of the expression being subtracted, not only of the first. . Worth 2 points.

Gathers the like terms of the result, leaving nothing further to combine. . Worth 1 point.

Part C 3 points

Locates the rise in the fact that subtracting a whole expression reverses the sign of each of its terms, so a negative constant arrives positive. . Worth 2 points. needs an explanation, not just an answer

States the constant that the truncated subtraction would have produced. . Worth 1 point.
2. One number, two spellings, and a sum that only looks impossible . 10 points. Question 2 of 10.

A radical and a fractional exponent are two ways of writing the same number, and which spelling you choose decides how much arithmetic you end up doing.
1. Part A.
  Evaluate $216^{2/3}$ and $\left(\dfrac{1}{32}\right)^{-1/5}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Simplify $\sqrt[3]{40} + \sqrt[3]{135}$ as far as it will go.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  The exponent $-\tfrac{1}{5}$ in part A carries three separate instructions: a minus sign, a numerator and a denominator. Say what each one contributes, and explain why the value came out as a whole number greater than $1$ rather than as a negative number or as a fraction below $1$ .
  
  Carry your own answer forward Account for the second value you produced in part A, whatever it was. The credit is for saying what each of the three pieces of the exponent does, not for reproducing one particular number.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer
Part A

$216^{2/3} = 36$ and $\left(\dfrac{1}{32}\right)^{-1/5} = 2$ .
- $36$ and $2$ only; $144$ for the first has multiplied the base by the exponent, computing $216 \times \tfrac{2}{3}$ , and $\tfrac{1}{2}$ for the second has left the base the way up it was given. Cubing first and rooting afterwards is not an error, only more arithmetic: $\sqrt[3]{216^2}$ is $36$ as well
Part B

$5\sqrt[3]{5}$ .
- $5\sqrt[3]{5}$ may be written $5 \cdot 5^{1/3}$ ; $\sqrt[3]{175}$ is what comes out if the radicands are added, which no rule permits
Part C

The minus sign inverts the base, turning $\tfrac{1}{32}$ into $32$ ; the denominator $5$ takes a fifth root; the numerator $1$ is the power. The result is positive because a negative exponent moves a power across the bar without touching a sign, and above $1$ because inverting a positive number below $1$ gives one above it.
Worked solution

Part A

In each exponent the denominator is the index of the root and the numerator is the power, and a minus sign turns the base upside down before either happens. Rooting first keeps the arithmetic small.

$216^{2/3} = \left(\sqrt[3]{216}\right)^2 = 6^2 = 36, \qquad \left(\frac{1}{32}\right)^{-1/5} = 32^{1/5} = \sqrt[5]{32} = 2.$

Part B

The index is $3$ , so the factor to pull out of each radicand is a perfect cube: $40 = 8 \times 5$ and $135 = 27 \times 5$ .

$\sqrt[3]{40} + \sqrt[3]{135} = 2\sqrt[3]{5} + 3\sqrt[3]{5} = (2 + 3)\sqrt[3]{5} = 5\sqrt[3]{5}.$

Once simplified the two terms share both index and radicand, so $\sqrt[3]{5}$ is a common factor and the distributive law gathers them.

Part C

Take the three instructions in turn.

$\left(\frac{1}{32}\right)^{-1/5} = \left(\frac{32}{1}\right)^{1/5} = \sqrt[5]{32} = 2.$

The minus sign says to reciprocate the base, so $\tfrac{1}{32}$ becomes $32$ . The denominator $5$ is the index of the root. The numerator $1$ is the power, and a first power leaves its number alone.

A negative exponent is not a negative answer. It relocates a power from one side of the fraction bar to the other, and $32$ , $\sqrt[5]{32}$ and $2$ are all positive.

The value exceeds $1$ because reciprocating a positive number smaller than $1$ produces a number larger than $1$ , and a root of a number larger than $1$ is still larger than $1$ .

In one line

$216^{2/3} = 36$ and $\left(\dfrac{1}{32}\right)^{-1/5} = 2$ , and $\sqrt[3]{40} + \sqrt[3]{135} = 2\sqrt[3]{5} + 3\sqrt[3]{5} = 5\sqrt[3]{5}$ . In the exponent $-\tfrac{1}{5}$ the minus sign inverts the base, the denominator $5$ takes a fifth root and the numerator $1$ is the power. The value is positive because a negative exponent moves a power across the fraction bar rather than changing a sign, and it is above $1$ because inverting a positive number below $1$ produces a number above it.

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 the denominator of each exponent as a root index and the numerator as a power, and inverts the base where the exponent is negative. . Worth 2 points.

Reports two plain positive numbers, neither of them negative. . Worth 1 point.

Part B 3 points

Pulls a perfect-cube factor out of each radicand, matching the factor to the index rather than to a square. . Worth 2 points.

Adds the coefficients of the matching radical and leaves the radicand itself untouched. . Worth 1 point.

Part C 4 points

Gives the minus sign, the numerator and the denominator a distinct job each, rather than describing the calculation as one indivisible step. . Worth 2 points. needs an explanation, not just an answer

Explains both why the result is positive and why it is greater than $1$ , arguing from what a negative exponent does to the base. . Worth 2 points. needs an explanation, not just an answer
3. Out and back along the same road . 10 points. Question 3 of 10.

Expanding turns a product into a sum; factoring turns a sum into a product. The two parts below travel that road in opposite directions.
1. Part A.
  Expand $(2x + 7)(x - 3)$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Factor $30x^3 + 42x^2$ completely.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Part A produced four products before any gathering, and part B left a bracket with exactly two terms inside. Explain, from the distributive law alone, what fixed each of those two counts, and say why the four products of part A finished as three terms rather than four.
  
