Algebra 3 Quiz: Polynomial Division

12 questions · exam conditions

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Polynomial DivisionQuestion 1 of 12

What is the result when $3x^3 - 2x^2 + x - 7$ is divided by $x - 2$ ?

3x^2+4x+9

, remainder

-11

3x^2+4x+9

, remainder

11

3x^2-8x+17

, remainder

-41

3x^2-8x+17

, remainder

41

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Algebra 3 Quiz

Algebra 3 Quiz: Polynomial Division

Practice Polynomial Division in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Polynomial Division, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the result when $3x^3 - 2x^2 + x - 7$ is divided by $x - 2$ ?

$3x^2+4x+9$ , remainder $-11$
$3x^2+4x+9$ , remainder $11$ (correct answer)
$3x^2-8x+17$ , remainder $-41$
$3x^2-8x+17$ , remainder $41$

Explanation: Whenever you see a polynomial divided by a binomial like

x - 2

, you can use synthetic division—it's faster and less error-prone than long division for this format. Write the coefficients of the dividend

3,-2,1,-7

, and bring down the leading 3 to start. Multiply by 2 to get 6, add to -2 to get 4. Multiply 4 by 2 to get 8, add to 1 to get 9. Multiply 9 by 2 to get 18, add to

-7

to get 11. The quotient coefficientes are

3,4,9

, giving

3x^2 + 4x + \ ومن then 9

, and the final 11 is the remainder. This matches the correct choice: **

3x^2+4x+9

,remainder 11**. The choice with remainder $-11$ comes from a sign slip—forgetting that when you multiply and add, the subtrahend's sign flips correctly; you always add the product in synthetic division, so remainders rarely stay negative from careless arithmetic. The choices **

3x^2-8x+17

** with remainders $\pm 41$ share common errors. The

-8x

coefficient suggests misunderstanding the multiplication step: perhaps adding

6

-2

incorrectly as

-8

(treating subtraction as us subtracting the product

6

instead of adding it). The

+ 17

coefficient follows from repeating that mistake, and the remainders

\pm 41

result from carrying that error through and then mishandling the final sign. Your study takeaway: in synthetic division, keep a tidy row of coefficients, always add column-wise (never subtract), and write the final number as "remainder

r

" not as part of the quotient. One quick sign check at the end—plug

x=2

into the original polynomial—should give

3(8)-2(4)+2-7 = 11

, confirming the remainder. Use that check to catch arithmetic slips before you choose.

Question 2

A polynomial $P(x)$ has remainder $5$ when divided by $x-2$ and remainder $-3$ when divided by $x+1$ . What is the remainder when $P(x)$ is divided by $(x-2)(x+1)$ ?

$\frac{8}{3}x+\frac{1}{3}$
$\frac{8}{3}x-\frac{1}{3}$ (correct answer)
$2x+1$
$-2x+9$

Explanation: When you see polynomial division tied to known values, think Remainder Theorem:

P(c)

is the remainder when

P(x)

is divided by

x-c

. Since

(x-2)(x+1)

is degree 2, the remainder here must be linear:

R(x)=ax+b

. From the given information,

P(2)=5

and

P(-1)=-3

. The remainder must match those outputs, so:

2a+b=5,\qquad -a+b=-3.

Solving gives

a=\frac{8}{3}

and

b=-\frac{1}{3}

, so the remainder is

\frac{8}{3}x-\frac{1}{3}

. The traps show common mistakes.

\frac{8}{3}x+\frac{1}{3}

has the correct slope but the wrong intercept: at

x=-1

, it gives

-\frac{7}{3}

, not

-3

2x+1

only fits

R(2)=5

; at

x=-1

, it gives

-1

, so it misses the second condition.

-2x+9

also only fits

R(2)=5

, and it gives

11

x=-1

, far from

-3

. Strategy: when a remainder question gives values at specific

x

's, write the unknown remainder as a polynomial one degree less than the divisor, then plug those

x

-values into the remainder itself. That turns polynomial division into a simple system of equations.

