Multiplying Polynomials: Free Response

1. A garden bed and its enlargement . Application, 10 points. Question 1 of 5.

A rectangular garden bed measures $(x + 8)$ feet by $(x + 3)$ feet. The gardener plans to enlarge it by extending both the length and the width by $2$ feet, keeping the shape rectangular, so the enlarged bed measures $(x + 10)$ feet by $(x + 5)$ feet.

Part A.
Write the ORIGINAL garden bed's area as a single polynomial in standard form, showing the four products that produce it.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part B.
Write the ENLARGED bed's area as a single polynomial in standard form, the same way.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part C.
It is tempting to expect the enlargement to add exactly $2 \times 2 = 4$ square feet, on the reasoning that only the corner where the two $2$ -foot extensions overlap really changes. Using your two areas from parts A and B, find how much area the enlargement actually adds, and explain, in terms of the strips and corner the extensions create, how the actual increase compares to that expectation.

Carry your own answer forward Use the two areas you found in parts A and B; the point here is explaining the size of the increase, not reproducing a particular pair of expressions.

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
Every area here is a product of two side lengths, so this whole question is the distributive property used twice: each side length is itself a sum of two parts.
Hint 2 of 4 · Part A
Before multiplying, notice the original bed's dimensions are $(x + 8)$ and $(x + 3)$ : form the four products these two binomials produce.
Hint 3 of 4 · Part B
The enlarged bed's dimensions add $2$ feet to EACH of the original two side lengths; write those two new binomials before you multiply them.
Hint 4 of 4 · Part C
Subtract the two areas you already have to get the actual increase, then split that increase into the pieces the box model predicts: two new strips and one new corner square.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

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 the four products from the two side lengths before combining anything. . Worth 2 points.

Combines the two middle products into a single linear term and reports the area with its units. . Worth 1 point.

Part B 3 points

Forms the four products from the enlarged pair of side lengths. . Worth 2 points.

Combines the two middle products into a single linear term. . Worth 1 point.

Part C 4 points

Finds the actual increase in area by comparing the two areas from parts A and B. . Worth 2 points.

Explains the increase using the strips and corner the extension creates, rather than the corner alone. . Worth 2 points. needs an explanation, not just an answer

2. Term times term, then across a whole polynomial . Foundational, 8 points. Question 2 of 5.

Every longer product in this lesson is built from two smaller skills: multiplying a single term by a single term, and distributing one term across every term of a polynomial.

Part A.
Multiply $(4x^3)(-5x^2)$ , and separately $(-2x)(-3x^4)$ . For each, show the exponent addition that produces the power of $x$ .

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part B.
Distribute $3x^2$ across $(2x^3 - 5x + 4)$ , and write the result in standard form.

Write the expression An equation or an expression is enough here. Show how you built it. 2 points
Part C.
Explain, in terms of what $x^2 \cdot x^3$ actually means as repeated multiplication, why multiplying two powers of $x$ ADDS their exponents rather than multiplying them. Then use that reasoning to evaluate $x^4 \cdot x^4$ .

Explain why it works A sentence or two. Reasons, not steps. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
This whole question rests on one small skill repeated: to multiply any two terms, multiply their coefficients and add their exponents, and nothing else changes when a whole polynomial is involved.
Hint 2 of 4 · Part A
Handle the sign of each coefficient first, then combine the two numbers, then add the powers of $x$ separately.
Hint 3 of 4 · Part B
The single factor $3x^2$ has to reach EVERY one of the three terms inside the parentheses, including the constant term at the end.
Hint 4 of 4 · Part C
Write $x^2$ and $x^3$ out as strings of factors of $x$ side by side, and count how many $x$ 's are now multiplied together in total.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Multiply the coefficients and add the exponents of the like base in each case.

$(4x^3)(-5x^2) = (4\cdot(-5))\,x^{3+2} = -20x^5.$

Both factors of the second product are negative, so their product is positive:

$(-2x)(-3x^4) = ((-2)\cdot(-3))\,x^{1+4} = 6x^5.$

Part B

The lone factor $3x^2$ reaches every term inside the parentheses.

