Algebra 3 Quiz: Zeros And Multiplicity

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Zeros And MultiplicityQuestion 1 of 12

Given that $x=2$ is a zero of $p(x)=x^3-5x^2+8x-4$ , which statement gives all zeros of $p$ with their multiplicities?

2

, multiplicity 2;

1

, multiplicity 1

2

, multiplicity 1;

1

, multiplicity 2

2

, multiplicity 2;

-1

, multiplicity 1

-2

, multiplicity 2;

1

, multiplicity 1

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Algebra 3 Quiz

Algebra 3 Quiz: Zeros And Multiplicity

Practice Zeros And Multiplicity 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 Zeros And Multiplicity, 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

Given that $x=2$ is a zero of $p(x)=x^3-5x^2+8x-4$ , which statement gives all zeros of $p$ with their multiplicities?

$2$ , multiplicity 2; $1$ , multiplicity 1 (correct answer)
$2$ , multiplicity 1; $1$ , multiplicity 2
$2$ , multiplicity 2; $-1$ , multiplicity 1
$-2$ , multiplicity 2; $1$ , multiplicity 1

Explanation: Whenever you see a zero given for a polynomial, your first move should be to factor it using that zero — this is a direct application of the Factor Theorem. Since

x=2

is a zero,

x-2

is a factor. Divide

p(x)=x^3-5x^2+8x-4

x-2

using synthetic division: you get

x^2-3x+2

. That quadratic factors further as

(x-1)(x-2)

. So the full factorization is

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

Counting linear factors tells you the zeros and their multiplicities:

x=2

appears twice, so it has multiplicity 2, and

x=1

appears once, so it has multiplicity 1. The correct pairing is therefore

2

, multiplicity 2;

1

, multiplicity 1. The statement that says

2

, multiplicity 1;

1

, multiplicity 2 reverses the two multiplicities — not supported by the factorization. The statement listing

-1

instead of

1

is a sign-error trap:

(x+1)

would correspond to zero

x=-1

, and substituting

-1

into

p(x)

gives

-18

, not

0

. Likewise, the statement listing

-2

with multiplicity 2 mistakes the factor

x-2

for

x+2

; substituting

-2

gives

-48

, not

0

. Study tip: once you have one zero, always divide and factor the quotient completely. The exponent on each linear factor is the multiplicity — that's the entire story.

Question 2

A polynomial of least possible degree with real coefficients crosses the $x$ -axis at $x=-4$ , touches but does not cross at $x=1$ , and crosses at $x=3$ . Which factored form could represent $p(x)$ ?

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

Explanation: Whenever you see a polynomial crossing vs. touching the x-axis, think multiplicity: crossing means odd multiplicity, and touching means even multiplicity. Here,

x=-4

and

x=3

are crossing points, so

(x+4)

and

(x-3)

must appear to odd powers; the least odd power is

1

. At

x=1

, the graph touches but doesn't cross, so

(x-1)

must appear to an even power; the least even power is

2

. Therefore, the least-degree pattern is

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

for some nonzero real number

a

. Among the choices, only

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

matches those exponents. The form

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

is wrong because it has exponent

1

(x-1)

, which would make the graph cross at

x=1

instead of touching, and exponent

2

(x-3)

,which would make it touch at

x=3

instead of crossing. The choice

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

reverses the required parity at both

-4

and

1

: the squared factor makes

x=-4

a touch point, and the linear factor makes

x=1

a crossing point. The form

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

has the correct multiplicity at

-4

and

1

, but

(x-3)^2

makes

x=3

a touch point instead of a crossing point; adding that extra squared factor also makes the degree larger than necessary. Study takeaway: for each x-axis intercept, decide whether it should be odd (cross) or even (touch), then match those parities with the smallest possible exponents.

Question 3

Given $p(x)=x^5+x^4-x^3+x^2+4x+2$ , what is the multiplicity of the zero $x=-1$ ?