  Carry your own answer forward Explain the counts your own parts A and B produced, even if either of them was not the expected one. The credit is for tying each count to what the distributive law does, not for a particular expression.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer
Part A

$2x^2 + x - 21$ .
- $2x^2 + 1x - 21$ is the same expression written with the invisible coefficient shown; $2x^2 - 21$ has lost the two middle products
Part B

$6x^2(5x + 7)$ .
- $6x^2(5x + 7)$ only; $2x^2(15x + 21)$ and $6x(5x^2 + 7x)$ are true equalities that have each left a shared factor behind inside the bracket
Part C

In part A the law pairs every term of one bracket with every term of the other, and two against two makes four. In part B the same law, read backwards, leaves one term inside for each term of the original sum, and there were two. The four became three only afterwards, when two of them turned out to be like terms and were gathered.
Worked solution

Part A

Every term of the first bracket multiplies every term of the second, which produces four products, and two of them turn out to be like terms.

$(2x + 7)(x - 3) = 2x^2 - 6x + 7x - 21 = 2x^2 + x - 21.$

Part B

The numerical part of the factor is the greatest common divisor of $30$ and $42$ , which is $6$ ; the variable part is $x$ at the lowest power present, which is $x^2$ .

$30x^3 + 42x^2 = 6x^2 \cdot 5x + 6x^2 \cdot 7 = 6x^2(5x + 7).$

Inside the bracket, $5x$ and $7$ share no factor beyond $1$ , so nothing more can come out.

Part C

What fixed the four. The distributive law, used twice, pairs each term of the first bracket with each term of the second, so the count is one factor's number of terms times the other's:

$(2x + 7)(x - 3): \quad 2 \times 2 = 4 \text{ products}.$

What fixed the two. Read the same law from right to left, $ab + ac = a(b + c)$ . The shared factor comes out front and exactly one term stays inside for each term of the original sum. The sum $30x^3 + 42x^2$ has two terms, so the bracket has two.

Why four finished as three. Gathering is a separate step that happens once the products exist. Two of them, $-6x$ and $7x$ , share a variable part and collapse into one; the other two have nothing to pair with. The multiplying fixed the count of products, and the gathering reduced the count of terms.

In one line

$(2x + 7)(x - 3) = 2x^2 + x - 21$ and $30x^3 + 42x^2 = 6x^2(5x + 7)$ . Four products appear in the first because the distributive law pairs each of two terms with each of two terms; two terms remain inside the bracket in the second because the same law, read backwards, leaves one term for each term of the original sum. Gathering is a later step, and it is what turns four products into three terms.

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

Forms all four products and carries the sign of the $-3$ into both products it appears in. . Worth 2 points.

Gathers the two middle products into a single term with the correct sign. . Worth 1 point.

Part B 3 points

Builds the factor from the greatest common divisor of the coefficients together with the lowest power of $x$ present in every term. . Worth 2 points.

Leaves a bracket whose terms share no factor beyond $1$ . . Worth 1 point.

Part C 4 points

Ties the number of products to the pairing of every term with every term, rather than to a mnemonic or to the shape of the factors. . Worth 2 points. needs an explanation, not just an answer

Ties the number of terms left inside the bracket to the number of terms in the original sum, and keeps the gathering step separate from the multiplying step. . Worth 2 points. needs an explanation, not just an answer
4. Two bars joined into one, and the values the join has to remember . 11 points. Question 4 of 10.

The expression

$F = \frac{3}{x} + \frac{x}{x + 2}$

is built from two fractions with nothing in common underneath.
1. Part A.
  Write $F$ as a single fraction, leaving the denominator as a product.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Evaluate $F$ at $x = 1$ , once from the printed pair of fractions and once from your single fraction.
  
  Carry your own answer forward Use the single fraction you produced in part A, whatever it was, and report honestly whether the two routes agree. The credit here is for evaluating both forms correctly and for saying what the comparison shows.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Two values of $x$ are barred from $F$ . Name them, say which of the two printed fractions is responsible for each, and decide whether your single fraction from part A bars exactly the same two values, arguing from its denominator.
  
  Carry your own answer forward Argue from the denominator of your own part A fraction, even if it was not the expected one, and say honestly whether it bars the same values as the printed pair.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer
Part A

$\dfrac{x^2 + 3x + 6}{x(x + 2)}$ .
- the denominator may be written $x^2 + 2x$ ; what is not the same is $\frac{3 + x}{x(x + 2)}$ , which has added the numerators without scaling either of them
Part B

Both give $\dfrac{10}{3}$ : the printed pair comes to $3 + \tfrac{1}{3}$ , and the single fraction to $\tfrac{10}{3}$ .
- $\frac{10}{3}$ may be written $3\tfrac{1}{3}$ or $3.33\ldots$ ; the two routes must land on the same number, whatever form it is written in
Part C

$x = 0$ , barred by the first fraction, and $x = -2$ , barred by the second. The combined denominator is the product $x(x + 2)$ , and a product is zero only when one of its factors is zero, so it vanishes at exactly those two values and at no others.
Worked solution

Part A

The two denominators share no factor, so the least common denominator is their product $x(x + 2)$ . Scale each fraction up to it by whatever its own denominator is missing, then add the numerators.

$\frac{3}{x} + \frac{x}{x + 2} = \frac{3(x + 2)}{x(x + 2)} + \frac{x \cdot x}{x(x + 2)} = \frac{3x + 6 + x^2}{x(x + 2)} = \frac{x^2 + 3x + 6}{x(x + 2)}.$

The three terms on top are pairwise unlike, so the numerator gathers no further.

Part B

Substitute $1$ for $x$ in each form and follow the order of operations.

$\frac{3}{1} + \frac{1}{1 + 2} = 3 + \frac{1}{3} = \frac{10}{3}, \qquad \frac{1^2 + 3(1) + 6}{1(1 + 2)} = \frac{10}{3}.$

The two agree, which is what a correct combination must do at every value the original allows. Agreement at one value is evidence rather than proof; a disagreement would have been proof of a slip.

