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Multiplying Polynomials: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The hidden factor

Find the single term $T(x)$ such that $(-2x^3)T(x)=14x^8$ for every real $x$ .
1. Hint 1
  The coefficient and power must each produce the corresponding part of the given product.
2. Hint 2
  The coefficients multiply to $14$ , and the exponents add to $8$ .
Answer

$T(x)=-7x^5$ .

Full solution

The coefficient must satisfy $-2c=14$ , giving $c=-7$ .
The exponent must satisfy
$3+n=8$
so $n=5$ .
Thus
$T(x)=-7x^5.$

Check by multiplying:
$(-2x^3)(-7x^5)=14x^8$
which is the given product.

Answer

$T(x)=-7x^5$ .

Key idea

A missing term in a product can be recovered by reversing coefficient multiplication and exponent addition.
Problem 1: Mark complete
Problem 2 One term across three

Expand $-5x^3(x^2+4x-7)$ and write the result in standard form.
1. Hint 1
  A single term reaches every term inside the parentheses, not only the first.
2. Hint 2
  For each product, multiply the coefficients and add the exponents, and keep the minus sign that sits on $-7$ .
Answer

$-5x^5-20x^4+35x^3$ .

Full solution

Distribute $-5x^3$ across all three terms, keeping each sign with its term.
The three products are
$\begin{aligned}(-5x^3)(x^2)&=-5x^5\\(-5x^3)(4x)&=-20x^4\\(-5x^3)(-7)&=35x^3\end{aligned}$

A negative times a negative gives the positive term $35x^3$ .
No two products are like terms, so standard form is
$-5x^5-20x^4+35x^3$

At $x=1$ the trinomial equals $-2$ , so the product is $10$ , which $-5-20+35$ also gives.

Answer

$-5x^5-20x^4+35x^3$ .

Key idea

Distributing one term gives one product for every term inside, each carrying its own sign.
Problem 2: Mark complete
Problem 3 The incomplete record

A record of partial products for $(x-2)(x^2+3x-4)$ lists $x^3$ , $3x^2$ , $-4x$ , $-2x^2$ , and $8$ . Exactly one partial product is missing. Find it.
1. Hint 1
  Each term in one factor must be paired with every term in the other.
2. Hint 2
  The products involving $x$ are all present; check the three involving $-2$ .
Answer

$-6x$ .

Full solution

The term $x$ contributes $x^3$ , $3x^2$ , and $-4x$ , all listed.
The term $-2$ contributes $-2x^2$ , $(-2)(3x)$ , and $8$ .

The missing contribution is
$(-2)(3x)=-6x$
Including it gives all six pairings, with no omitted term.

Answer

$-6x$ .

Key idea

A complete product record includes every pairing even when some contributions later combine.
Problem 3: Mark complete
Problem 4 The display border

A rectangular display has outer dimensions $x+4$ cm by $3x+2$ cm, where $x>0$ . Its uncovered inner rectangle has dimensions $x$ cm by $3x$ cm. Find the covered area as a polynomial in standard form, in square cm.
1. Hint 1
  The covered portion is the outer area with the inner area removed.
2. Hint 2
  Form all four products for the outer rectangle before subtracting the inner area.
Answer

$14x+8$ square cm.

Full solution

The outer product has partial products $3x^2$ , $2x$ , $12x$ , and $8$ .
Combining them gives
$A_{\text{outer}}=3x^2+14x+8$
in square cm.

The inner area is $3x^2$ square cm.
Subtracting gives
$A_{\text{covered}}=14x+8$
in square cm.

For example, at $x=1$ the areas are $25$ and $3$ square cm, so the covered area is $22$ square cm, agreeing with $14(1)+8$ .

Answer

$14x+8$ square cm.

Key idea

Expanding both dimensions accounts for every area contribution before an inner region is removed.
Problem 4: Mark complete
Problem 5 The two-stage processor

A processor first replaces a real input $x$ with $3x-1$ . It then squares that result and adds twice the same intermediate result. Write the final output in standard form and state its degree.
1. Hint 1
  The second stage acts on the entire output of the first stage.
2. Hint 2
  Write $(3x-1)^2+2(3x-1)$ and keep both cross products in the square.
Answer

$9x^2-1$ ; degree $2$ .

Full solution

The square expands as
$\begin{aligned}(3x-1)^2&=9x^2-6x+1\\2(3x-1)&=6x-2\end{aligned}$

Adding cancels the linear terms, leaving
$9x^2-1$
Its degree is $2$ .
At $x=1$ , the intermediate result is $2$ and the final output is $4+4=8$ , matching the polynomial.

Answer

$9x^2-1$ ; degree $2$ .

Key idea

A repeated intermediate expression must be substituted as a whole before its products are expanded.
Problem 5: Mark complete
Problem 6 The workshop's leftover pieces

A workshop makes $x+5$ batches with $x^2+x+4$ pieces in each batch, where $x$ is a positive whole number. It sets aside $2x+3$ pieces. Find the number left as a polynomial in standard form.
1. Hint 1
  Count all pieces from the batches before removing the reserved pieces.
2. Hint 2
  The batch count has two terms and the batch size has three, so include six partial products.
Answer

$x^3+6x^2+7x+17$ pieces.