Question 3

If $P(x)$ is divided by $x-2$ , the quotient is $Q(x)$ and the remainder is $5$ . What is the remainder when $xP(x)$ is divided by $x-2$ ?

$5$
$2$
$10$ (correct answer)
$0$

Explanation: When you see a divisor like

x-2

, the Remainder Theorem should come to mind immediately: the remainder when a polynomial is divided by

x-a

is just the polynomial's value at

x=a

. Here, since

P(x)

divided by

x-2

leaves remainder

5

, that tells you directly that

P(2)=5

. Now you are asked about

xP(x)

divided by

x-2

. The same theorem applies: evaluate

xP(x)

x=2

. That gives

2 \cdot P(2) = 2 \cdot 5 = 10

. So the remainder is

10

. Why the others are traps: choosing

5

means you forgot the extra factor of

x

in front of

P(x)

— the remainder changes because you're evaluating

xP(x)

, not

P(x)

. Choosing

2

confuses the remainder with the value of

x

itself at

x=2

, but the remainder is the whole expression's value, not just the divisor's root. Choosing

0

might come from thinking multiplying by

x

somehow makes the polynomial divisible by

x-2

, but since

xP(2)=10

, not

0

, there is definitely a nonzero remainder. Study tip: whenever you multiply a polynomial by a simple factor, use the Remainder Theorem directly — substitute the root into the entire new expression. This avoids redoing polynomial long division entirely.

Question 4

If $P(x)$ has remainder $3$ when divided by $x-2$ , what is the remainder when $P(x)+x^2$ is divided by $x-2$ ?

$7$ (correct answer)
$3$
$4$
$1$

Explanation: Whenever you see a question about remainders when dividing by a linear factor like

x-2

, think of the Remainder Theorem: the remainder is the value of the polynomial at

x=2

. So

P(2)=3

. Now for the new polynomial

P(x)+x^2

, the remainder when divided by

x-2

is found by plugging in

2

again:

P(2)+2^2 = 3+4=7.

So the remainder is

7

. Why are the others traps?

$3$ is the remainder of $P(x)$ alone. It forgets that the problem asks about $P(x)+x^2$ , so the extra $x^2$ term must also be evaluated and added.
$4$ is just the value of $x^2$ at $x=2$ . It ignores $P(2)=3$ entirely.
$1$ might come from subtracting $4-3$ , but the polynomial is a sum, not a difference, so the values must be added.

A good study takeaway: whenever you divide by

x-c

, the remainder is simply the polynomial's value at

c

. If the polynomial has several terms added together, just evaluate all of them at

c

and add the results. This small idea appears often on algebra exams, so make it automatic.

Question 5

What is the quotient when $4x^3+2x^2-3x+1$ is divided by $2x-1$ ?

$2x^2+2x-\frac{1}{2}$ , remainder $-\frac{1}{2}$
$4x^2+4x-1$ , remainder $\frac{1}{2}$
$2x^2+2x-\frac{1}{2}$ , remainder $\frac{1}{2}$ (correct answer)
$2x^2+2x+\frac{1}{2}$ , remainder $\frac{1}{2}$

Explanation: Whenever you see polynomial division, the goal is to find a quotient and remainder that satisfy: divisor × quotient + remainder = dividend. You can use long division, but keep careful track of signs and place values. Here, divide

4x^3+2x^2-3x+1

2x-1

. Start with the leading terms:

4x^3 \div 2x = 2x^2

. Multiply

2x^2

2x-1

to get

4x^3-2x^2

, subtract to get

4x^2-3x+1

. Next,

4x^2 \div 2x = 2x

, multiply to get

4x^2-2x

, subtract to get

-x+1

. Finally,

-x \div 2x = -\frac{1}{2}

, multiply to get

-x+\frac{1}{2}

, subtract to get remainder

\frac{1}{2}

. So the quotient is

2x^2+2x-\frac{1}{2}

with remainder

\frac{1}{2}

. The choice that says quotient

2x^2+2x-\frac{1}{2}

but remainder

-\frac{1}{2}

flips the sign of the remainder — a classic subtraction error. The choice with quotient