$3x^2(2x^3 - 5x + 4) = (3x^2)(2x^3) + (3x^2)(-5x) + (3x^2)(4).$

Each small product multiplies coefficients and adds exponents:

$3x^2(2x^3 - 5x + 4) = 6x^5 - 15x^3 + 12x^2.$

Part C

Write each power out as a repeated product. $x^2$ is $x\cdot x$ and $x^3$ is $x\cdot x\cdot x$ , so their product strings all five factors of $x$ together:

$x^2\cdot x^3 = (x\cdot x)(x\cdot x\cdot x) = x^5.$

There are five $x$ 's multiplied, not six, because lining up two strings of factors end to end only concatenates them; it does not change how many factors are in each string. That is why the rule adds the exponents rather than multiplying them.

The same reasoning applies to any pair of powers of $x$ . For $x^4\cdot x^4$ , four factors meet four more factors, eight in total:

$x^4\cdot x^4 = x^{4+4} = x^8.$

In one line

$(4x^3)(-5x^2) = -20x^5$ and $(-2x)(-3x^4) = 6x^5$ ; distributing $3x^2$ across $(2x^3-5x+4)$ gives $6x^5-15x^3+12x^2$ ; and exponents add rather than multiply because $x^2\cdot x^3$ strings five factors of $x$ together, the same reasoning that gives $x^4\cdot x^4 = x^8$ .

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

Adds the exponents rather than multiplying them, in both products. . Worth 2 points.

Multiplies the coefficients with the correct sign in both products. . Worth 1 point.

Part B 2 points

Distributes the term across all three parts of the polynomial, not just the first. . Worth 1 point.

Keeps every term's own power of $x$ on its product, including the constant term's. . Worth 1 point.

Part C 3 points

Explains the add-not-multiply rule by counting how many factors of $x$ each power actually represents. . Worth 2 points. needs an explanation, not just an answer

Applies the same reasoning to evaluate the second product correctly. . Worth 1 point.

3. Expanding a binomial times a trinomial . Foundational, 9 points. Question 3 of 5.

FOIL names four products: First, Outer, Inner, Last. That is a nickname for exactly one case, a binomial times another binomial. This question asks what happens when the second factor has more than two terms.

Part A.
Expand $(x + 5)(x^2 - 3x + 2)$ , writing out every individual product before combining any of them, then combine like terms into standard form.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part B.
Pairing terms by First, Outer, Inner, Last on $(x + 5)(x^2 - 3x + 2)$ names exactly four products: $x\cdot x^2$ , $x\cdot2$ , $5\cdot x^2$ , and $5\cdot2$ . Two of the products you found in part A are missing from that list. Identify both missing products and give their sum.

Carry your own answer forward Use the products you listed in part A; the task here is spotting which two of them a four-slot FOIL pairing has no room for, not re-deriving the expansion.

Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 3 points
Part C.
Using the term counts of $(x + 5)$ (two terms) and $(x^2 - 3x + 2)$ (three terms), explain why FOIL's four products can never be the whole answer for this multiplication, and state how many products the multiplication actually needs.

Justify your claim State the claim, then give the reason it has to be true. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
A trinomial has three terms, not two, so any pairing built for exactly two terms per side is going to come up short somewhere.
Hint 2 of 4 · Part A
Distribute the binomial's two terms one at a time across all three terms of the trinomial before you combine anything.
Hint 3 of 4 · Part B
Look specifically for the trinomial's middle term among the products you listed in part A: which of your products involve it, and did the four named products include either one?
Hint 4 of 4 · Part C
Count the terms in each factor separately, then think about how many total pairings that count of terms actually forces.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Distribute $x$ across the trinomial, then $5$ across it, six products in all.

