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Special Factorizations: 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, 95 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. Comparing two squared binomials . 8 points. Question 1 of 10.

A squared sum and a squared difference are built from the same two pieces and differ only in one sign. Expand both, then track exactly which parts of the result depend on that sign and which do not.
1. Part A.
  Expand $(4x + 5)^2$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Expand $(4x - 5)^2$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
3. Part C.
  Name the two terms that came out identical in both expansions, and explain why only the middle term's sign changes, tracing it to the sign of the cross-products.
  
  Carry your own answer forward Compare your own two expansions from parts A and B, whatever they came out to; the relationship between them is the point, not the specific numbers.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
The answer

Part A

$(4x + 5)^2 = 16x^2 + 40x + 25$ .

Part B

$(4x - 5)^2 = 16x^2 - 40x + 25$ .

Part C

The first term $16x^2$ and the last term $25$ are identical in both. Only the middle term changes sign, because the two cross-products carry the sign of the second term: both positive in the sum, both negative in the difference.

Worked solution

Part A

Square the whole first term, double the product of the two terms, and square the last term.

$(4x + 5)^2 = (4x)^2 + 2(4x)(5) + 5^2 = 16x^2 + 40x + 25$

Part B

Apply the squared-difference pattern to the same two terms; only the middle term's sign changes.

$(4x - 5)^2 = (4x)^2 - 2(4x)(5) + 5^2 = 16x^2 - 40x + 25$

Part C

Line the results up:

$16x^2 + 40x + 25 \qquad \text{and} \qquad 16x^2 - 40x + 25.$

The first term and the last term never involve the sign of the second piece, since $5^2$ and $(-5)^2$ are both $25$ . The middle term is the sum of the two cross-products $4x\cdot 5$ , both positive in $(4x + 5)^2$ and both negative in $(4x - 5)^2$ , so only it flips.

In one line

$(4x + 5)^2 = 16x^2 + 40x + 25$ and $(4x - 5)^2 = 16x^2 - 40x + 25$ ; both share the first term $16x^2$ and last term $25$ , and only the middle term's sign changes, because the cross-products carry the sign of the second 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 3 points

Squares the whole first term $4x$ to $16x^2$ and uses the doubled middle term $2ab$ , rather than squaring only the variable or halving the middle. . Worth 2 points.

Computes all three terms correctly, keeping the last term positive. . Worth 1 point.

Part B 2 points

Uses the squared-difference pattern on the whole terms $4x$ and $5$ . . Worth 1 point.

Computes all three terms correctly, keeping the last term positive and only the middle term negative. . Worth 1 point.

Part C 3 points

Names the first term and the last term as the two shared by both expansions. . Worth 1 point.

Explains the sign change by tracing it to the sign of the two cross-products, not merely stating that the signs differ. . Worth 2 points. needs an explanation, not just an answer
2. One root, then two . 10 points. Question 2 of 10.

A radical in a denominator is cleared by multiplying by a carefully chosen form of $1$ . Which form to use depends on whether the denominator is a single root or a two-term expression.
1. Part A.
  Rationalize the denominator and simplify: $\dfrac{10}{\sqrt{5}}$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Rationalize the denominator and simplify: $\dfrac{2}{3 + \sqrt{5}}$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  In part B you cleared the root by multiplying by the conjugate. Explain why the conjugate works where multiplying by another copy of the denominator would not.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$2\sqrt{5}$ .

Part B

$\dfrac{3 - \sqrt{5}}{2}$ .

Part C

The conjugate has the opposite middle sign, so the two cross-products cancel and leave $9 - 5$ , a rational number. A copy of $3 + \sqrt{5}$ squares it, giving $9 + 6\sqrt{5} + 5$ , whose middle terms add instead of cancelling, so a root survives.

Worked solution

Part A

Multiply the numerator and denominator by $\sqrt{5}$ , so the denominator becomes $5$ , then reduce.

$\frac{10}{\sqrt{5}} = \frac{10}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}$

Part B

Multiply by the conjugate $3 - \sqrt{5}$ . The denominator becomes a difference of squares, $9 - 5 = 4$ , and the fraction reduces by $2$ .

$\frac{2}{3 + \sqrt{5}} = \frac{2(3 - \sqrt{5})}{9 - 5} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$

Part C

The conjugate product is a difference of squares:

$(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4,$

because the cross-products $-3\sqrt{5}$ and $+3\sqrt{5}$ are opposites and cancel. Multiplying by a copy instead squares the expression:

$(3 + \sqrt{5})^2 = 9 + 6\sqrt{5} + 5,$

where both cross-products are $+3\sqrt{5}$ and add, so the radical stays. Only opposite signs force the cancellation.

In one line

$\dfrac{10}{\sqrt{5}} = 2\sqrt{5}$ and $\dfrac{2}{3 + \sqrt{5}} = \dfrac{3 - \sqrt{5}}{2}$ ; the conjugate works because its opposite middle sign makes the cross-products cancel into $a^2 - b$ , while a repeated copy squares the denominator and leaves a radical.

