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Adding and Subtracting 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 term cards

Three cards describe the terms of a polynomial in $x$ : coefficient $-4$ with degree $2$ , coefficient $6$ with degree $0$ , and coefficient $1$ with degree $5$ . Write the polynomial in standard form.
1. Hint 1
  Each coefficient belongs with the power specified on its card.
2. Hint 2
  Write the highest power first and the constant last.
Answer

$x^5-4x^2+6$ .

Full solution

The cards give $-4x^2$ , $6x^0$ , and $x^5$ .
The degree-zero term is the constant $6$ .
Sorting by decreasing degree gives
$x^5-4x^2+6$

The result has exactly the three stated coefficients and powers.

Answer

$x^5-4x^2+6$ .

Key idea

A term record determines a polynomial by pairing each coefficient with its power before ordering the terms.
Problem 1: Mark complete
Problem 2 The sign slots

Each box holds either a plus sign or a minus sign. Fill the boxes so that $-(x^2\,\square\,4x\,\square\,7)$ equals $-x^2+4x-7$ for every real $x$ .
1. Hint 1
  The outside minus reverses the sign of every term inside.
2. Hint 2
  Work backward from the signs of the $x$ term and constant in the required result.
Answer

First box: minus. Second box: plus.

Full solution

To obtain $+4x$ after negation, the inside term must be $-4x$ .
To obtain $-7$ , the inside term must be $+7$ .
Thus
$-(x^2-4x+7)=-x^2+4x-7$

Both filled signs produce the required terms.

Answer

First box: minus. Second box: plus.

Key idea

An outside minus reverses each inside sign, so signs can be recovered by reversing that action.
Problem 2: Mark complete
Problem 3 The empty columns

Write $(5x^3+2x^2-7)-(3x^3-4x+1)$ in standard form.
1. Hint 1
  Subtracting a polynomial subtracts each of its terms, so every sign inside the second parentheses turns over.
2. Hint 2
  Two of the four powers appear in only one of the polynomials, so those terms have nothing to combine with.
Answer

$2x^3+2x^2+4x-8$ .

Full solution

Distributing the minus sign flips all three signs of the second polynomial, so its terms become $-3x^3$ , $+4x$ , and $-1$ .

Combining like terms one power at a time gives
$\begin{aligned}x^3&: 5-3=2\\x^2&: 2-0=2\\x&: 0+4=4\\x^0&: -7-1=-8\end{aligned}$

Written in descending order, the difference is
$2x^3+2x^2+4x-8$

Checking at $x=2$ , the two polynomials read $41$ and $17$ , and their difference $24$ matches
$2(8)+2(4)+8-8=24.$

Answer

$2x^3+2x^2+4x-8$ .

Key idea

A power present in only one polynomial carries down unchanged once the subtraction's signs are distributed.
Problem 3: Mark complete
Problem 4 The canceling entries

A record lists the terms $2x^4$ , $-3x$ , $7$ , $-2x^4$ , and $5x^2$ . Combine the record into standard form, then state its degree and leading coefficient.
1. Hint 1
  Terms with the same power share one coefficient in the final record.
2. Hint 2
  Check the two degree- $4$ entries before choosing the leading term.
Answer

$5x^2-3x+7$ ; degree $2$ ; leading coefficient $5$ .

Full solution

The degree- $4$ terms cancel:
$2x^4-2x^4=0$

The other powers have one entry each, so standard form is
$5x^2-3x+7$

Its surviving terms have degrees $2$ , $1$ , and $0$ .
The largest is $2$ , and its coefficient is $5$ .

Answer

$5x^2-3x+7$ ; degree $2$ ; leading coefficient $5$ .

Key idea

The degree and leading coefficient come from the terms remaining after like terms combine.
Problem 4: Mark complete
Problem 5 The sensor offset

Two sensors report $A(x)=3x^2-2x+6$ and $B(x)=x^2+4x-1$ for a real input $x$ . A display reports the first reading minus the second, then removes a fixed offset of $5$ . Write the display rule $D(x)$ in standard form and find its reading at $x=-1$ .
1. Hint 1
  Follow the order of the stated changes to the two readings.
2. Hint 2
  Subtract every term of $B(x)$ , then include the separate offset.
Answer

$D(x)=2x^2-6x+2$ ; $D(-1)=10$ .

Full solution

The rule is $D=A-B-5$ .
The coefficients combine as
$\begin{aligned}3-1&=2\\-2-4&=-6\\6-(-1)-5&=2\end{aligned}$
Hence
$D(x)=2x^2-6x+2$

At the requested input,
$D(-1)=2+6+2$
so the reading is $10$ .
Checking separately, $A(-1)=11$ and $B(-1)=-4$ , giving
$11-(-4)-5=10.$

Answer

$D(x)=2x^2-6x+2$ ; $D(-1)=10$ .

Key idea

A sequence of changes to polynomial readings becomes a sum of signed contributions at each power.
Problem 5: Mark complete
Problem 6 The reversed difference

For real $x$ , two quantities satisfy $U(x)-V(x)=2x^3-x+4$ . A record needs $V(x)-U(x)+3x-6$ . Write that expression in standard form and state its degree.
1. Hint 1
  Reversing a subtraction negates the entire difference.
2. Hint 2
  Negate the given polynomial before adding the two new terms.
Answer

$-2x^3+4x-10$ ; degree $3$ .