$x(x^2 - 3x + 2) = x^3 - 3x^2 + 2x, \qquad 5(x^2 - 3x + 2) = 5x^2 - 15x + 10.$

Add the two rows and combine like powers:

$x^3 - 3x^2 + 2x + 5x^2 - 15x + 10 = x^3 + 2x^2 - 13x + 10.$

Part B

The four named products use only the first and last term of the trinomial ( $x^2$ and $2$ ), pairing each with a term of the binomial. That leaves the trinomial's middle term, $-3x$ , with no slot at all.

The two products that involve $-3x$ are

$x\cdot(-3x) = -3x^2, \qquad 5\cdot(-3x) = -15x,$

and their sum is $-3x^2 - 15x$ . A pairing built for two two-term factors simply has nowhere to put a third term.

Part C

The general rule is each term of the first factor times each term of the second, so a factor with $m$ terms times a factor with $n$ terms produces $m\times n$ products, no more and no fewer. Here $m = 2$ and $n = 3$ , so

$2\times3 = 6$

products are required. FOIL supplies exactly four products because it was built for the case $m = n = 2$ ; it has no fifth or sixth slot to offer, whatever the factors actually contain. FOIL is not wrong on a binomial times a trinomial, it is simply the wrong tool: the count it hands over is fixed at four while the multiplication asks for six.

In one line

$(x + 5)(x^2 - 3x + 2) = x^3 + 2x^2 - 13x + 10$ ; the FOIL pairing misses $x\cdot(-3x)$ and $5\cdot(-3x)$ , summing to $-3x^2 - 15x$ , both involving the trinomial's middle term; and the multiplication needs $2\times3 = 6$ products, since every term of the first factor must multiply every term of the second.

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 every product between a term of the binomial and a term of the trinomial before combining anything. . Worth 2 points.

Combines the two $x^2$ terms and the two $x$ terms correctly. . Worth 1 point.

Part B 3 points

Names both missing products correctly. . Worth 2 points.

Connects the omission specifically to the trinomial's middle term. . Worth 1 point.

Part C 3 points

Derives the required product count directly from the two factors' own term counts, rather than asserting a number. . Worth 2 points. needs an explanation, not just an answer

States plainly that FOIL's fixed four products cannot match what this multiplication actually requires. . Worth 1 point.

4. Comparing the degree of a sum and the degree of a product . Reasoning, 13 points. Question 4 of 5.

Here is a claim: 'Whenever you combine two polynomials, the degree of the result follows the same pattern whether you add or multiply: add the two degrees for a product, or take the larger of the two degrees for a sum.' This question checks each half of that claim in turn.

Part A.
Let $P(x) = 4x^3 - x + 2$ and $Q(x) = -2x^2 + 5x - 1$ . Without expanding the whole product, state the degree of $P(x)\cdot Q(x)$ and identify its leading term.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part B.
Give ONE specific pair of polynomials, both of the SAME degree, whose sum has a smaller degree than either one, and show the sum to prove it.

Construct a counterexample Give one specific case, and show it breaks the claim. 5 points
Part C.
Contrast what you found in part A with what you found in part B. State, in GUARDED language, what is always true about the degree of a sum or difference of two polynomials, and explain why the analogous statement for a product needs no such guard.

Carry your own answer forward Draw on the pair you built in part B and the leading-term argument from part A; the point here is the contrast between the two rules, not reproducing either computation.

Compare the two methods Say what each one costs you, and when you would reach for it. 5 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
The two halves of the stem's claim behave very differently: one of them is provably safe, and the other can be broken by a single well-chosen pair of polynomials.
Hint 2 of 4 · Part A
You do not need every term of the full expansion, only the single highest-power term, and that comes from exactly one pairing of terms.
Hint 3 of 4 · Part B
Look for two polynomials of the SAME degree whose leading coefficients are exact opposites, so that the highest power is guaranteed to vanish when you add them.
Hint 4 of 4 · Part C
State the sum rule with its exception attached, then ask what a leading coefficient would have to do to make a product's top term vanish the same way, and whether that is ever possible.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

The leading term of a product comes from multiplying the two leading terms, since every other pairing involves at least one factor of smaller degree.