4x^2+4x-1

ignores the factor of 2 in the divisor; it divides by

x-1

instead of

2x-1

. The choice with quotient

2x^2+2x+\frac{1}{2}

puts a plus on the constant term, which would come from adding instead of subtracting during the last step. Study tip: Always verify your answer by plugging back:

(2x-1)(2x^2+2x-\frac{1}{2})+\frac{1}{2}

should give the original polynomial. Watch sign errors in the final subtraction — they're the most common trap on this exam.

Question 6

A student wants to use synthetic division to divide $3x^4-x^2+7$ by $x-2$ . Which setup is correct?

divisor $-2$ ; coefficients $3,0,-1,0,7$
divisor $2$ ; coefficients $3,-1,7$
divisor $2$ ; coefficients $3,0,-1,7$
divisor $2$ ; coefficients $3,0,-1,0,7$ (correct answer)

Explanation: Whenever you set up synthetic division, focus on two things: the root of the divisor and the complete list of coefficients. For division by

x-2

, solving

x-2=0

gives

x=2

, so the divisor number must be

2

, not

-2

. Also, the dividend

3x^4-x^2+7

has missing terms: there is no

x^3

term and no

x

term. You must write a zero placeholder for each missing power, giving coefficients

3,0,-1,0,7

. Therefore, the correct setup is divisor

2

with coefficients

3,0,-1,0,7

. The setup with divisor

-2

uses the correct coefficients but the wrong sign:

-2

would correspond to dividing by

x+2

, not

x-2

. The setup with coefficients

3,-1,7

skips all zero placeholders, which collapses the degree and loses the

x^3

and

x

terms. The setup with coefficients

3,0,-1,7

correctly includes the zero for

x^3

, but it omits the zero for the

x

term, so it represents a degree-3 polynomial instead of degree-4. On this exam, synthetic division questions often test whether you remember both the sign change and the zero placeholders. Always write the dividend in descending order, fill in zeros for missing powers, and use the root of the divisor. That habit eliminates nearly every setup error.

Question 7

What is the remainder when $x^4-3x^3+x-2$ is divided by $x^2+1$ ?

$4x+1$
$-4x-1$
$-4x+1$
$4x-1$ (correct answer)

Explanation: When you see a polynomial divided by a quadratic like

x^2+1

, the remainder must have degree less than 2, so it will be of the form

ax+b

. A fast way to find it is to use the divisor as a simplification rule: since

x^2+1=0

, we have

x^2=-1

. Then

x^4=(x^2)^2=(-1)^2=1

, and

x^3=x\cdot x^2=-x

. Substituting into the dividend gives

1-3(-x)+x-2=1+3x+x-2=4x-1.

So the remainder is

4x-1

. Each wrong choice comes from a small substitution or sign error.

4x+1

uses the correct

1+3x+x

but forgets to include the constant

-2

-4x-1

makes two sign errors: treating

x^3

x

instead of

-x

, and treating the original

+x

term as

-x

, while keeping the

-2

1-3x-x-2=-4x-1.

-4x+1

uses those same two sign errors but drops the

-2

, giving

1-3x-x=-4x+1.

So the trap here is not the polynomial division itself, but careless sign and constant tracking during substitution. Study tip: whenever the divisor is

x^2+1

, replace

x^2

with

-1

, reduce powers, and simplify carefully. You can always check your remainder by substituting

x=i

and verifying it matches

ai+b

Question 8

What is the result when $2x^4 - 5x^3 + 3x^2 - 7x + 10$ is divided by $x+1$ ?