Part C

Each printed fraction bars the value that empties its own denominator:

$x = 0 \text{ from } \frac{3}{x}, \qquad x = -2 \text{ from } \frac{x}{x + 2}.$

The combined denominator is the product $x(x + 2)$ . A product of two numbers is zero exactly when at least one of them is zero, so this denominator vanishes when $x = 0$ or when $x + 2 = 0$ , and at no other value.

The single fraction therefore bars the same two values, which is what it has to do. A combination that quietly gained a permitted value, or lost one, would not be naming the same expression.

In one line

$\dfrac{3}{x} + \dfrac{x}{x + 2} = \dfrac{x^2 + 3x + 6}{x(x + 2)}$ , and at $x = 1$ both forms come to $\tfrac{10}{3}$ . The barred values are $x = 0$ , from the first printed fraction, and $x = -2$ , from the second. The combined denominator $x(x + 2)$ is a product, and a product is zero only when a factor is, so it bars those two values and no others.

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

Uses a denominator that both originals divide, and scales each numerator by exactly what its own denominator was missing. . Worth 2 points.

Expands and gathers the numerator correctly, leaving three unlike terms over the common denominator. . Worth 2 points.

Part B 3 points

Evaluates both forms at the stated value with the order of operations intact. . Worth 2 points.

Says what agreement between the two values does and does not establish. . Worth 1 point.

Part C 4 points

Names both barred values and attaches each to the fraction whose denominator produces it. . Worth 2 points. needs an explanation, not just an answer

Argues from a product being zero only when a factor is zero that the combined denominator bars exactly the same values, rather than testing the two values one at a time. . Worth 2 points. needs an explanation, not just an answer
5. A claim that survives the first two things you try . 11 points. Question 5 of 10.

Consider the claim that

$(a + b)^2 = a^2 + b^2$

holds for every pair of numbers $a$ and $b$ .
1. Part A.
  Evaluate both sides at $a = 3$ , $b = 0$ , and again at $a = 0$ , $b = 7$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Produce one pair of numbers on which the claim fails, and give the value of each side on that pair.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 3 points
3. Part C.
  Expand $(a + b)^2$ as the ordinary product it is, and use the result to say exactly which pairs make the claim true. Then explain why the two pairs in part A were bound to agree.
  
  Carry your own answer forward Read your part A pairs against whatever expansion you produce here, and say honestly why each of them agreed. The credit is for identifying the term the two sides differ by and for describing which pairs make it vanish.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

At $a = 3$ , $b = 0$ both sides give $9$ . At $a = 0$ , $b = 7$ both sides give $49$ . Neither pair separates the two sides.

Part B

Take $a = 2$ and $b = 5$ . The left side is $(2 + 5)^2 = 49$ and the right side is $2^2 + 5^2 = 29$ , so the two sides differ and the claim is false.

Part C

$(a + b)^2 = a^2 + 2ab + b^2$ , so the two sides differ by exactly $2ab$ and are equal precisely when $2ab = 0$ , that is when $a$ is zero or $b$ is zero. Both pairs in part A carried a zero, which sent that middle term to zero, so both were bound to agree.

Worked solution

Part A

Square the sum before comparing, not after.

$(3 + 0)^2 = 9 \text{ and } 3^2 + 0^2 = 9; \qquad (0 + 7)^2 = 49 \text{ and } 0^2 + 7^2 = 49.$

Both pairs agree, so neither of them refutes anything.

Part B

Choose a pair with no zero in it.

$(2 + 5)^2 = 7^2 = 49, \qquad 2^2 + 5^2 = 4 + 25 = 29.$

The sides come to $49$ and $29$ , so they are not equal here. A claim made about every pair is contradicted outright by one pair, so nothing further needs checking.

Part C

Multiply the square out as the product it is, every term against every term:

$(a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2.$

The two sides of the claim differ by exactly the middle term $2ab$ , so they are equal precisely when $2ab = 0$ . A product of numbers is zero only when a factor is, so that happens when $a = 0$ or when $b = 0$ . The description is exact in both directions: if either letter is zero the middle term vanishes and the sides do agree, and if neither is zero the middle term is not zero and they do not.

Part A used $b = 0$ and then $a = 0$ . Each pair killed the middle term, so each was guaranteed to agree, and neither could ever have exposed the fault.

In one line

$(a + b)^2 = a^2 + 2ab + b^2$ , so the claim is short by the middle term $2ab$ and holds exactly when $a = 0$ or $b = 0$ . It fails at $a = 2$ , $b = 5$ , where the sides come to $49$ and $29$ . Both pairs in part A carried a zero, which sent the middle term to zero, so both were bound to agree and neither could have exposed the fault.

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

Evaluates both sides at each pair, squaring the sum itself rather than the terms separately. . Worth 2 points.

Reports that both pairs agree without treating that agreement as settling the claim. . Worth 1 point.

Part B 3 points

Supplies a specific pair and evaluates both sides on it, reaching two different numbers. . Worth 2 points.

Concludes that one disagreeing pair refutes a claim made about every pair. . Worth 1 point.

Part C 5 points

Expands the square into three terms and identifies the middle term as the whole of the difference between the two sides. . Worth 3 points. needs an explanation, not just an answer

Describes exactly which pairs make the claim true, in both directions, and shows that the two pairs of part A are among them. . Worth 2 points. needs an explanation, not just an answer
6. Two baskets from one price list . 12 points. Question 6 of 10.

A shop sells notebooks at $n$ dollars each and pens at $n - 3$ dollars each, where $n$ is greater than $3$ . On Monday a customer buys $5$ notebooks and $4$ pens. On Tuesday a different customer buys $2$ notebooks and $9$ pens.
1. Part A.
  Write, in simplest form, an expression in $n$ for Monday's total and an expression in $n$ for Tuesday's total.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Write, in simplest form, an expression for how much more Tuesday's customer paid than Monday's.
  