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 numerator and denominator by $\sqrt{5}$ , the same root in the denominator. . Worth 1 point.

Uses $\sqrt{5}\cdot\sqrt{5} = 5$ and reduces $\tfrac{10}{5}$ so no common factor is left. . Worth 2 points.

Part B 4 points

Multiplies numerator and denominator by the conjugate $3 - \sqrt{5}$ , not by a copy of the denominator. . Worth 2 points.

Evaluates the denominator as $9 - 5 = 4$ and reduces the fraction completely. . Worth 2 points.

Part C 3 points

Explains that the conjugate's opposite-signed cross-products cancel to leave a rational $a^2 - b$ , arguing the mechanism rather than asserting it. . Worth 2 points. needs an explanation, not just an answer

Contrasts this with squaring a copy, whose equal-signed cross-products add and keep the root. . Worth 1 point.
3. Split it, solve it, then swap the split . 10 points. Question 3 of 10.

The AC method rewrites the middle term of a non-monic trinomial as two pieces so the four terms can be grouped. Once it is factored, the equation it came from falls to the zero-product property.
1. Part A.
  Factor $3x^2 + 14x + 8$ by the AC method: find the two numbers, split the middle term, then group.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Use your factorization to solve $3x^2 + 14x + 8 = 0$ . Report every solution.
  
  Carry your own answer forward Solve using whichever two factors your part A produced, even if they are not the expected pair.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Redo the split from part A with the same two numbers written in the opposite order, group again, and say whether the order the split is written in changes the final factorization, and why.
  
  Carry your own answer forward Use the same two numbers your part A search found, written in the reverse order; the comparison holds whichever pair you found.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 3 points
The answer

Part A

$3x^2 + 14x + 8 = (x + 4)(3x + 2)$ .

Part B

$x = -4$ or $x = -\dfrac{2}{3}$ .

Part C

No. Splitting as $2x + 12x$ groups to $(3x + 2)(x + 4)$ , the same two factors. The order does not matter, because addition does not depend on order, so the four-term polynomial being grouped is the same expression either way.

Worked solution

Part A

With $a = 3$ , $b = 14$ , $c = 8$ , the method needs two numbers with product $ac = 24$ and sum $14$ : those are $12$ and $2$ . Split and group.

$3x^2 + 14x + 8 = 3x^2 + 12x + 2x + 8 = 3x(x + 4) + 2(x + 4) = (x + 4)(3x + 2)$

Part B

Set each factor equal to zero.

$x + 4 = 0 \ \Rightarrow\ x = -4, \qquad 3x + 2 = 0 \ \Rightarrow\ x = -\frac{2}{3}$

Part C

Writing the split in the other order gives

$3x^2 + 2x + 12x + 8 = x(3x + 2) + 4(3x + 2) = (3x + 2)(x + 4),$

the same factorization as part A. The middle term $14x$ is the sum $12x + 2x$ , and addition gives the same total no matter which piece is written first, so the four-term line has the same value either way and must group to the same result.

In one line

$3x^2 + 14x + 8 = (x + 4)(3x + 2)$ , giving $x = -4$ and $x = -\dfrac{2}{3}$ ; splitting the middle term in the opposite order regroups to the identical factorization, because addition does not depend on which piece is written first.

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 with product $ac = 24$ and sum $14$ , satisfying both conditions at once. . Worth 2 points.

Splits the middle term with that pair, groups, and factors each pair to the complete factorization. . Worth 2 points.

Part B 3 points

Sets each factor from part A equal to zero separately. . Worth 1 point.

Reports both solutions, including the fractional one, as the complete solution set. . Worth 2 points.

Part C 3 points

Actually redoes the split and grouping in the reversed order and reaches the identical factorization, rather than only asserting it. . Worth 2 points.

Explains that the reason is addition not depending on order, so the four-term line is unchanged in value. . Worth 1 point. needs an explanation, not just an answer
4. Which ones factor, and which only look like they should . 9 points. Question 4 of 10.

Two-term expressions can be a difference of squares, a sum or difference of cubes, or none of these. Telling them apart, and being precise about what 'does not factor' means, is the whole job here.
1. Part A.
  Classify and factor $x^2 - 100$ and $x^3 - 125$ , naming which special form each one is.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Does $x^2 + 9$ factor? Answer precisely, and justify your answer from what a product of two real binomials can produce.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
3. Part C.
  The difference of cubes $x^3 - 8$ factors as $(x - 2)(x^2 + 2x + 4)$ . Show that its trinomial factor $x^2 + 2x + 4$ does not factor further over the integers.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$x^2 - 100 = (x + 10)(x - 10)$ , a difference of squares; $x^3 - 125 = (x - 5)(x^2 + 5x + 25)$ , a difference of cubes.

Part B

$x^2 + 9$ does not factor over the real numbers. It is a sum of squares, and the only candidate, $(x + a)(x - a)$ , expands to $x^2 - a^2$ , a difference, which can never equal $x^2 + 9$ for any real $a$ .