Full solution

Multiplying the $x$ term through the batch size gives
$x(x^2+x+4)=x^3+x^2+4x$

Multiplying the $5$ term supplies $5x^2+5x+20$ .
Combining all six contributions gives
$N=x^3+6x^2+9x+20$

Remove $2x+3$ to obtain
$N_{\text{left}}=x^3+6x^2+7x+17$
pieces.
At $x=1$ , six batches of six pieces leave $36-5=31$ pieces, which the polynomial also gives.

Answer

$x^3+6x^2+7x+17$ pieces.

Key idea

A binomial times a trinomial needs six contributions before further additions or subtractions are made.
Problem 6: Mark complete
Problem 7 The coefficient setting

Choose the real constant $k$ so that the coefficient of $x$ in $(x-3)(x^2+kx+2)$ is $-7$ . Give $k$ and the full product in standard form.
1. Hint 1
  Identify the partial products that reach degree $1$ .
2. Hint 2
  They are $x(2)$ and $(-3)(kx)$ ; the remaining products have other degrees.
Answer

$k=3$ ; product $x^3-7x-6$ .

Full solution

The coefficient of $x$ is $2-3k$ , so the requirement reads
$2-3k=-7$
giving $k=3$ .

The product is then built from
$\begin{aligned}x(x^2+3x+2)&=x^3+3x^2+2x\\-3(x^2+3x+2)&=-3x^2-9x-6\end{aligned}$

Combining gives
$x^3-7x-6$
At $x=0$ , both the product and its expansion equal $-6$ , checking the constant term.

Answer

$k=3$ ; product $x^3-7x-6$ .

Key idea

To control one coefficient of a product, collect every partial product with that power.
Problem 7: Mark complete
Problem 8 Mara's comparison

Mara says a binomial times a trinomial must have at least three nonzero terms after like terms are combined. Decide whether she is right. Justify your decision either by explaining why every such product has at least three nonzero terms, or by giving a binomial and a trinomial, each with all its written coefficients nonzero, whose product has fewer.
1. Hint 1
  The number of partial products need not equal the number of surviving terms.
2. Hint 2
  Consider whether several pairs of partial products can cancel.
Answer

No; for example, $(x-1)(x^2+x+1)=x^3-1$ .

Full solution

Choose $(x-1)(x^2+x+1)$ .
Distributing gives the contributions $x^3$ , $x^2$ , $x$ , $-x^2$ , $-x$ , and $-1$ .

The two pairs at powers $2$ and $1$ cancel, so
$(x-1)(x^2+x+1)=x^3-1$

Both factors have all their written coefficients nonzero, yet only two terms remain, so the claim is false.

Answer

No; for example, $(x-1)(x^2+x+1)=x^3-1$ .

Key idea

Partial products can cancel, so a factor term count does not determine the final number of terms.
Problem 8: Mark complete
Problem 9 The shared factor

Let $P(x)=(x^2+3)(x-2)$ and $Q(x)=(x^2+3)(-x-2)$ . Determine whether $P(x)+Q(x)$ has lower degree than $P$ and $Q$ themselves. Explain, and give the degree of $P(x)+Q(x)$ .
1. Hint 1
  A sum of two polynomials can lose its leading term, while a product of two nonzero polynomials cannot.
2. Hint 2
  Both products carry the factor $x^2+3$ , so add $x-2$ and $-x-2$ first.
Answer

It does; $P(x)+Q(x)=-4x^2-12$ , degree $2$ .

Full solution

Each of $P$ and $Q$ multiplies a degree- $2$ factor by a degree- $1$ factor, so each has degree $2+1=3$ .

The shared factor $x^2+3$ can be taken back out of the sum, and the other two factors add to
$(x-2)+(-x-2)=-4$

So the sum is $(x^2+3)(-4)$ , which equals
$-4x^2-12$

Its degree is $2$ , below the degree $3$ of each product, so $P(x)+Q(x)$ does have lower degree than $P$ and $Q$ .
The cubic terms $x^3$ and $-x^3$ cancel, which a sum can do and a product cannot.
At $x=1$ , $P(1)=-4$ and $Q(1)=-12$ , and their sum $-16$ agrees with $-4-12$ .

Answer

It does; $P(x)+Q(x)=-4x^2-12$ , degree $2$ .

Key idea

A sum of two polynomials can fall below the degree of each one, while a product of nonzero polynomials never can.
Problem 9: Mark complete
Problem 10 The two checks

The products $x(x+1)$ and $(2x)x$ agree at $x=0$ and at $x=1$ . Jo says they must therefore agree for every real input. Expand both products and decide whether Jo is right. If the two expansions differ, give an input at which the outputs differ.
1. Hint 1
  Agreement at selected inputs does not establish agreement of every polynomial coefficient.
2. Hint 2
  Expand the products, then try an input different from the two given ones.
Answer

No; the forms are $x^2+x$ and $2x^2$ . At $x=2$ they give $6$ and $8$ .

Full solution

Distribution gives
$\begin{aligned}x(x+1)&=x^2+x\\(2x)x&=2x^2\end{aligned}$

At $x=2$ , the first output is $4+2=6$ and the second is $2(4)=8$ .
They differ, despite agreeing at the two specified inputs, so Jo is wrong.

Answer

No; the forms are $x^2+x$ and $2x^2$ . At $x=2$ they give $6$ and $8$ .

Key idea

Selected numerical checks can expose an incorrect polynomial identity but do not by themselves prove it.
Problem 10: Mark complete

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