$P$ has leading term $4x^3$ and $Q$ has leading term $-2x^2$ , so

$(4x^3)(-2x^2) = -8x^{3+2} = -8x^5.$

No other product in the full expansion can reach degree $5$ , and the coefficient is not zero, so the product genuinely has degree $5$ with that leading term.

Part B

Pick two degree- $2$ polynomials whose leading terms are opposites, so they are guaranteed to cancel.

$S(x) = x^2 + 4x - 1, \qquad T(x) = -x^2 + 3x + 5.$

Both have degree $2$ . Add them:

$S(x) + T(x) = (x^2 - x^2) + (4x + 3x) + (-1 + 5) = 7x + 4.$

The $x^2$ terms cancel completely, leaving a degree- $1$ result. Since $1$ is smaller than the $2$ the unguarded claim demands, this single pair refutes it.

Part C

Part B shows the guard is necessary: two degree- $2$ polynomials with opposite leading coefficients summed to something of degree $1$ , so 'the degree of a sum is the larger of the two' fails whenever the top terms are equal and opposite. The correct statement carries that exception: the degree of a sum or difference is at most the larger of the two degrees, and equals it unless the two polynomials share that degree and their leading terms cancel.

Part A shows why a product never needs the same guard. Adding two leading coefficients can hit $0$ , but multiplying them cannot:

$ab = 0 \quad\text{only if } a = 0 \text{ or } b = 0,$

and a leading coefficient is never $0$ by definition. So the top term of a product,

$ab\,x^{m+n},$

always survives, with nothing else of that degree to cancel it against. Sums have a genuine escape hatch that products structurally cannot.

In one line

$P(x)\cdot Q(x)$ has degree $5$ and leading term $-8x^5$ ; the pair $S(x)=x^2+4x-1$ , $T(x)=-x^2+3x+5$ refutes the unguarded sum claim, since $S+T=7x+4$ has degree $1$ ; and the degree of a sum or difference is at most the larger of the two degrees, equal to it unless the two share that degree and their leading terms cancel, while a product needs no such guard because a nonzero coefficient times a nonzero coefficient is never zero.

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

Identifies the two leading terms and multiplies them instead of expanding the whole product. . Worth 2 points.

Adds the exponents correctly and finds the correct leading coefficient by multiplying the two leading coefficients together. . Worth 1 point.

Part B 5 points

Chooses two same-degree polynomials whose leading coefficients are opposites. . Worth 2 points.

Adds them correctly and shows the leading terms cancel. . Worth 2 points.

States plainly that the resulting degree contradicts the unguarded claim. . Worth 1 point.

Part C 5 points

States the sum/difference rule in its guarded form, not as an unqualified 'larger of the two'. . Worth 2 points. needs an explanation, not just an answer

Explains the structural reason a product's leading term can never cancel the same way. . Worth 2 points. needs an explanation, not just an answer

Frames the answer as an explicit contrast between the two rules rather than treating them separately. . Worth 1 point.

5. Squaring $(3x - 4)$, with no work shown . Reasoning, 10 points. Question 5 of 5.

Here is a computation, given with no supporting work:

$(3x - 4)^2 = 9x^2 + 16.$

Part A.
State exactly what assumption this computation makes about squaring a binomial, and show, by expanding $(3x - 4)^2$ as $(3x - 4)(3x - 4)$ , what it should equal instead.

Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
Part B.
Verify by substituting $x = 2$ into the given line's result, $9x^2 + 16$ , and separately into the corrected expansion from part A, then compare both to a DIRECT evaluation of $(3x - 4)^2$ at $x = 2$ .

Carry your own answer forward Use whichever corrected expansion you found in part A; the point of this part is testing both expressions against a direct evaluation, not reproducing a particular pair of coefficients.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part C.
Explain, in terms of the four products every binomial-times-binomial multiplication owes, exactly which products the given line left out, and why those particular products are the ones a shortcut like this always misses.