Full solution

The reversed difference is
$V(x)-U(x)=-2x^3+x-4$

Adding $3x-6$ combines the linear terms and constants, giving
$-2x^3+4x-10$

The leading coefficient is $-2$ , which is nonzero, so the degree is $3$ .
At $x=0$ , the original difference is $4$ and the requested quantity is $-4-6=-10$ , agreeing with the constant.

Answer

$-2x^3+4x-10$ ; degree $3$ .

Key idea

Reversing a difference changes every sign before any further terms are combined.
Problem 6: Mark complete
Problem 7 The total counter

A counter totals three contributions. Their first two together equal $4x^3+x^2-6$ , while their last two together equal $x^3-2x+5$ . The middle contribution is $2x^3-3x+1$ . Find the total of all three in standard form.
1. Hint 1
  Adding the two given pair totals counts the middle contribution twice.
2. Hint 2
  Remove one copy of the middle contribution from the sum of the pair totals.
Answer

$3x^3+x^2+x-2$ .

Full solution

Call the contributions $A$ , $B$ , and $C$ .
The total is $(A+B)+(B+C)-B$ , and each power is counted on its own row:
$\begin{aligned}x^3&: 4+1-2=3\\x^2&: 1+0-0=1\\x&: 0-2-(-3)=1\\x^0&: -6+5-1=-2\end{aligned}$

Written in descending order,
$A+B+C=3x^3+x^2+x-2$

At $x=0$ , the pair totals are $-6$ and $5$ , and the shared entry is $1$ , so the total is $-6+5-1=-2$ , matching the result.

Answer

$3x^3+x^2+x-2$ .

Key idea

When overlapping totals are added, remove the shared contribution once to count every part once.
Problem 7: Mark complete
Problem 8 The shared coefficient

Polynomials $A$ and $B$ each have degree $3$ and leading coefficient $4$ . Nia says $A-B$ must have degree $2$ . Is she right? If not, give polynomials $A$ and $B$ meeting these conditions whose nonzero difference does not have degree $2$ .
1. Hint 1
  The leading terms cancel, but the remaining powers depend on the other terms.
2. Hint 2
  Choose two cubics whose degree- $2$ coefficients also agree.
Answer

No; for example, $A=4x^3+x$ and $B=4x^3$ , giving degree $1$ .

Full solution

The shared leading coefficient guarantees that the cubic terms cancel, but it does not guarantee a quadratic term.
Choose $A=4x^3+x$ and $B=4x^3$ .
Then
$A-B=x$

Both inputs satisfy the degree and coefficient conditions, while their nonzero difference has degree $1$ , so the claim is false.

Answer

No; for example, $A=4x^3+x$ and $B=4x^3$ , giving degree $1$ .

Key idea

Canceling leading terms gives an upper bound on the remaining degree, not a guarantee of the next degree.
Problem 8: Mark complete
Problem 9 The degree report

A nonzero polynomial $P$ has degree $5$ . Another polynomial $Q$ has degree at most $3$ . Leo says the degree of $P+Q$ is the same as the degree of $P-Q$ . Is he right? Explain.
1. Hint 1
  Check which powers $Q$ can change in either operation.
2. Hint 2
  Neither adding nor subtracting $Q$ changes the degree- $5$ coefficient.
Answer

Yes; both degrees are $5$ .

Full solution

Let $a\ne0$ be the coefficient of $x^5$ in $P$ .
In the sum and difference, the coefficients at that power are
$\begin{aligned}a+0&=a\\a-0&=a\end{aligned}$

Both coefficients stay nonzero, and no higher power appears.
Thus both degrees are $5$ , so Leo is right.

Answer

Yes; both degrees are $5$ .

Key idea

A lower-degree polynomial cannot change a higher-degree leading coefficient through addition or subtraction.
Problem 9: Mark complete
Problem 10 Two canceling quartics

Create two different polynomials, each with exactly three nonzero terms and degree $4$ , whose sum is a nonzero constant. Give the polynomials and their sum, and explain your construction.
1. Hint 1
  Adding polynomials adds coefficients power by power, so the sum is a constant exactly when every coefficient above $x^0$ adds to zero.
2. Hint 2
  Use opposite coefficients on two shared positive powers and constants that do not add to zero.
Answer

One choice: $P=x^4+2x+3$ , $Q=-x^4-2x+1$ ; $P+Q=4$ .

Full solution

Choose the degree- $4$ and degree- $1$ terms in opposite pairs, but choose constants $3$ and $1$ .
Then
$\begin{aligned}x^4-x^4&=0\\2x-2x&=0\\3+1&=4\end{aligned}$

Thus $P=x^4+2x+3$ and $Q=-x^4-2x+1$ each have three nonzero terms and degree $4$ , yet their sum is the nonzero constant $4$ .
Many choices of opposite nonconstant terms work.

Answer

One choice: $P=x^4+2x+3$ , $Q=-x^4-2x+1$ ; $P+Q=4$ .

Key idea

A constant sum can result when every nonconstant coefficient cancels and the constants do not.
Problem 10: Mark complete

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