Algebra 3 Quiz: Factor And Remainder Theorems

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Factor And Remainder TheoremsQuestion 1 of 12

Let $f(x)=x^4-3x^3+ax^2-7x+6$ . If $x-2$ is a factor of $f(x)$ , what is the quotient when $f(x)$ is divided by $x-2$ ?

x^3 - x^2 + 2x - 3

x^3 - x^2 + 2x + 3

x^3 - x^2 - 2x - 3

x^3 - x^2 - 2x + ￥ 3

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Algebra 3 Quiz

Algebra 3 Quiz: Factor And Remainder Theorems

Practice Factor And Remainder Theorems 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 Factor And Remainder Theorems, 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

Let $f(x)=x^4-3x^3+ax^2-7x+6$ . If $x-2$ is a factor of $f(x)$ , what is the quotient when $f(x)$ is divided by $x-2$ ?

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

Explanation: When you see "

x-2

is a factor," use the factor theorem:

f(2)

must equal

0

. That tells you the unknown coefficient before you divide. Compute

f(2)=16-24+4a-14+6=4a-16

, so

4a-16=0

and

a=4

. Thus

f(x)=x^4-3x^3+4x^2-7x+6

. Now divide by

x-2

using synthetic division with

2

: bring down

1

;

2\cdot 1+(-3)=-1

;

2\cdot(-1)+4=2

;

2\cdot 2+(-7)=-3

;

2\cdot(-3)+6=0

. The quotient is built from coefficients

1,-1,2,-3

, giving

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

The choices ending in

+3

—

x^3-x^2+2x+3

and

x^3-x^2-2x+3

— treat the final term with the wrong sign: the remainder is

0

, so the last quotient term must be

-3

, not

+3

. The choice

x^3-x^2-2x-3

has the wrong

x

-coefficient: the step

2\cdot(-1)+4

gives

2

, not

-2

. A quick check:

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

multiplies back to

x^4-3x^3+4x^2-7x+6

. On this exam, when a factor is given, first plug the root into

f

to find any unknown coefficient, then use synthetic division. The coefficients of the quotient come directly from the bottom row.

Question 2

Which of the following is the complete factorization of $g(x)=x^4-5x^3+6x^2+4x-8$ over the integers?

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

Explanation: When you see a quartic polynomial to factor over the integers, your first move should be the Rational Root Theorem: test factors of the constant term,

-8

, so

\pm1,\pm2,\pm4,\pm8

. Here,

g(2)=0

and

g(-1)=0

, so

x-2

and

x+1

are factors. Dividing

g(x)

x-2

gives

x^3-3x^2+4

, and dividing that cubic by

x-2

again gives

x^2-x-2

. This quadratic factors as

(x-2)(x+1)

. Pulling it all together,

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

. That is the complete integer factorization. Now look at the wrong choices.

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

would require

x=1

and

x=-2

to be roots, but

g(1)=-2

and

g(-2)\neq0

, so this is just a mismatch of roots.

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

includes the correct

(x+1)

and two

(x-2)

factors, but the extra

(x+2)

is not a factor, and you would be missing one

(x-2)

; checking

g(-2)\neq0

rules it out. Finally,

(x-1)(x-2)^3

gets the repeated

(x-2)

right but replaces

(x+1)

with

(x-1)

, and

g(1)\neq0

proves

(x-1)

is not a factor. Your takeaway: for factoring polynomials over the integers, systematically test rational roots and divide until you reach a quadratic. This prevents both missing repeated factors and inventing roots that don't actually satisfy the polynomial.

Question 3

A degree-4 polynomial $f$ satisfies $f(2)=0$ and $f(-2)\neq0$ . Which statement must be true?

Both $x-2$ and $x+2$ are factors of $f(x)$ .
$x+2$ is a factor of $f(x)$ , while $x^2-4$ is not a factor.
$x-2$ is a factor of $f(x)$ , but $x+2$ is not. (correct answer)
$x^2-4$ is a factor of $f(x)$ .