Justify your claim State the claim, then give the reason it has to be true. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
Squaring a binomial is multiplying it by itself, so it still owes the same four products as any other binomial-times-binomial, First, Outer, Inner, and Last.
Hint 2 of 4 · Part A
Write $(3x - 4)^2$ out as $(3x - 4)(3x - 4)$ before you touch anything, and form all four products from that.
Hint 3 of 4 · Part B
Plug $x = 2$ into the ORIGINAL squared expression directly first, before you touch either of the two expanded forms, so you have something honest to compare them against.
Hint 4 of 4 · Part C
Ask which two of the four named products, First, Outer, Inner, Last, involve one term from each side rather than a term multiplied by itself.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

The given line keeps only the square of each term, $(3x)^2 = 9x^2$ and $(-4)^2 = 16$ , as though squaring a difference squares each part separately and adds them. That is the assumption: treating $(a - b)^2$ as though it were $a^2 + b^2$ , with no cross term at all.

Expand it properly as a product of two binomials, four products in all:

$(3x - 4)(3x - 4) = (3x)(3x) + (3x)(-4) + (-4)(3x) + (-4)(-4).$

$= 9x^2 - 12x - 12x + 16 = 9x^2 - 24x + 16.$

Part B

Evaluate all three at $x = 2$ .

Directly, $3(2) - 4 = 2$ , so

$(3(2) - 4)^2 = 2^2 = 4.$

The given line's expression:

$9(2)^2 + 16 = 36 + 16 = 52.$

The corrected expansion from part A:

$9(2)^2 - 24(2) + 16 = 36 - 48 + 16 = 4.$

Only the corrected expansion matches the direct evaluation, confirming which of the two is right.

Part C

A binomial times a binomial owes four products: First, Outer, Inner, Last. The given line kept only the First and Last, $(3x)(3x) = 9x^2$ and $(-4)(-4) = 16$ , and supplied no Outer or Inner product at all.

Those two missing products are

$(3x)(-4) = -12x, \qquad (-4)(3x) = -12x,$

which combine into the missing middle term.

This is the product every such shortcut misses, because squaring a binomial IS multiplying it by itself: the same two terms play both roles, so the Outer and Inner products both exist and, since the factor is repeated, they are equal to each other. Treating a square as though each term only meets itself is what erases them.

In one line

The given line assumes squaring a binomial squares each term separately with no cross term; the correct expansion is $(3x-4)^2 = 9x^2 - 24x + 16$ , confirmed by substituting $x=2$ into all three expressions; and the missing pair is the Outer and Inner products, which coincide because the same two terms are being multiplied by themselves.

Rubric

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

Part A 4 points

Names the specific incorrect assumption the given computation makes about squaring a two-term expression. . Worth 2 points. needs an explanation, not just an answer

Expands the square into four products and combines the two middle terms into a single linear term. . Worth 2 points.

Part B 3 points

Evaluates all three expressions correctly at $x = 2$ . . Worth 2 points.

States which expression matches the direct evaluation and which does not. . Worth 1 point.

Part C 3 points

Identifies the missing products among the four named, and combines them into a single term. . Worth 2 points. needs an explanation, not just an answer

Explains why squaring in particular is prone to this omission, since the same two terms multiply each other twice. . Worth 1 point.

1. A garden bed and its enlargement . Application, 10 points. Question 1 of 5.

2. Term times term, then across a whole polynomial . Foundational, 8 points. Question 2 of 5.

3. Expanding a binomial times a trinomial . Foundational, 9 points. Question 3 of 5.

4. Comparing the degree of a sum and the degree of a product . Reasoning, 13 points. Question 4 of 5.

5. Squaring $(3x - 4)$, with no work shown . Reasoning, 10 points. Question 5 of 5.

Marking rubrics

1. A garden bed and its enlargement 10 points

2. Term times term, then across a whole polynomial 8 points

3. Expanding a binomial times a trinomial 9 points

4. Comparing the degree of a sum and the degree of a product 13 points

5. Squaring $(3x - 4)$, with no work shown 10 points