Part C

An integer factorization $(x + p)(x + q)$ would need $pq = 4$ and $p + q = 2$ . The integer pairs multiplying to $4$ are $1$ and $4$ (sum $5$ ) and $2$ and $2$ (sum $4$ ); negatives give a negative sum. None sums to $2$ , so it does not factor over the integers.

Worked solution

Part A

The first is a difference of squares, $x^2 - 10^2$ , with opposite-sign factors. The second is a difference of cubes, $x^3 - 5^3$ , factored by SOAP with a single- $ab$ trinomial.

$x^2 - 100 = (x + 10)(x - 10), \qquad x^3 - 125 = (x - 5)(x^2 + 5x + 25)$

Part B

A sum of squares has no real linear factorization. The natural candidate is a product of two real binomials $(x + a)(x - a)$ , but that expands to a difference:

$(x + a)(x - a) = x^2 - a^2,$

whose constant term is $-a^2 \le 0$ and can never equal the $+9$ in $x^2 + 9$ . So the claim must be stated with its domain: $x^2 + 9$ does not factor over the real numbers.

Part C

To factor $x^2 + 2x + 4$ as $(x + p)(x + q)$ with integers you would need $pq = 4$ and $p + q = 2$ . List the integer pairs with product $4$ and check their sums:

$1 \times 4 \ (\text{sum } 5), \qquad 2 \times 2 \ (\text{sum } 4), \qquad (-1)(-4), \ (-2)(-2) \ (\text{negative sums}).$

None of them sums to $2$ , so no integer pair works and $x^2 + 2x + 4$ does not factor over the integers.

In one line

$x^2 - 100 = (x + 10)(x - 10)$ and $x^3 - 125 = (x - 5)(x^2 + 5x + 25)$ ; $x^2 + 9$ does not factor over the real numbers, since $(x + a)(x - a) = x^2 - a^2$ can never build a positive constant; and $x^2 + 2x + 4$ does not factor over the integers, since no integer pair multiplies to $4$ and sums to $2$ .

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

Names each form correctly (difference of squares, difference of cubes) rather than only factoring. . Worth 1 point.

Factors both correctly, with opposite signs for the squares and the SOAP signs and single $ab$ for the cubes. . Worth 2 points.

Part B 3 points

Argues from what $(x + a)(x - a) = x^2 - a^2$ can produce that no real factorization gives $x^2 + 9$ , rather than asserting it. . Worth 2 points. needs an explanation, not just an answer

States the verdict with its domain named: does not factor over the real numbers. . Worth 1 point.

Part C 3 points

Sets the correct test: an integer pair with product $4$ and sum $2$ at once. . Worth 1 point.

Checks the integer pairs, rules out negatives, and concludes it does not factor over the integers. . Worth 2 points. needs an explanation, not just an answer
5. Four pieces of one square . 9 points. Question 5 of 10.

A square plot of side length $x + 6$ is paved with four rectangular pieces arranged in a two-by-two block: one $x$ by $x$ piece, two $x$ by $6$ strips, and one $6$ by $6$ piece. The area can be found two ways, and they must agree.
1. Part A.
  Write the plot's total area as a single trinomial by squaring its side length.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
2. Part B.
  Add the areas of the four separate pieces and confirm the total matches your trinomial from part A, term by term.
  
  Carry your own answer forward Check the sum against your own trinomial from part A, whatever it came out to; what matters is whether the two routes agree term by term.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
3. Part C.
  Suppose only ONE $x$ by $6$ strip had been used instead of two. Name the area that would be missing, and explain in general why the middle term of a squared binomial always needs two equal strips, not one.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$(x + 6)^2 = x^2 + 12x + 36$ square units.

Part B

The four areas are $x^2$ , $6x$ , $6x$ , and $36$ , and their sum $x^2 + 12x + 36$ matches part A. The single piece gives the first square, the corner piece gives the last square, and the two strips together give the middle term.

Part C

The missing area is one strip, $6x$ . In general the middle term of $(a + b)^2$ is $2ab$ because a square of side $a + b$ splits into two equal $a$ by $b$ rectangles, one along the top and one along the side; leaving out either loses a full $ab$ .

Worked solution

Part A

The whole plot is a square of side $x + 6$ , so its area is $(x + 6)^2$ . Expand with the squared-sum pattern.

$(x + 6)^2 = x^2 + 2(x)(6) + 6^2 = x^2 + 12x + 36$

Part B

Each piece contributes one area: $x\cdot x = x^2$ , then $x\cdot 6 = 6x$ twice, then $6\cdot 6 = 36$ .

$x^2 + 6x + 6x + 36 = x^2 + 12x + 36$

That is exactly part A, matched term by term: the two $6x$ strips are what combine into the middle term $12x$ .

Part C

With one strip, the three pieces total $x^2 + 6x + 36$ , short of the correct $x^2 + 12x + 36$ by one $6x$ strip.