1
2
3 (correct answer)
4

Explanation: Whenever a question asks for the multiplicity of a zero, you are really asking: how many times does the factor

(x+1)

appear in the factorization of

p(x)

? A zero of multiplicity

m

means

(x+1)^m

divides

p(x)

, but

(x+1)^{m+1}

does not. Start by testing

x=-1

p(-1)= -1+1+1+1-4+2=0,

x=-1

is a zero. Now divide

p(x)

x+1

using synthetic division to get

x^4-x^2+2x+2

. This quotient is still zero at

x=-1

, so divide again:

(x+1)^2

leaves

x^3-x^2+2

. That also vanishes at

x=-1

, and a third division by

x+1

gives

x^2-2x+2

. At

x=-1

, this equals

1+2+2=5\neq 0

, so the factor stops after three copies. Thus the multiplicity is 3. If you chose 1, you likely confirmed the zero but did not repeat the division to see how many times the factor occurs. Choosing 2 means you stopped after the second division even though

x^3-x^2+2

still had

x=-1

as a root. Choosing 4 would require

x^2-2x+2

to be divisible by

x+1

, but its remainder at

x=-1

is 5. Study tip: keep dividing by the same factor until the remainder is nonzero; the number of successful divisions is the multiplicity.

Question 4

Let $f(x)=-2(x-3)^2(x+1)^4(x^2+9)$ . Which statement about the graph of $f$ is true?

It crosses the $x$ -axis at $x=3$ and touches at $x=-1$ .
It touches the $x$ -axis at $x=3$ and at $x=-1$ ; it never crosses the $x$ -axis. (correct answer)
It touches at $x=3$ and crosses at $x=-1$ .
It crosses at $x=3$ and $x=-1$ , and it also has complex zeros $3i$ and $-3i$ .

Explanation: Whenever you see a question about where a polynomial graph crosses or touches the

x

-axis, focus on the multiplicity of each real linear factor. A root with odd multiplicity makes the graph cross the axis; a root with even multiplicity makes it touch and turn around. ? Here,

f(x)=-2(x-3)^2(x+1)^4(x^2+9)

. The factor

(x-3)^2

gives

x=3

with multiplicity

2

, an even number, so the graph touches (not crosses) at

x=3

. Similarly,

(x+1)^4

gives

x=-1

with multiplicity

4

, also even, so the graph touches at

x=-1

. Since both real zeros have even multiplicity, the graph never crosses the

x

-axis. The factor

x^2+9

has no real zeros — its roots are complex

\pm 3i

— so it contributes no $x$ -intercepts at all. Thus the correct statement is: It touches at

x=3

and at

x=-1

; it never crosses. Why the others fail? "It crosses at

x=3

and touches at

x=-1

" reverses the behavior at

3

."It touches at

x=3

and crosses at

x=-1

" reverses the behavior at

-1

."It crosses at

x=3

and

x=-1

,and it also has complex zeros

3i

and

-3i

" gets the complex zeros right, but incorrectly says both real intercepts are crossings. Complex zeros are roots of the polynomial, but they do not appear as points on the real

x

-axis. The key takeaway: before deciding "cross"vs."touch," factor the polynomial and check each real zero's multiplicity. Even means touch, odd means cross. Complex factors never create

x

-intercepts.

Question 5

Let $g(x)=(x^2+2x+1)(x^2+4)^2$ . Which statement about the zeros of $g$ is true?

$-1$ , multiplicity 2; $2$ and $-2$ , each multiplicity 2
$-1$ , multiplicity 2; $2i$ and $-2i$ , each multiplicity 1
$1$ , multiplicity 2; $2i$ and $-2i$ , each multiplicity 2
$-1$ , multiplicity 2; $2i$ and $-2i$ , each multiplicity 2 (correct answer)

Explanation: When you see a factored polynomial and are asked about zeros, your job is to solve each factor equal to zero and count the exponent of each factor as its multiplicity. Start with the first factor:

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

. Setting this equal to zero gives

x=-1

, and the square means the multiplicity is 2. Now look at

(x^2+4)^2

. Setting

x^2+4=0

gives

x^2=-4

, so the zeros are

2i

and

-2i

. Because the whole factor is squared, each of these complex zeros has multiplicity 2. Therefore the true statement is:

-1

multiplicity 2;

2i

and

-2i

each multiplicity 2. The choice that lists

2

and

-2

as zeros comes from treating

x^2+4=0

x^2=4

, which mistakenly gives real roots instead of imaginary ones. The choice that lists

2i

and

-2i

each with multiplicity 1 ignores the exponent 2 on that factor. The choice listing

1

instead of

-1

comes from misreading

(x+1)^2

(x-1)^2

; always check the sign inside the factor. A good habit: factor completely, set each distinct factor to zero, and let the exponent tell you the multiplicity. Also remember that complex zeros come in conjugate pairs, and their multiplicities match the power on their factor.