Explanation: When you see a polynomial value like

f(2)=0

, your first thought should be the Factor Theorem: if

f(c)=0

, then

x-c

is a factor. Here,

c=2

, so

x-2

is definitely a factor. At the same time,

f(-2)\neq 0

tells you that

x+2

is not a factor. Therefore the statement that must be true is exactly: "

x-2

is a factor of

f(x)

, but

x+2

is not." The degree being 4 simply means there are two more factors you don't know about; it doesn't force either of the missing factors. Now look at the wrong answers. The claim that both

x-2

and

x+2

are factors cannot be true, because

f(-2)\neq 0

directly rules out

x+2

as a factor. Similarly, the statement that "

x+2

is a factor, while

x^2-4

is not" fails because the first part is false. Finally, "

x^2-4

is a factor" is impossible, since

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

; for that to be a factor, both

x-2

and

x+2

would have to be factors, and you already know

x+2

is not. Your study tip: whenever a question gives you a value like

f(a)=0

f(a)\neq 0

, immediately write down which factor is confirmed and which factor is ruled out. Then translate compound factors like

x^2-4

back into their linear pieces before deciding whether they can be factors. This avoids the common trap of assuming a degree-4 polynomial must include every nice-looking factor.

Question 4

What is the remainder when $x^{100}-2x^{49}+4$ is divided by $x+1$ ?

$5$
$3$
$7$ (correct answer)
$1$

Explanation: Whenever you see "remainder when a polynomial is divided by

x+1

," think of the Remainder Theorem: the remainder is just the polynomial evaluated at the root of the divisor. Since

x+1=0

gives

x=-1

, you need to compute

P(-1)

for

P(x)=x^{100}-2x^{49}+4

. Now evaluate carefully.

(-1)^{100}=1

, because 100 is even.

(-1)^{49}=-1

, because 49 is odd. So

P(-1)=1-2(-1)+4=1+2+4=7.

That is the remainder, so the correct answer is

7

. The other choices come from common sign or evaluation slips. The choice

3

is what you get if you mistakenly plug in

x=1

, since

1-2+4=3

; remember,

x+1

means

x=-1

, not

x=1

. The choice

5

likely comes from thinking

(-1)^{100}=-1

, which would give

-1+2+4=5

; but any even power of

-1

is positive

1

. The choice

1

may come from only handling the first term, or forgetting that

-2(-1)

adds

2

and then

+4

adds another

4

. Your study tip: for division by

x-c

, evaluate the polynomial at

c

. For

x+1

, use

c=-1

. Always check the parity of the exponents — even powers of

-1

are

1

, odd powers are

-1

. This will make remainder questions quick and reliable.

Question 5

The remainder when $f(x)=2x^3+bx^2-3x+5$ is divided by $x+1$ is equal to the remainder when $g(x)=x^2+5x+2$ is divided by $x-2$ . What is the value of $b$ ?

$10$ (correct answer)
$16$
$12$
$2$

Explanation: Whenever you see "remainder when a polynomial is divided by

x-c

," use the Remainder Theorem: the remainder is just

p(c)

. Here,

x+1

means

c=-1

, so the first remainder is

f(-1)

. Likewise, dividing by

x-2

means the second remainder is

g(2)

. Compute:

f(-1)=2(-1)^3+b(-1)^2-3(-1)+5=-2+b+3+5=b+6

and

g(2)=2^2+5(2)+2=4+10+2=16.

Since the remainders are equal,

b+6=16 \implies b=10.

Now, why are the other choices traps? 16 is exactly

g(2)

, but it is not

b

; you must remember that

f(-1)

already contributes

+6

before

b

. 12 comes from plugging

x=1

into

f

instead of

x=-1

;

x+1

has root

-1

, not

1

, so that gives

b+4=16

. 2 comes from evaluating

g(1)=8

instead of

g(2)=16

, mistaking the root of

x-2

1

rather than

2

. Your study tip: before substituting, write each divisor as

x-c

explicitly. For

x+1

, rewrite it as

x-(-1)

; for

x-2

, it is already

x-2

. This small step prevents the most common sign and root errors on remainder questions.

Question 6

The polynomial $p(x)=x^5-2x^4-3x^3+4x^2+4x$ has repeated factors. How many distinct real zeros does it have?