This is general:

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

The middle term $2ab$ exists because a square of side $a + b$ genuinely splits into TWO equal $a$ by $b$ rectangles, not one. Counting only one leaves the area short by exactly one $ab$ , whatever $a$ and $b$ are.

In one line

The plot's area is $(x + 6)^2 = x^2 + 12x + 36$ , matching the four piece areas $x^2 + 6x + 6x + 36$ term by term; using only one strip would leave the total short by $6x$ , because a squared binomial's middle term comes from two equal cross-rectangles, not 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 2 points

Sets the area up as $(x + 6)^2$ , the square of the side length, before expanding. . Worth 1 point.

Expands to a trinomial with the correct first, middle, and last terms. . Worth 1 point.

Part B 4 points

Computes the correct area for all four pieces and adds them. . Worth 2 points.

Matches the sum to part A term by term, noting the middle term comes from the two strips together. . Worth 2 points.

Part C 3 points

Identifies the missing area as one strip's worth. . Worth 1 point.

Explains in general, not just for this plot, why the middle term needs two equal regions rather than one. . Worth 2 points. needs an explanation, not just an answer
6. Clearing two roots, and the reason it works . 10 points. Question 6 of 10.

Clearing a two-term radical denominator uses the conjugate, and the reason it works is a pattern from earlier in this chapter. This question uses it twice, then asks you to name it.
1. Part A.
  Rationalize the denominator and simplify: $\dfrac{4}{\sqrt{7} - \sqrt{3}}$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Rationalize the denominator and simplify: $\dfrac{6}{\sqrt{5} + \sqrt{2}}$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Explain why multiplying a denominator $\sqrt{a} + \sqrt{b}$ by its conjugate always produces the rational number $a - b$ , and name the identity from earlier in this chapter that this is.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$\sqrt{7} + \sqrt{3}$ .

Part B

$2\sqrt{5} - 2\sqrt{2}$ .

Part C

$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b$ : this is the difference-of-squares identity $A^2 - B^2$ with $A = \sqrt{a}$ and $B = \sqrt{b}$ . The cross-terms $-\sqrt{ab}$ and $+\sqrt{ab}$ cancel, and each squared root becomes a whole number, so no radical remains.

Worked solution

Part A

Multiply by the conjugate $\sqrt{7} + \sqrt{3}$ . The denominator becomes $7 - 3 = 4$ , which cancels the numerator's $4$ .

$\frac{4}{\sqrt{7} - \sqrt{3}} = \frac{4(\sqrt{7} + \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} + \sqrt{3})}{4} = \sqrt{7} + \sqrt{3}$

Part B

Multiply by the conjugate $\sqrt{5} - \sqrt{2}$ . The denominator becomes $5 - 2 = 3$ .

$\frac{6}{\sqrt{5} + \sqrt{2}} = \frac{6(\sqrt{5} - \sqrt{2})}{5 - 2} = \frac{6(\sqrt{5} - \sqrt{2})}{3} = 2\sqrt{5} - 2\sqrt{2}$

Part C

The conjugate product is exactly the difference-of-squares pattern with the two roots as $A$ and $B$ :

$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b.$

The two cross-terms $-\sqrt{ab}$ and $+\sqrt{ab}$ are opposites and cancel, and squaring a square root removes the radical, so the result $a - b$ is rational for any nonnegative $a$ and $b$ .

In one line

$\dfrac{4}{\sqrt{7} - \sqrt{3}} = \sqrt{7} + \sqrt{3}$ and $\dfrac{6}{\sqrt{5} + \sqrt{2}} = 2\sqrt{5} - 2\sqrt{2}$ ; the conjugate clears the root because $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$ is the difference-of-squares identity, whose cross-terms cancel and whose squared roots are rational.

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 numerator and denominator by the conjugate $\sqrt{7} + \sqrt{3}$ . . Worth 2 points.

Evaluates the denominator as $7 - 3 = 4$ and cancels to leave no fraction. . Worth 1 point.

Part B 4 points

Multiplies numerator and denominator by the conjugate $\sqrt{5} - \sqrt{2}$ . . Worth 2 points.

Evaluates the denominator as $5 - 2 = 3$ and reduces both numerator terms by it. . Worth 2 points.

Part C 3 points

Argues the mechanism: the cross-terms cancel and each squared root becomes rational, giving $a - b$ , rather than asserting it. . Worth 2 points. needs an explanation, not just an answer

Names this as the difference-of-squares identity $A^2 - B^2$ applied to the two roots. . Worth 1 point.
7. Taking a cubic apart . 11 points. Question 7 of 10.

A three-term expression with a leading coefficient is still a factoring problem in two moves: remove what every term shares, then treat what is left with the AC method. The factored form then solves the equation.
1. Part A.
  Factor completely: $12x^3 + 2x^2 - 4x$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Use your complete factorization to solve $12x^3 + 2x^2 - 4x = 0$ . Report every solution.
  
  Carry your own answer forward Solve using your own complete factorization from part A, including every one of its factors.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  How many distinct solutions does the equation have, and why does the monomial factor pulled out in part A contribute a solution while the bare constant inside it contributes none? Argue from the zero-product property.
  