$2x^3-7x^2+10x-17$ , remainder $27$ (correct answer)
$2x^3-7x^2+10x-17$ , remainder $-27$
$2x^3-3x^2+0x-7$ , remainder $3$
$2x^3-3x^2+0x-7$ , remainder $-3$

Explanation: Whenever you divide a polynomial by a linear divisor like

x+1

, the key is to set the divisor equal to zero:

x+1=0

, so

x=-1

. Use synthetic division with

-1

. Write the coefficients of the dividend:

2, -5, 3, -7, 10

. Bring down the

2

; multiply by

-1

to get

-2

,and add to

-5

,giving

-7

. Then multiply

-7

-1

to get

7

,add to

3

,giving

10

. Multiply

10

-1

to get

-10

,add to

-7

,giving

-17

. Finally, multiply

-17

-1

to get

17

,and add to

10

,giving remainder

27

. So the quotient is

2x^3-7x^2+10x-17

and the remainder is

27

. Thistbe also matches the Remainder Theorem: substituting

x=-1

into the original polynomial gives

27

. The choice with quotient

2x^3-7x^2+10x-17

but remainder

-27

flips the sign of the remainder; using the theorem hastily with

-1

might lead to sign errors, but the true value is positive

27

. The choice with quotient

2x^3-3x^2+0x-7

and remainder

3

is what you would get if you divided by

x-1

instead of

x+1

– its root is

+1

, not

-1

. The choice withthe same quotient but remainder

-3

repeats that divisor-sign confusion and then also miscomputes the remainder. Watch the sign in the divisor:

x+1

means use

-1

, not

+1

. A good habit is to check your answer visually by rewriting:

(x+1)(2x^3-7x^2+10x-17)+27

expands exactly to the original polynomial. Synthetic division with the correct root is faster and safer than long division – just be meticulous with signs at every step.

Question 9

A student correctly performs synthetic division of a cubic polynomial $P(x)$ by $x+3$ and obtains the bottom row $2,-1,4,-5$ . Which statement is true?

$P(x)=(x+3)(2x^2-x+4)+5$
$P(x)=(x+3)(2x^2-x+4)-5$ (correct answer)
$P(x)=(x-3)(2x^2-x+4)-5$
$P(x)=(x+3)(2x^3-x^2+4x-5)$

Explanation: When you see synthetic division, recall that dividing by

x+3

means the divisor is

x - (-3)

, so the number in the box is

-3

. The bottom row from synthetic division always gives the quotient coefficients followed by the remainder. Here, the last number

-5

is the remainder, and the first three numbers

2, -1, 4

are the coefficients of a quadratic quotient, so the quotient is

2x^2 - x + 4

. Thus

P(x) = (x+3)(2x^2 - x + 4) + (-5)

, which simplifies to

(x+3)(2x^2-x+4) - 5

. That matches the correct choice. Now check each distractor. The option with

+5

instead of

-5

misreads the remainder sign—the remainder is already

-5

, so you add

-5

, not

+5

. The option using

(x-3)

incorrectly flips the sign of the divisor; since the divisor is

x+3

, the factor must be

x+3

. Finally, the option with

(2x^3 - x^2 + 4x - 5)

treats the bottom row as a cubic quotient, but dividing a cubic by a linear factor yields a quadratic quotient—the bottom row's first three entries give a degree-2 quotient, not degree-3. For the exam, always identify the remainder as the last number in the bottom row, then write

P(x) = (\text{divisor})(\text{quotient}) + \text{remainder}

. Double-check the sign of the constant in the divisor—

x+3

becomes

(x - (-3))

. And verify the quotient degree is one less than the dividend's degree.

Question 10

Given that $x=3$ is a root of $P(x)=2x^3-3x^2-11x+6$ , which of the following is the complete factorization of $P(x)$ ?