  Carry your own answer forward Subtract whichever two totals you produced in part A, in the order the question asks. The credit here is for sending the minus sign onto both terms of the total being subtracted, and for gathering the result.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Your part B expression still contains an $n$ , so the gap between the two baskets is not the same in every shop. Say how that gap responds to the notebook price, and find the notebook price at which the two customers paid exactly the same.
  
  Carry your own answer forward Read your own part B expression, even if it was not the expected one, and answer both questions from it. The credit here is for saying how the gap responds to the notebook price and for finding the price that closes it.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
The answer

Part A

Monday's total is $9n - 12$ dollars and Tuesday's total is $11n - 27$ dollars.
Part B

$2n - 15$ dollars.
- $2n - 15$ only; $2n - 39$ is what comes out if the minus sign in front of Monday's total reaches only its first term
Part C

The gap widens by $2$ dollars for every dollar the notebook price rises, and below the price that closes it the gap is negative, meaning Monday's customer paid more. The two paid the same where $2n - 15 = 0$ , at a notebook price of $7.50$ dollars.
Worked solution

Part A

Price each kind of item as a count times its price, then add.

$\text{Monday: } 5n + 4(n - 3) = 5n + 4n - 12 = 9n - 12,$

$\text{Tuesday: } 2n + 9(n - 3) = 2n + 9n - 27 = 11n - 27.$

The $3$ comes off once for every pen, so it is multiplied by the number of pens rather than removed once from the basket.

Part B

Subtract the whole of Monday's total, which reverses the sign of both of its terms.

$(11n - 27) - (9n - 12) = 11n - 27 - 9n + 12 = 2n - 15.$

The $-12$ arrives as $+12$ , giving the constant $-27 + 12 = -15$ .

Part C

The difference is $2n - 15$ , so each extra dollar on the notebook price adds $2$ dollars to Tuesday's lead. Tuesday's basket carries $11$ notebook prices to Monday's $9$ , and it is those two extra copies that respond to every rise. A comparison whose difference were a constant would be the same in every shop; this one is not.

The two customers paid the same when the difference is zero:

$2n - 15 = 0 \quad \text{gives} \quad 2n = 15, \quad n = 7.5.$

At a notebook price of $7.50$ dollars a pen costs $4.50$ dollars, and both baskets come to $55.50$ dollars. Below that price the difference is negative, so Monday's customer paid more; above it, Tuesday's did.

In one line

Monday's total is $9n - 12$ dollars and Tuesday's is $11n - 27$ dollars, so Tuesday's customer paid $2n - 15$ dollars more. That gap depends on the shop: it widens by $2$ dollars for every dollar the notebook price rises, and it closes at $2n - 15 = 0$ , a notebook price of $7.50$ dollars, where both baskets come to $55.50$ dollars.

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

Prices the pens as a count times the pen price, so the discount is taken once for each pen and not once for the whole basket. . Worth 2 points.

Distributes and gathers each total into one variable term and one constant. . Worth 1 point.

States each total as an amount of money. . Worth 1 point.

Part B 3 points

Reverses the sign of both terms of the total being subtracted. . Worth 2 points.

Gathers the difference into one variable term and one constant. . Worth 1 point.

Part C 5 points

Says how the gap responds to a change in the notebook price, reading the rate from the coefficient rather than from a couple of sample prices. . Worth 2 points.

Sets the difference to zero and solves for the price at which the two totals are equal. . Worth 2 points.

Gives that price as an amount of money and reads it back against the situation. . Worth 1 point.
7. A bracket that has to be earned before anything can be crossed out . 13 points. Question 7 of 10.

Consider the algebraic fraction

$G = \frac{14x^3 - 21x^2}{35x^2 - 42x}.$

Nothing on the top is written as a factor of the whole top, and nothing on the bottom is written as a factor of the whole bottom.
1. Part A.
  Factor the numerator of $G$ completely, and factor its denominator completely, treating each one on its own.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Simplify $G$ completely, and state every value of $x$ that the original expression excludes.
  
  Carry your own answer forward Cancel from whichever factored forms you produced in part A, and read the excluded values off the original denominator rather than off your own answer. The credit here is for cancelling a factor of the whole top and the whole bottom, and for reporting every value the original rules out.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Cancelling the shared $7$ alone would give $\dfrac{2x^3 - 3x^2}{5x^2 - 6x}$ , which is a true equality and yet has not been taken as far as it goes. Say what is still available in that form, state the test that decides when a fraction of this kind is finished, and say why multiplying such a form back out cannot settle that question.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

The numerator is $7x^2(2x - 3)$ and the denominator is $7x(5x - 6)$ .
- $7x^2(2x - 3)$ and $7x(5x - 6)$ only; $7x(2x^2 - 3x)$ is a true equality for the numerator that has left an $x$ inside the bracket
Part B

$\dfrac{x(2x - 3)}{5x - 6}$ , and the original excludes $x = 0$ and $x = \dfrac{6}{5}$ .
- the numerator may be left expanded as $2x^2 - 3x$ ; what is not the same is dropping the restriction $x \neq 0$ , which the simplified denominator no longer objects to
Part C

An $x$ is still a factor of the whole of each part, so it can be pulled out of both and cancelled, which returns the part B answer. A fraction of this kind is finished when the numerator and the denominator, each factored completely, share no factor other than $1$ .
Worked solution

Part A

Take each expression separately. For $14x^3 - 21x^2$ the greatest common divisor of $14$ and $21$ is $7$ and the lowest power of $x$ present is $x^2$ . For $35x^2 - 42x$ the greatest common divisor of $35$ and $42$ is $7$ and the lowest power of $x$ present is $x$ .

$14x^3 - 21x^2 = 7x^2(2x - 3), \qquad 35x^2 - 42x = 7x(5x - 6).$

Neither bracket holds a shared factor, so each expression is fully factored.