  Carry your own answer forward Count from your own factorization and solutions in parts A and B, whatever they were.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

$12x^3 + 2x^2 - 4x = 2x(3x + 2)(2x - 1)$ .

Part B

$x = 0$ , $x = -\dfrac{2}{3}$ , or $x = \dfrac{1}{2}$ .

Part C

Three distinct solutions: $0$ , $-\dfrac{2}{3}$ , and $\dfrac{1}{2}$ . The zero-product property sets each factor containing $x$ to zero, and $2x = 0$ forces $x = 0$ ; but the constant $2$ is never zero, so it can contribute no solution.

Worked solution

Part A

Every term shares $2x$ , so pull it out first, then apply the AC method to what remains.

$12x^3 + 2x^2 - 4x = 2x(6x^2 + x - 2).$

For $6x^2 + x - 2$ , $ac = -12$ and the split pair with sum $1$ is $4$ and $-3$ :

$6x^2 + 4x - 3x - 2 = 2x(3x + 2) - 1(3x + 2) = (3x + 2)(2x - 1).$

Keeping the common factor, the complete factorization is $2x(3x + 2)(2x - 1)$ .

Part B

Set each variable factor of $2x(3x + 2)(2x - 1)$ equal to zero.

$2x = 0 \ \Rightarrow\ x = 0, \qquad 3x + 2 = 0 \ \Rightarrow\ x = -\frac{2}{3}, \qquad 2x - 1 = 0 \ \Rightarrow\ x = \frac{1}{2}$

Part C

The complete factorization has three factors that contain $x$ : $2x$ , $3x + 2$ , and $2x - 1$ . The zero-product property makes the product zero exactly when one of these is zero, giving three distinct values.

$2x(3x + 2)(2x - 1) = 0 \ \Rightarrow\ x = 0,\ -\frac{2}{3},\ \frac{1}{2}$

The factor $2x$ contributes $x = 0$ because it contains $x$ ; the bare constant $2$ inside it is never zero, so it adds no equation and no solution.

In one line

$12x^3 + 2x^2 - 4x = 2x(3x + 2)(2x - 1)$ , reached by pulling out $2x$ and then applying the AC method; the equation has three distinct solutions $x = 0$ , $-\dfrac{2}{3}$ , and $\dfrac{1}{2}$ , one from each factor that contains $x$ , while the constant $2$ contributes none.

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

Pulls out the common factor $2x$ before applying the AC method, rather than trying to factor the trinomial with the common factor still in it. . Worth 2 points.

Applies the AC method correctly to the remaining trinomial and keeps the common factor in the final answer. . Worth 2 points.

Part B 3 points

Applies the zero-product property to every factor, including the leading $2x$ , not only the two binomials. . Worth 1 point.

Reports all three solutions, including the fractional ones. . Worth 2 points.

Part C 4 points

States three distinct solutions, tied to the three factors that contain $x$ . . Worth 2 points.

Explains from the zero-product property why $2x$ contributes a solution but the constant $2$ does not. . Worth 2 points. needs an explanation, not just an answer
8. Cubing a sum is not adding two cubes . 8 points. Question 8 of 10.

A claim states that $a^3 + b^3 = (a + b)^3$ , so that a sum of cubes could be factored just by cubing the sum. This question tests that claim and then applies the real identity.
1. Part A.
  Disprove the claim $a^3 + b^3 = (a + b)^3$ : choose one specific pair of numbers, evaluate both sides, and show they differ.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 3 points
2. Part B.
  Factor $x^3 + 125$ using the correct sum-of-cubes identity.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
3. Part C.
  Expand $(a + b)^3$ in full, and state exactly which terms it carries beyond $a^3 + b^3$ , so that the difference found in part A is accounted for.
  
  Carry your own answer forward Account for the specific gap you found in part A, whatever pair you used, using the two extra terms.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 3 points
The answer

Part A

With $a = 1$ and $b = 2$ : $a^3 + b^3 = 1 + 8 = 9$ , but $(a + b)^3 = 3^3 = 27$ . Since $9 \neq 27$ , the claim is false.

Part B

$x^3 + 125 = (x + 5)(x^2 - 5x + 25)$ .

Part C

$(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$ , so beyond $a^3 + b^3$ it carries two extra terms, $3a^2 b$ and $3ab^2$ . With $a = 1$ , $b = 2$ those extra terms are $6 + 12 = 18$ , exactly the gap between $9$ and $27$ .

Worked solution

Part A

A claim about all numbers falls to a single pair on which it fails, so evaluate both sides on a small pair.

$a^3 + b^3 = 1^3 + 2^3 = 9, \qquad (a + b)^3 = (1 + 2)^3 = 27$

Since $9 \neq 27$ , the pair $a = 1$ , $b = 2$ refutes the claim.

Part B

This is a sum of cubes with $a = x$ and $b = 5$ . By SOAP the binomial keeps the plus sign, the trinomial's middle term takes the opposite sign, and the last term is $b^2$ .