$(x-3)(2x+1)(x-2)$
$(x+3)(2x-1)(x+2)$
$(x-3)(2x-1)(x+2)$ (correct answer)
$(x-3)(2x^2+3x-2)$

Explanation: Since

x=3

is a root, the Factor Theorem tells you

x-3

must be one factor. Use synthetic division with

3

on the coefficients

2,-3,-11,6

: bring down

2

, multiply by

3

to get

6

, add to

3

, multiply by

3

to get

9

, add to

-2

, multiply by

3

to get

-6

, add to

0

. The quotient is

2x^2+3x-2

, so

P(x)=(x-3)(2x^2+3x-2)

. But "complete factorization" means you must keep going: factor

2x^2+3x-2

(2x-1)(x+2)

, since

(2x-1)(x+2)=2x^2+4x-x-2

. Therefore the complete factorization is

(x-3)(2x-1)(x+2)

. Why the others miss the mark? The choice

(x-3)(2x^2+3x-2)

equals

P(x)

but stopstoo early—it is a factorization, not the complete factorization. The choice

(x-3)(2x+1)(x-2)

comes from sign errors in factoring the quadratic; expandingit gives

2x^3-9x^2+7x+6

, not

P(x)

. The choice

(x+3)(2x-1)(x+2)

uses

x+3

instead of

x-3

; since

3

is a root, the factor must be

x-3

, and expanding that choice gives different signs altogether. Study tip: whenever you divide by a known root, immediately test whether the quotient factors further—especially a quadratic. If it does, "complete factorization" requires writing all linear factors. A quick multiplication check catches sign errors before you commit.

Question 11

When $x^3+2x^2-7x+k$ is divided by $x-2$ , the quotient is $x^2+4x+1$ and the remainder is zero. What is $k$ ?

$2$
$-1$
$1$
$-2$ (correct answer)

Explanation: Whenever a division problem gives you the quotient and the remainder, remember the core relationship: dividend = divisor × quotient + remainder. Here the remainder is zero, so multiply

x-2

x^2+4x+1

to reconstruct the original polynomial.

(x-2)(x^2+4x+1)=x^3+4x^2+x-2x^2-8x-2=x^3+2x^2-7x-2

Now compare that result to

x^3+2x^2-7x+k

. The constant terms must match, so

k=-2

. The wrong choices each come from a common slip. Choosing 2 means you forgot the negative sign in

-2\cdot 1

, turning the product's constant into

+2

. Choosing -1 often happens in synthetic division: the last quotient coefficient is

1

, and if you mistake that for the remainder, you solve

k+2=1

and get

-1

. Choosing 1 is just grabbing the constant term from the quotient, but

k

is the dividend's constant, which only appears after you multiply the divisor and quotient together. Your quickest check: set

x=2

in the original polynomial and require the result to be

0

. That gives

8+8-14+k=0

, so

k=-2

. On test day, whenever you see a division setup, multiply the divisor and quotient to find the dividend—then compare term by term, especially the constant.

Question 12

When $P(x)=x^3+kx^2-2x+4$ is divided by $x+1$ , the remainder is $10$ . What is $k$ ?

$5$ (correct answer)
$-5$
$7$
$13$

Explanation: Whenever you see a division-by-a-linear-factor question, the Remainder Theorem is your tool: the remainder when a polynomial

P(x)

is divided by

x-a

P(a)

. Here the divisor is

x+1

, which is

x-(-1)

, so you should evaluate

P(-1)

, not

P(1)

. Plug

x=-1

into

P(x)=x^3+kx^2-2x+4

P(-1)=(-1)^3+k(-1)^2-2(-1)+4=-1+k+2+4=k+5.

The problem says the remainder is

10

, so set

k+5=10

, giving

k=5

. Now let's see why the other choices are traps. Choosing

7

usually comes from evaluating at

x=1

instead of

x=-1

; that gives

P(1)=1+k-2+4=k+3

, and

k+3=10

leads to

k=7

. Choosing

-5

means you set

P(-1)=0

, treating the remainder as if the polynomial were divisible by

x+1

. Choosing

13

often comes from a sign error on the constant term: if you accidentally compute

P(-1)=-1+k+2-4=k-3

, then

k-3=10

gives

k=13

. Your takeaway: for a divisor like

x+1

, always use

x=-1

, and be careful with every sign when substituting. A quick synthetic division check can confirm your remainder before moving on.