Part B

The two factored forms share $7x$ , and that quantity is a factor of the whole numerator and of the whole denominator, so it cancels.

$G = \frac{7x^2(2x - 3)}{7x(5x - 6)} = \frac{x(2x - 3)}{5x - 6}.$

The original denominator is $7x(5x - 6)$ , which is zero when $x = 0$ and when $5x - 6 = 0$ . So the excluded values are $x = 0$ and $x = \tfrac{6}{5}$ . Both belong to the original expression and travel with the simplified form, even though the simplified denominator no longer objects at $x = 0$ .

Part C

Factor the intermediate form and the remaining shared factor becomes visible:

$\frac{2x^3 - 3x^2}{5x^2 - 6x} = \frac{x^2(2x - 3)}{x(5x - 6)}.$

An $x$ divides the whole of the top and the whole of the bottom, so cancelling it returns the answer of part B. The intermediate form is equal to $G$ but has stopped early.

The test is this: factor the numerator completely, factor the denominator completely, and ask whether they still share a factor other than $1$ . If they do, there is work left; if they do not, there is none.

Multiplying an unfinished answer back out cannot detect the problem. A fraction that cancelled too little is still equal to the one it came from, so that check returns clean whether the work stopped early or not.

In one line

$14x^3 - 21x^2 = 7x^2(2x - 3)$ and $35x^2 - 42x = 7x(5x - 6)$ , so $G = \dfrac{x(2x - 3)}{5x - 6}$ , and the original excludes $x = 0$ and $x = \tfrac{6}{5}$ . Cancelling only the $7$ leaves an $x$ still shared by the whole of each part, so that form has stopped early: a fraction of this kind is finished when the fully factored top and bottom share nothing beyond $1$ , which is a question multiplying back out can never answer.

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

Builds each factor from the greatest common divisor of that expression's own coefficients and the lowest power of $x$ present in all of its terms. . Worth 2 points.

Leaves each bracket holding terms that share no factor beyond $1$ . . Worth 2 points.

Part B 4 points

Cancels a factor shared by the whole numerator and the whole denominator, leaving the remaining factors of each intact. . Worth 2 points.

Reads both excluded values off the original denominator, including the one the simplified form no longer objects to. . Worth 2 points.

Part C 5 points

Identifies the factor still shared by the whole of each part of the intermediate form and shows what cancelling it produces. . Worth 3 points. needs an explanation, not just an answer

States a completeness test in terms of what the fully factored numerator and denominator still share, and says why multiplying back out cannot settle it. . Worth 2 points. needs an explanation, not just an answer
8. A setting turned right down, and an output that does not follow it . 11 points. Question 8 of 10.

The flow rate $P$ of a pump, in litres per minute, is fixed by its drive setting $R$ through

$P = 12R^{3/2},$

where $R$ is positive.
1. Part A.
  Find the flow rate when the drive setting is $R = 16$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find the flow rate when the drive setting is $R = \dfrac{1}{9}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Between the two settings the drive setting was multiplied by $\tfrac{1}{144}$ . Work out the factor by which the flow rate changed, and explain from the exponent alone why it is not $\tfrac{1}{144}$ itself, and why no single multiplier could carry a setting's factor to the flow rate's factor for every pair of settings.
  
  Carry your own answer forward Compare whichever two flow rates you found in parts A and B, and account for the factor your own numbers produce. The credit here is for the account of how a factor on the setting reaches the flow rate, not for one 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

$768$ litres per minute.
Part B

$\dfrac{4}{9}$ litres per minute.
- $\frac{4}{9}$ may be written $\frac{12}{27}$ before reducing, or as roughly $0.44$ ; what is not the same is $\frac{4}{3}$ , which has used the setting as it stands, taking neither the root nor the cube
Part C

The flow rate was multiplied by $\tfrac{1}{1728}$ , which is $\left(\tfrac{1}{144}\right)^{3/2}$ . A setting factor $k$ reaches the flow rate as $k^{3/2}$ , so the multiplier relating the two is $\sqrt{k}$ , and that changes with $k$ : no one constant serves every pair of settings.
Worked solution

Part A

The denominator of the exponent is the index of the root and the numerator is the power, so take the square root first and cube afterwards.

$P = 12 \cdot 16^{3/2} = 12\left(\sqrt{16}\right)^3 = 12(4)^3 = 12(64) = 768.$

Rooting first works with $4$ instead of with $16^3 = 4096$ .

Part B

A root of a quotient is the quotient of the roots, so root the top and the bottom of the setting separately, then cube.

$P = 12\left(\frac{1}{9}\right)^{3/2} = 12\left(\sqrt{\frac{1}{9}}\right)^3 = 12\left(\frac{1}{3}\right)^3 = \frac{12}{27} = \frac{4}{9}.$

Part C

First the factor itself, from the two flow rates found above:

$\frac{4/9}{768} = \frac{4}{6912} = \frac{1}{1728}.$

Now the reason. Write the new setting as $k$ times the old one and let the rule for a power of a product separate the two factors:

$12(kR)^{3/2} = 12\,k^{3/2}R^{3/2} = k^{3/2} \cdot 12R^{3/2}.$

So a factor applied to the setting reaches the flow rate raised to the same exponent, never as itself. With $k = \tfrac{1}{144}$ , the square root gives $\tfrac{1}{12}$ and the cube gives $\tfrac{1}{1728}$ , exactly the factor the two answers show.

Nor is there a multiplier that would do the job in general. Comparing $k^{3/2}$ with $k$ gives

$\frac{k^{3/2}}{k} = k^{1/2} = \sqrt{k},$

so the number relating the two factors is itself a function of the setting change. Here it happens to be $\tfrac{1}{12}$ ; at a different pair of settings it would be something else, which is why the relationship cannot be captured by any single constant.