$x^3 + 125 = x^3 + 5^3 = (x + 5)(x^2 - 5x + 25)$

Part C

Expand the cubed sum fully:

$(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3.$

Comparing with $a^3 + b^3$ , the cubed sum carries two extra middle terms, $3a^2 b$ and $3ab^2$ , that the sum of cubes never has. On the pair from part A they total $3(1)(2) + 3(1)(4) = 18$ , which is precisely $27 - 9$ .

In one line

The pair $a = 1$ , $b = 2$ gives $a^3 + b^3 = 9$ but $(a + b)^3 = 27$ , so the claim is false; the correct factorization is $x^3 + 125 = (x + 5)(x^2 - 5x + 25)$ ; and $(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$ carries the two extra terms $3a^2 b$ and $3ab^2$ that account for the gap.

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

Chooses a specific numerical pair rather than arguing in general terms. . Worth 1 point.

Evaluates both sides correctly on that pair and reaches two different values. . Worth 2 points.

Part B 2 points

Identifies the cube roots $x$ and $5$ and keeps the binomial's plus sign. . Worth 1 point.

Forms the trinomial with a single $ab$ middle term of the opposite sign and the correct last term. . Worth 1 point.

Part C 3 points

Expands $(a + b)^3$ correctly into its four terms. . Worth 2 points.

Names the two extra terms $3a^2 b$ and $3ab^2$ and ties them to the numerical gap from part A. . Worth 1 point.
9. Correct is not the same as complete . 11 points. Question 9 of 10.

A factorization can multiply back to the original expression and still not be finished, because one of its factors hides another pattern. Telling correct from complete is the point of this question.
1. Part A.
  A proposed factorization of $x^4 - 16$ is $(x^2 + 4)(x^2 - 4)$ . Decide whether it is complete, and if not, finish it.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
2. Part B.
  Factor $3x^3 - 3x$ completely.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  State the general two-step habit for factoring completely, and use $x^4 - 16$ to explain how a factorization can rebuild the original expression and still not be complete.
  
  Carry your own answer forward Argue from the expression and the proposed factorization in part A, whatever you concluded about it.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

Not complete. $(x^2 + 4)(x^2 - 4)$ does multiply back to $x^4 - 16$ , but $x^2 - 4$ is itself a difference of squares. The complete factorization is $(x^2 + 4)(x + 2)(x - 2)$ ; $x^2 + 4$ is a sum of squares and stops over the real numbers.

Part B

$3x^3 - 3x = 3x(x + 1)(x - 1)$ .

Part C

Pull out any common factor first, then re-check every resulting factor for another pattern. $(x^2 + 4)(x^2 - 4)$ multiplies back to $x^4 - 16$ , so it is a correct product, but not complete, because $x^2 - 4$ is still a difference of squares that factors again.

Worked solution

Part A

Check each factor for another pattern. The product is genuinely correct, since $(x^2 + 4)(x^2 - 4) = x^4 - 16$ , but $x^2 - 4$ is a difference of squares and factors again:

$x^2 - 4 = (x + 2)(x - 2).$

The factor $x^2 + 4$ is a sum of squares and does not factor over the real numbers, so the work stops there. The complete factorization is $(x^2 + 4)(x + 2)(x - 2)$ .

Part B

Pull out the common factor $3x$ first, which uncovers a difference of squares, then factor that too.

$3x^3 - 3x = 3x(x^2 - 1) = 3x(x + 1)(x - 1)$

Part C

The habit has a fixed order: remove any common factor first, then re-check every factor that remains for another special pattern, and only stop when no factor matches one.

Expanding proves a product correct, but it cannot prove it complete:

$(x^2 + 4)(x^2 - 4) = x^4 - 16, \qquad x^2 - 4 = (x + 2)(x - 2).$

The product expands to $x^4 - 16$ , yet completeness is a separate question, answered only by testing each factor again. Here $x^2 - 4$ still factors, so the correct product was not the finished one.

In one line

$(x^2 + 4)(x^2 - 4)$ is correct but incomplete, since $x^2 - 4$ factors again; complete is $(x^2 + 4)(x + 2)(x - 2)$ . Also $3x^3 - 3x = 3x(x + 1)(x - 1)$ . The habit is common factor first, then re-check every factor, because expanding proves a product correct but never proves it complete.

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

Recognizes the proposed product is correct yet incomplete because $x^2 - 4$ factors further. . Worth 2 points.

Finishes the factorization and correctly stops at the sum of squares, naming that it does not factor over the real numbers. . Worth 2 points.

Part B 4 points

Pulls out the common factor $3x$ before checking for a pattern. . Worth 2 points.

Factors the remaining difference of squares and keeps the common factor in the answer. . Worth 2 points.

Part C 3 points

States the two-step habit in the correct order: common factor first, then re-check every resulting factor. . Worth 1 point.