$5$
$3$ (correct answer)
$2$
$1$

Explanation: Whenever a polynomial has repeated factors, the key is to separate multiplicity from distinct zeros. Factor completely:

p(x)=x(x^4-2x^3-3x^2+4x+4)

Testing the quartic,

x=-1

works, so

p(x)=x(x+1)(x^3-3x^2+4)

Then

x=2

works in the cubic, giving

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

Now set each distinct factor to zero:

x=0

x=-1

, and

x=2

. The squared factors make

-1

and

2

repeated roots, but a repeated root is still just one real zero. Multiplicity tells you how many times a root appears, not how many new zeros it creates. So the correct number of distinct real zeros is

3

. The answer "

5

" is the trap of counting zeros with multiplicity:

1+2+2=5

. The answer "

2

" likely comes from counting only the repeated roots

-1

and

2

, forgetting the simple root

0

. The answer "

1

" may come from thinking that "repeated factors" means only repeated roots count, or from an incomplete factorization. All of these confuse multiplicity with distinctness. Study tip: after factoring completely, list the distinct roots first, then note each multiplicity separately. For "number of distinct real zeros," every different factor contributes exactly one root, no matter how many times it is squared or cubed.

Question 7

Which polynomial has $x+1$ as a factor but does NOT have $x-1$ as a factor?

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

Explanation: Whenever you see "has a factor" on a polynomial question, your first move should be the factor theorem: if

x-c

is a factor, then plugging

c

into the polynomial gives

0

. Here you need

P(-1)=0

for

x+1

to be a factor, and

P(1)\neq 0

for

x-1

not to be a factor. Test the polynomial

x^3+x^2+x+1

. At

x=-1

, you get

-1+1-1+1=0

, so

x+1

is a factor. At

x=1

, you get

1+1+1+1=4

, not

0

, so

x-1

is not a factor. That matches the requirement. Now look at the others.

x^3+x^2-x-1

gives

0

at both

-1

and

1

, so it has both factors. The same is true for

x^3-x^2-x+1

, which also gives

0

at both. The remaining choice,

x^3-x^2+x-1

, gives

-4

-1

and

0

1

, so it has

x-1

but not

x+1

— the opposite of what is asked. The quick takeaway: on factor questions, don't start factoring the whole polynomial. Plug the candidate roots into the polynomial. A zero means the factor exists; a nonzero value means it doesn't. Checking

P(-1)

and

P(1)

alone will separate these choices instantly.

Question 8

Suppose $x+1$ and $x-2$ are factors of $p(x)=x^3+ax^2+bx-6$ . What is the value of $p(3)$ ?

$24$ (correct answer)
$0$
$-24$
$6$

Explanation: Whenever you see linear factors of a polynomial, use the factor theorem: if

x-c

is a factor, then

p(c)=0

. So

p(-1)=0

and

p(2)=0

. Since

p(x)

is a monic cubic and already has two linear factors, the third factor must be

x+k

. Multiplying the constant terms gives

(1)(-2)(k)=-6

, so

-2k=-6

, meaning

k=3

. Thus

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

Now evaluate:

p(3)=(3+1)(3-2)(3+3)=4\cdot 1\cdot 6=24.

The value

0

would be correct only if

x-3

were a factor, but the third factor is

x+3

, not

x-3

. The value

-24

is a sign error: at

x=3

, all three factors are positive, so

p(3)

must be positive; the negative constant term does not make the value negative. The value

6

is just the constant term, or its absolute value, and it ignores the contributions of the cubic and quadratic terms when evaluating

x^3+2x^2-5x-6

. A good habit: once you know the factors, multiply them out or compare constants to find the missing factor before evaluating.

Question 9

A cubic polynomial $p$ has zeros $-2$ , $1$ , and $4$ . If $p(0)=16$ , which of the following could be $p(x)$ ?

$2(x-2)(x+1)(x-4)$
$-2(x+2)(x-1)(x-4)$
$2(x+2)(x+1)(x-4)$
$2(x+2)(x-1)(x-4)$ (correct answer)

Explanation: When you see a cubic with given zeros, remember each zero

r

corresponds to a factor

(x-r)

. So zeros

-2,1,4

mean the factored form must contain

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

. The multiplier in front is then determined by

p(0)

. The choice

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

has exactly the correct factors. Checking

x=0

2(2)(-1)(-4)=16

, which matches

p(0)=16

. So this is the correct polynomial. Now look at why the others fail. The choice

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

has zeros

2,-1,4

, not

-2,1,4

, so it's not even the right root set. The choice

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

has the correct zeros, but the leading multiplier is

-2

, so

p(0)=-16

, not

16

. The choice

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

gives zeros

-2,-1,4

, missing the zero

1

and incorrectly including

-1

. So each wrong choice either misidentifies a sign in a factor or misapplies the multiplier. A strong strategy: first write the factors straight from the zeros, then evaluate at

0

to solve for the leading coefficient. Also watch signs carefully: a zero of

-2

requires

(x+2)

, not

(x-2)

. This avoids both root errors and constant errors.