In one line

At $R = 16$ the pump delivers $768$ litres per minute, and at $R = \tfrac{1}{9}$ it delivers $\tfrac{4}{9}$ litres per minute, a factor of $\tfrac{1}{1728}$ . That is because a factor applied to the setting splits out of the power and reaches the flow rate carrying the same exponent: $\left(\tfrac{1}{144}\right)^{3/2}$ is $\tfrac{1}{12}$ cubed, which is $\tfrac{1}{1728}$ . No single multiplier would serve in general either, since a setting factor $k$ and its flow factor $k^{3/2}$ are related by $\sqrt{k}$ , which changes with the setting.

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

Takes the square root before the cube, using the denominator of the exponent as the index and the numerator as the power. . Worth 2 points.

Reports the flow rate with its unit. . Worth 1 point.

Part B 3 points

Roots the numerator and the denominator of the setting separately, then cubes, then reduces the result. . Worth 2 points.

Reports the flow rate with its unit. . Worth 1 point.

Part C 5 points

Computes the factor between the two flow rates from the two values found earlier. . Worth 2 points.

Explains that a factor on the setting reaches the flow rate raised to the same exponent, uses that to reproduce the factor found, and says why no one multiplier would serve for every pair of settings. . Worth 3 points. needs an explanation, not just an answer
9. Three rewritings, and the reasons offered for them . 12 points. Question 9 of 10.

Each line below rewrites the expression on its left as the expression on its right.

$\text{(i)} \quad 5 + (3 + y) = (5 + 3) + y$

$\text{(ii)} \quad 6(y + 2) = 6y + 12$

$\text{(iii)} \quad 4 - (y - 1) = 4 - y - 1$
1. Part A.
  For each of the three lines, decide whether it is true for every value of $y$ , and name the law behind each of the ones that are.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
2. Part B.
  For the line you judged false, give one value of $y$ at which the two sides differ, and give both of the values they take there.
  
  Carry your own answer forward Test whichever line you judged false in part A, even if it was not the expected one, and report both sides honestly. The credit here is for producing a value that separates the two sides, not for choosing any particular value.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 3 points
3. Part C.
  The line you judged false reaches for a law that does hold. Name that law, write the line as it should have read, and explain why one value settled it in part B while no number of agreeing values could have settled the lines you judged true.
  
  Carry your own answer forward Work from your own verdicts in part A and from the value you used in part B. The credit is for naming the law the false line was reaching for, for writing that line correctly, and for the account of why refuting and establishing need different amounts of evidence.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

Line (i) is true for every $y$ , by the associative law of addition. Line (ii) is true for every $y$ , by the distributive law. Line (iii) is not true for any value of $y$ .

Part B

Take $y = 0$ in line (iii). The left side is $4 - (0 - 1) = 4 - (-1) = 5$ and the right side is $4 - 0 - 1 = 3$ , so the two sides differ.

Part C

It reaches for the distributive law, with the hidden factor $-1$ that a leading minus sign stands for, so it should read $4 - (y - 1) = 4 - y + 1 = 5 - y$ . One value refutes a universal claim outright. No list of agreeing values proves one, because infinitely many stay untested; only a derivation from the laws does that.

Worked solution

Part A

Line (i). Only the bracketing of three quantities has changed, which is exactly what the associative law of addition permits. Both sides come to $8 + y$ .

Line (ii). The outside factor reaches both terms inside, which is the distributive law: $6(y + 2) = 6y + 12$ .

Line (iii). Subtracting a bracket means adding its opposite, and the opposite reverses both signs:

$4 - (y - 1) = 4 - y + 1 = 5 - y,$

while the printed right-hand side gathers to $3 - y$ . The two sides differ by $2$ at every value, so this line is true nowhere, let alone everywhere.

Part B

Substitute $0$ for $y$ on each side, keeping the bracket intact on the left, and follow the order of operations.

$4 - (0 - 1) = 4 - (-1) = 5, \qquad 4 - 0 - 1 = 3.$

One value at which the two sides disagree contradicts the word every outright, so the line is settled. Any other value would serve equally well here, since the two sides are $5 - y$ and $3 - y$ and differ by $2$ everywhere.

Part C

The law is the distributive law, applied to the factor $-1$ that a leading minus sign stands for:

$4 - (y - 1) = 4 + (-1)(y - 1) = 4 - y + 1 = 5 - y.$

That is line (iii) as it should have read. The printed version sent the minus to the first term and stopped there.

The logic of testing is not symmetric, and the asymmetry is the point. A claim that an equation holds for every number is a claim about all of them at once, so a single value where the two sides disagree contradicts it directly, which is why part B closed the matter with one substitution. Agreement can never do the same work: there are infinitely many numbers, and any list actually tested leaves infinitely many untested, so a run of successes could always break at the next value. Lines (i) and (ii) are established instead by deriving them from laws already known to hold for every number.

In one line

Line (i) is the associative law of addition and line (ii) is the distributive law, so both hold for every $y$ . Line (iii) does not: subtracting the bracket reverses both of its signs, so it should have read $4 - (y - 1) = 5 - y$ rather than $3 - y$ , and at $y = 0$ the two sides come to $5$ and $3$ . One value refutes a claim made about every number, while no list of agreeing values establishes one, which is why lines (i) and (ii) rest on the laws instead of on testing.

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

Reaches the right verdict on all three lines, supporting each rather than asserting it. . Worth 2 points. needs an explanation, not just an answer

Names the law behind each true line, telling a regrouping apart from a distribution. . Worth 2 points. needs an explanation, not just an answer

Part B 3 points

Substitutes a value into both sides, keeping the bracket intact rather than removing it first. . Worth 2 points.

Reaches two different numbers and treats that disagreement as settling the line. . Worth 1 point.

Part C 5 points

Names the law behind the intended rewriting, identifies the hidden factor of $-1$ , and writes the line correctly. . Worth 3 points. needs an explanation, not just an answer

Explains the asymmetry between one refuting value and any number of agreeing ones, and says what does establish a claim about every number. . Worth 2 points. needs an explanation, not just an answer
10. Everything the terms hold in common, and nothing they do not . 15 points. Question 10 of 10.