Explains that expanding proves a product correct but not complete, using $x^2 - 4$ still factoring as the evidence. . Worth 2 points. needs an explanation, not just an answer
10. The step where the reasoning first breaks . 9 points. Question 10 of 10.

Below is an attempt to factor $25x^2 - 34x + 9$ as a perfect square, written one step at a time.

Step 1 restates the trinomial: $25x^2 - 34x + 9$ .

Step 2 notes the outer terms are perfect squares: $\sqrt{25x^2} = 5x$ and $\sqrt{9} = 3$ .

Step 3 concludes it is a perfect square: $(5x - 3)^2$ .

Step 4 expands to check: $(5x - 3)^2 = 25x^2 - 30x + 9$ .

Step 5 concludes: $25x^2 - 34x + 9 = (5x - 3)^2$ .

The final claim is false, but the step where the reasoning first breaks is not the step where the false equation first appears.
1. Part A.
  Identify the first step that is not fully justified (not the step where the false equation first appears), and say exactly what check it skipped.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 3 points
2. Part B.
  Factor $25x^2 - 34x + 9$ correctly.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Explain in general why matching the two outer terms is not enough to conclude a perfect-square trinomial, and state the third check that actually decides it.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

Step 3. It concluded a perfect square from the two outer squares alone, never checking that the middle term equals $2(5x)(3) = 30x$ . The actual middle term is $34x$ , so the trinomial is not a perfect square.

Part B

$25x^2 - 34x + 9 = (25x - 9)(x - 1)$ .

Part C

Two outer perfect squares only fix the candidate roots $A$ and $B$ ; they say nothing about the middle. The middle term must independently equal $2AB$ . Without that third check, every trinomial with square outer terms would pass, which is exactly the slip in Step 3.

Worked solution

Part A

Test each step for what it establishes. Steps 1 and 2 are sound: the outer terms really are perfect squares. Step 3 is the first unjustified step: it names the trinomial a perfect square on the strength of the outer terms alone, skipping the middle-term test.

$2(5x)(3) = 30x \neq 34x$

Step 4 is correct arithmetic on the wrong claim (the false equation only surfaces at Step 5), which is why the reasoning breaks at Step 3, before the false line appears.

Part B

It is not a perfect square, so use the AC method: $ac = 225$ , and the split pair with sum $-34$ is $-25$ and $-9$ .

$25x^2 - 25x - 9x + 9 = 25x(x - 1) - 9(x - 1) = (25x - 9)(x - 1)$

Part C

Perfect-square outer terms $A^2$ and $B^2$ determine only the candidate roots $A$ and $B$ . They place no constraint on the middle term, so infinitely many trinomials share the same outer squares while having different middle terms. The pattern $(A \pm B)^2 = A^2 \pm 2AB + B^2$ holds only when the middle term is exactly $2AB$ , so that equality is the decisive third check.

$25x^2 - 34x + 9 \text{ has } 2AB = 30x \neq 34x, \text{ so it is not a perfect square.}$

In one line

Step 3 is the first unjustified step: it called the trinomial a perfect square from the outer squares alone, skipping the check that the middle term equals $2(5x)(3) = 30x$ , when it is $34x$ . The correct factorization is $(25x - 9)(x - 1)$ , and the decisive test for any perfect square is that the middle term equal $2AB$ .

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

Names Step 3 as the first unjustified step, not Step 5 where the false equation surfaces. . Worth 2 points.

States exactly the skipped check: the middle term must equal $2(5x)(3) = 30x$ , but it is $34x$ . . Worth 1 point. needs an explanation, not just an answer

Part B 3 points

Recognizes it is not a perfect square and applies a valid factoring method. . Worth 1 point.

Produces the correct factorization, verifiable by expansion. . Worth 2 points.

Part C 3 points

Explains that the outer squares fix only the candidate roots and leave the middle term free. . Worth 2 points. needs an explanation, not just an answer

States the decisive third check: the middle term must equal $2AB$ . . 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. Comparing two squared binomials 8 points

Part A 3

Squares the whole first term $4x$ to $16x^2$ and uses the doubled middle term $2ab$ , rather than squaring only the variable or halving the middle. 2
Computes all three terms correctly, keeping the last term positive. 1

Part B 2

Uses the squared-difference pattern on the whole terms $4x$ and $5$ . 1
Computes all three terms correctly, keeping the last term positive and only the middle term negative. 1

Part C 3

Names the first term and the last term as the two shared by both expansions. 1
Explains the sign change by tracing it to the sign of the two cross-products, not merely stating that the signs differ. 2

2. One root, then two 10 points

Part A 3

Multiplies numerator and denominator by $\sqrt{5}$ , the same root in the denominator. 1
Uses $\sqrt{5}\cdot\sqrt{5} = 5$ and reduces $\tfrac{10}{5}$ so no common factor is left. 2

Part B 4

Multiplies numerator and denominator by the conjugate $3 - \sqrt{5}$ , not by a copy of the denominator. 2
Evaluates the denominator as $9 - 5 = 4$ and reduces the fraction completely. 2