Question 10

If $2x+1$ is a factor of $p(x)=6x^3+ax^2-5x+2$ , what is the value of $a$ ?

$-1$
$15$
$-15$ (correct answer)
$1$

Explanation: When you see "factor of a polynomial," you're being tested on the factor theorem: if

2x+1

is a factor, then

2x+1=0

gives a root, so

x=-\frac12

must make the polynomial equal zero. Plug that into

p(x)

6\left(-\frac12\right)^3 + a\left(-\frac12\right)^2 -5\left(-\frac12\right)+2=0.

Simplify carefully:

6\left(-\frac18\right)+a\left(\frac14\right)+\frac52+2=0

-\frac34+\frac{a}{4}+\frac92=0.

Combine the numbers:

-\frac34+\frac92 = -\frac34+\frac{18}{4}=\frac{15}{4}

. So:

\frac{a}{4}+\frac{15}{4}=0 \Rightarrow a+15=0 \Rightarrow a=-15.

Now, why the others are traps. $15$ comes from solving

a+15=0

incorrectly — forgetting the negative sign. $-1$ might result from misreading the root as

x=-\frac13

or fumbling the fraction arithmetic. $1$ usually appears when you accidentally substitute

x=\frac12

instead of

-\frac12

, or when you lose track of signs in the coefficients. All of them would fail the factor theorem: only

a=-15

makes

p\left(-\frac12\right)=0

. Study tip: Whenever a linear factor like

cx+d

is given, immediately solve for the root

x=-\frac dc

and substitute it into the polynomial, setting it equal to zero. Then write out every fraction step — sign errors are the most common trap on this type of question. If you get a non-integer, don't panic; simplify methodically and you'll land on the right value.

Question 11

Suppose $x-3$ is a factor of $p(x)$ . Which of the following must be a factor of $p(2x+1)$ ?

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

Explanation: Whenever you're told that

x-3

is a factor of

p(x)

, the Factor Theorem is your key:

x-c

is a factor exactly when

p(c)=0

. So here you know

p(3)=0

. Now the polynomial in the question is

p(2x+1)

, so ask: what value of

x

makes the input equal to

3

? Solve

2x+1=3

, which gives

x=1

. Therefore

p(2(1)+1)=p(3)=0

, so

x-1

must be a factor of

p(2x+1)

. That is the correct choice. . The choice

x-3

is a tempting trap: you might copy the original factor unchanged. But plugging

x=3

into the composed expression gives

p(7)

, not

p(3)

, so it does not have to be a factor. The choice

x+1

would require

x=-1

, making the input

-1

, and you're not guaranteed that

p(-1)=0

. Similarly,

x-2

would require

x=2

, making the input

5

, and no information guarantees

p(5)=0

. Those could be factors only if

p

happened to have other specific roots, but they are not forced by the given fact. . Strategy: when a factor of

p(x)

is given and you're asked about

p(\text{linear expression})

, transform the known root through that expression: set the linear expression equal to the known root and solve for

x

. The root shifts, but the factor does not stay the same.

Question 12

If polynomial $P$ has remainder $-6$ when divided by $x-2$ , what is the remainder when $P(x)+3x^2$ is divided by $x-2$ ?

$-18$
$12$
$-6$
$6$ (correct answer)

Explanation: When you see "remainder when divided by

x-2

," the key is the Remainder Theorem: the remainder equals the polynomial evaluated at

x=2

. So

P(2)=-6

is given. For

P(x)+3x^2

, evaluate the whole expression at

x=2

P(2)+3(2)^2 = -6 + (3\cdot 4) = -6 +12 = 6.

Thus the remainder is

6

. Now, the traps: the choice

-6

treats the added

3x^2

as if it doesn't change the remainder; that would only be true if it were divisible by

x-2

, but at

x=2

3x^2

contributes

12

. The choice

12

correctly evaluates

3x^2

, but then forgets to add

P(2)=-6

, leaving out the original remainder. The choice

-18

might come from multiplying the original remainder by

3

, or from subtracting

12

instead of adding it; either way, it reverses the needed sign or overweights the given information. A useful habit: anytime a question asks for a remainder after division by a linear term, plug the root into the whole expression. For sums like

P(x)+Q(x)

, just evaluate

P(a)+Q(a)

. And be careful to include both contributions—do not forget the given

P(a)

value.