Consider the expression

$K = 18a^3b - 30a^2b^2 + 12a^2b.$
1. Part A.
  Factor $K$ completely, showing separately how the coefficients fix the numerical part of the factor you take out and how the powers fix its variable part.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Factor $-18a^3b + 30a^2b^2 - 12a^2b$ so that the first term inside the bracket is positive.
  
  Carry your own answer forward Build this from whichever factor you took out in part A, attaching a minus sign to it. The credit here is for pulling out a negative factor and for the effect that has on the sign of every term inside the bracket.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  The quantity $6a^2b^2$ divides the middle term of $K$ and looks like a plausible thing to take out. Decide whether it can be taken out of $K$ , defend the decision by testing it against each of the three terms, and state in general what a quantity must satisfy across the terms of an expression before it may be taken outside a bracket.
  
  Carry your own answer forward Compare the candidate with whichever factor you took out in part A, and test it against all three terms of $K$ . The credit here is for the term-by-term test and for the general requirement you draw from it.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$6a^2b(3a - 5b + 2)$ .
- $6a^2b(3a - 5b + 2)$ only; $3a^2b(6a - 10b + 4)$ and $6ab(3a^2 - 5ab + 2a)$ are true equalities that have each left a shared factor inside the bracket
Part B

$-6a^2b(3a - 5b + 2)$ .
- $-6a^2b(3a - 5b + 2)$ is the form asked for; $6a^2b(-3a + 5b - 2)$ is the same expression but opens the bracket with a negative term
Part C

It cannot. It divides $-30a^2b^2$ , but $18a^3b$ and $12a^2b$ each carry only one $b$ , so it fits neither. To come out, a quantity must divide every term without exception: its number must divide every coefficient, and each letter it uses must appear in every term to at least that power.
Worked solution

Part A

The numerical part is the greatest common divisor of $18$ , $30$ and $12$ , which is $6$ . The variable part takes each letter at the lowest power appearing in every term: $a$ appears as $a^3$ , $a^2$ and $a^2$ , so $a^2$ ; $b$ appears as $b$ , $b^2$ and $b$ , so $b$ . The factor is therefore $6a^2b$ .

$18a^3b - 30a^2b^2 + 12a^2b = 6a^2b(3a) + 6a^2b(-5b) + 6a^2b(2) = 6a^2b(3a - 5b + 2).$

Inside the bracket, $3a$ , $-5b$ and $2$ share no factor beyond $1$ , so nothing further comes out.

Part B

Every term here is the opposite of the matching term of $K$ , so this expression is $-K$ . Take the same factor out with a minus sign attached, and the bracket comes back with its first term positive.

$-18a^3b + 30a^2b^2 - 12a^2b = -6a^2b(3a) - 6a^2b(-5b) - 6a^2b(2) = -6a^2b(3a - 5b + 2).$

Multiplying back out returns each printed term, which is the check that the signs are right.

Part C

Test the candidate against each term in turn.

$\frac{-30a^2b^2}{6a^2b^2} = -5, \qquad \frac{18a^3b}{6a^2b^2} = \frac{3a}{b}, \qquad \frac{12a^2b}{6a^2b^2} = \frac{2}{b}.$

Only the first comes out as a term. The first and third terms of $K$ carry a single $b$ each, so a factor demanding two of them cannot be lifted out of either, and fitting one term out of three is not enough: the identity behind factoring, $ab + ac + ad = a(b + c + d)$ , needs the same quantity present in every term before anything can be taken outside.

The general requirement follows directly. A quantity may be taken outside a bracket exactly when it divides every term: its numerical part must divide every coefficient, and each letter it contains must appear in every term to at least the power the factor uses. That is why the greatest quantity meeting the requirement takes each letter at its lowest power across the terms, and why $6a^2b$ comes out of $K$ while $6a^2b^2$ does not.

In one line

$K = 6a^2b(3a - 5b + 2)$ , built from the greatest common divisor $6$ of the three coefficients together with each letter at its lowest power, $a^2$ and $b$ . The opposite expression factors as $-6a^2b(3a - 5b + 2)$ . The candidate $6a^2b^2$ cannot come out, because the first and third terms of $K$ carry only one $b$ each: a quantity may be taken outside a bracket only when it divides every term, which is exactly why the greatest common factor uses each letter at its lowest power.

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

Takes the numerical part to be the greatest common divisor of all three coefficients, not of two of them. . Worth 2 points.

Takes each letter at the lowest power appearing in every term, and includes no letter that is missing from a term. . Worth 2 points.

Leaves a three-term bracket whose terms share no factor beyond $1$ . . Worth 1 point.

Part B 4 points

Pulls out a factor carrying a minus sign, so the bracket opens with a positive term. . Worth 2 points.

Gets the sign of every term inside the bracket right, so that the printed expression is what the factorization multiplies back out to. . Worth 2 points.