Part C 3

Explains that the conjugate's opposite-signed cross-products cancel to leave a rational $a^2 - b$ , arguing the mechanism rather than asserting it. 2
Contrasts this with squaring a copy, whose equal-signed cross-products add and keep the root. 1

3. Split it, solve it, then swap the split 10 points

Part A 4

Finds a pair with product $ac = 24$ and sum $14$ , satisfying both conditions at once. 2
Splits the middle term with that pair, groups, and factors each pair to the complete factorization. 2

Part B 3

Sets each factor from part A equal to zero separately. 1
Reports both solutions, including the fractional one, as the complete solution set. 2

Part C 3

Actually redoes the split and grouping in the reversed order and reaches the identical factorization, rather than only asserting it. 2
Explains that the reason is addition not depending on order, so the four-term line is unchanged in value. 1

4. Which ones factor, and which only look like they should 9 points

Part A 3

Names each form correctly (difference of squares, difference of cubes) rather than only factoring. 1
Factors both correctly, with opposite signs for the squares and the SOAP signs and single $ab$ for the cubes. 2

Part B 3

Argues from what $(x + a)(x - a) = x^2 - a^2$ can produce that no real factorization gives $x^2 + 9$ , rather than asserting it. 2
States the verdict with its domain named: does not factor over the real numbers. 1

Part C 3

Sets the correct test: an integer pair with product $4$ and sum $2$ at once. 1
Checks the integer pairs, rules out negatives, and concludes it does not factor over the integers. 2

5. Four pieces of one square 9 points

Part A 2

Sets the area up as $(x + 6)^2$ , the square of the side length, before expanding. 1
Expands to a trinomial with the correct first, middle, and last terms. 1

Part B 4

Computes the correct area for all four pieces and adds them. 2
Matches the sum to part A term by term, noting the middle term comes from the two strips together. 2

Part C 3

Identifies the missing area as one strip's worth. 1
Explains in general, not just for this plot, why the middle term needs two equal regions rather than one. 2

6. Clearing two roots, and the reason it works 10 points

Part A 3

Multiplies numerator and denominator by the conjugate $\sqrt{7} + \sqrt{3}$ . 2
Evaluates the denominator as $7 - 3 = 4$ and cancels to leave no fraction. 1

Part B 4

Multiplies numerator and denominator by the conjugate $\sqrt{5} - \sqrt{2}$ . 2
Evaluates the denominator as $5 - 2 = 3$ and reduces both numerator terms by it. 2

Part C 3

Argues the mechanism: the cross-terms cancel and each squared root becomes rational, giving $a - b$ , rather than asserting it. 2
Names this as the difference-of-squares identity $A^2 - B^2$ applied to the two roots. 1

7. Taking a cubic apart 11 points

Part A 4

Pulls out the common factor $2x$ before applying the AC method, rather than trying to factor the trinomial with the common factor still in it. 2
Applies the AC method correctly to the remaining trinomial and keeps the common factor in the final answer. 2

Part B 3

Applies the zero-product property to every factor, including the leading $2x$ , not only the two binomials. 1
Reports all three solutions, including the fractional ones. 2

Part C 4

States three distinct solutions, tied to the three factors that contain $x$ . 2
Explains from the zero-product property why $2x$ contributes a solution but the constant $2$ does not. 2

8. Cubing a sum is not adding two cubes 8 points

Part A 3

Chooses a specific numerical pair rather than arguing in general terms. 1
Evaluates both sides correctly on that pair and reaches two different values. 2

Part B 2

Identifies the cube roots $x$ and $5$ and keeps the binomial's plus sign. 1
Forms the trinomial with a single $ab$ middle term of the opposite sign and the correct last term. 1

Part C 3

Expands $(a + b)^3$ correctly into its four terms. 2
Names the two extra terms $3a^2 b$ and $3ab^2$ and ties them to the numerical gap from part A. 1

9. Correct is not the same as complete 11 points

Part A 4

Recognizes the proposed product is correct yet incomplete because $x^2 - 4$ factors further. 2
Finishes the factorization and correctly stops at the sum of squares, naming that it does not factor over the real numbers. 2

Part B 4

Pulls out the common factor $3x$ before checking for a pattern. 2
Factors the remaining difference of squares and keeps the common factor in the answer. 2

Part C 3

States the two-step habit in the correct order: common factor first, then re-check every resulting factor. 1
Explains that expanding proves a product correct but not complete, using $x^2 - 4$ still factoring as the evidence. 2

10. The step where the reasoning first breaks 9 points

Part A 3

Names Step 3 as the first unjustified step, not Step 5 where the false equation surfaces. 2
States exactly the skipped check: the middle term must equal $2(5x)(3) = 30x$ , but it is $34x$ . 1

Part B 3

Recognizes it is not a perfect square and applies a valid factoring method. 1
Produces the correct factorization, verifiable by expansion. 2

Part C 3

Explains that the outer squares fix only the candidate roots and leave the middle term free. 2
States the decisive third check: the middle term must equal $2AB$ . 1

Multiple choice

Free response