Part C 6 points

Rejects the candidate by testing it against every term rather than only the one it fits, naming the terms it fails on and why. . Worth 3 points. needs an explanation, not just an answer

States the general requirement as dividing every term, covering both the numerical part and the power of each letter. . Worth 2 points. needs an explanation, not just an answer

Connects that requirement back to why the greatest common factor takes each letter at its lowest power. . 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 brackets opened, and one sign with further to travel than it looks 9 points

Part A 3

Multiplies the outside factor against both terms of each bracket, keeping the sign of the second term. 2
Leaves each answer as two terms, recognising that a variable term and a constant do not combine. 1

Part B 3

Reverses the sign of every term of the expression being subtracted, not only of the first. 2
Gathers the like terms of the result, leaving nothing further to combine. 1

Part C 3

Locates the rise in the fact that subtracting a whole expression reverses the sign of each of its terms, so a negative constant arrives positive. 2
States the constant that the truncated subtraction would have produced. 1

2. One number, two spellings, and a sum that only looks impossible 10 points

Part A 3

Uses the denominator of each exponent as a root index and the numerator as a power, and inverts the base where the exponent is negative. 2
Reports two plain positive numbers, neither of them negative. 1

Part B 3

Pulls a perfect-cube factor out of each radicand, matching the factor to the index rather than to a square. 2
Adds the coefficients of the matching radical and leaves the radicand itself untouched. 1

Part C 4

Gives the minus sign, the numerator and the denominator a distinct job each, rather than describing the calculation as one indivisible step. 2
Explains both why the result is positive and why it is greater than $1$ , arguing from what a negative exponent does to the base. 2

3. Out and back along the same road 10 points

Part A 3

Forms all four products and carries the sign of the $-3$ into both products it appears in. 2
Gathers the two middle products into a single term with the correct sign. 1

Part B 3

Builds the factor from the greatest common divisor of the coefficients together with the lowest power of $x$ present in every term. 2
Leaves a bracket whose terms share no factor beyond $1$ . 1

Part C 4

Ties the number of products to the pairing of every term with every term, rather than to a mnemonic or to the shape of the factors. 2
Ties the number of terms left inside the bracket to the number of terms in the original sum, and keeps the gathering step separate from the multiplying step. 2

4. Two bars joined into one, and the values the join has to remember 11 points

Part A 4

Uses a denominator that both originals divide, and scales each numerator by exactly what its own denominator was missing. 2
Expands and gathers the numerator correctly, leaving three unlike terms over the common denominator. 2

Part B 3

Evaluates both forms at the stated value with the order of operations intact. 2
Says what agreement between the two values does and does not establish. 1

Part C 4

Names both barred values and attaches each to the fraction whose denominator produces it. 2
Argues from a product being zero only when a factor is zero that the combined denominator bars exactly the same values, rather than testing the two values one at a time. 2

5. A claim that survives the first two things you try 11 points

Part A 3

Evaluates both sides at each pair, squaring the sum itself rather than the terms separately. 2
Reports that both pairs agree without treating that agreement as settling the claim. 1

Part B 3

Supplies a specific pair and evaluates both sides on it, reaching two different numbers. 2
Concludes that one disagreeing pair refutes a claim made about every pair. 1

Part C 5

Expands the square into three terms and identifies the middle term as the whole of the difference between the two sides. 3
Describes exactly which pairs make the claim true, in both directions, and shows that the two pairs of part A are among them. 2

6. Two baskets from one price list 12 points

Part A 4

Prices the pens as a count times the pen price, so the discount is taken once for each pen and not once for the whole basket. 2
Distributes and gathers each total into one variable term and one constant. 1
States each total as an amount of money. 1

Part B 3

Reverses the sign of both terms of the total being subtracted. 2
Gathers the difference into one variable term and one constant. 1

Part C 5

Says how the gap responds to a change in the notebook price, reading the rate from the coefficient rather than from a couple of sample prices. 2
Sets the difference to zero and solves for the price at which the two totals are equal. 2
Gives that price as an amount of money and reads it back against the situation. 1

7. A bracket that has to be earned before anything can be crossed out 13 points

Part A 4

Builds each factor from the greatest common divisor of that expression's own coefficients and the lowest power of $x$ present in all of its terms. 2
Leaves each bracket holding terms that share no factor beyond $1$ . 2

Part B 4

Cancels a factor shared by the whole numerator and the whole denominator, leaving the remaining factors of each intact. 2
Reads both excluded values off the original denominator, including the one the simplified form no longer objects to. 2

Part C 5

Identifies the factor still shared by the whole of each part of the intermediate form and shows what cancelling it produces. 3
States a completeness test in terms of what the fully factored numerator and denominator still share, and says why multiplying back out cannot settle it. 2

8. A setting turned right down, and an output that does not follow it 11 points

Part A 3

Takes the square root before the cube, using the denominator of the exponent as the index and the numerator as the power. 2
Reports the flow rate with its unit. 1

Part B 3

Roots the numerator and the denominator of the setting separately, then cubes, then reduces the result. 2
Reports the flow rate with its unit. 1

Part C 5

Computes the factor between the two flow rates from the two values found earlier. 2
Explains that a factor on the setting reaches the flow rate raised to the same exponent, uses that to reproduce the factor found, and says why no one multiplier would serve for every pair of settings. 3

9. Three rewritings, and the reasons offered for them 12 points

Part A 4

Reaches the right verdict on all three lines, supporting each rather than asserting it. 2
Names the law behind each true line, telling a regrouping apart from a distribution. 2

Part B 3

Substitutes a value into both sides, keeping the bracket intact rather than removing it first. 2
Reaches two different numbers and treats that disagreement as settling the line. 1

Part C 5

Names the law behind the intended rewriting, identifies the hidden factor of $-1$ , and writes the line correctly. 3
Explains the asymmetry between one refuting value and any number of agreeing ones, and says what does establish a claim about every number. 2

10. Everything the terms hold in common, and nothing they do not 15 points

Part A 5

Takes the numerical part to be the greatest common divisor of all three coefficients, not of two of them. 2
Takes each letter at the lowest power appearing in every term, and includes no letter that is missing from a term. 2
Leaves a three-term bracket whose terms share no factor beyond $1$ . 1

Part B 4

Pulls out a factor carrying a minus sign, so the bracket opens with a positive term. 2
Gets the sign of every term inside the bracket right, so that the printed expression is what the factorization multiplies back out to. 2

Part C 6

Rejects the candidate by testing it against every term rather than only the one it fits, naming the terms it fails on and why. 3
States the general requirement as dividing every term, covering both the numerical part and the power of each letter. 2
Connects that requirement back to why the greatest common factor takes each letter at its lowest power. 1

Multiple choice

Free response