Question 6

A student claims: There is a degree-5 polynomial with real coefficients whose zeros, including multiplicity, are $-2$ (multiplicity 2), $3i$ , and $4-i$ . Which is the best reason this claim is false?

A polynomial with real coefficients cannot have two different nonreal zeros.
The conjugates $-3i$ and $4+i$ are also forced, giving six zeros counting multiplicity. (correct answer)
Every nonreal zero must have multiplicity 2; therefore the degree would be at least 8.
A degree-5 polynomial with real coefficients must have exactly one real zero, but $-2$ already has multiplicity 2.

Explanation: Whenever you see zeros of a polynomial with real coefficients, immediately think about conjugate pairs. Nonreal zeros cannot appear alone: if

a+bi

is a zero, then

a-bi

must also be a zero. Here the listed zeros are

-2

(multiplicity 2),

3i

, and

4-i

. Since the polynomial must have real coefficients,

3i

forces

-3i

, and

4-i

forces

4+i

. That gives:

-2

counted twice, then

3i, -3i, 4-i, 4+i

— six zeros counting multiplicity. A degree-5 polynomial cannot have six zeros counting multiplicity, so the claim is false. That is exactly the best reason. The other choices miss the real issue. "A polynomial with real coefficients cannot have two different nonreal zeros" is false — it can, as long as they come in conjugate pairs. "Every nonreal zero must have multiplicity 2; therefore the degree would be at least 8" is also false: nonreal zeros have multiplicity 1 in this situation, and the forced conjugates already bring the count to six, not eight. "A degree-5 polynomial with real coefficients must have exactly one real zero" is false as well — a degree-5 real polynomial can have three real zeros and two nonreal conjugates; the real zeros just have to sum in multiplicity to an odd number. Your study takeaway: whenever a polynomial has real coefficients, count every nonreal zero together with its conjugate. That check alone would catch this claim immediately.

Question 7

Let $f(x)=(x^2-4x+4)(x^2+6x+9)(x^2+1)$ . Which of the following lists all zeros of $f$ with their multiplicities?

$2$ , multiplicity 2; $-3$ , multiplicity 2; $i$ and $-i$ , each multiplicity 1 (correct answer)
$-2$ , multiplicity 2; $3$ , multiplicity 2; $i$ and $-i$ , each multiplicity 1
$2$ , multiplicity 2; $-3$ , multiplicity 2
$2$ , multiplicity 1; $-3$ , multiplicity 1; $i$ and $-i$ , each multiplicity 1

Explanation: When you see a polynomial given in factored form like this, your job is to read each factor as a statement about zeros. A factor of the form

(x - r)^k

tells you that

r

is a zero with multiplicity

k

. The factor

(x^2+1)

is not a difference of squares over the reals, but over the complex numbers it factors as

(x - i)(x + i)

, giving zeros

i

and

-i

, each with multiplicity

1

. Now examine each quadratic:

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

, so

2

has multiplicity

2

x^2 + 6x + 9 = (x + 3)^2 = (x - (-3))^2

, so

-3

has multiplicity

2

. The correct answer lists exactly these:

2

multiplicity

2

-3

multiplicity

2

, and

i, -i

each multiplicity

1

. Now look at the traps. The choice that lists

-2

and

3

instead of

2

and

-3

confuses signs—check the constant terms:

x^2 - 4x + 4

gives

+2

, not

-2

. The choice that omits the complex zeros forgets that

x^2+1

has no real solutions but does have imaginary ones; the question asks for all zeros, not just real zeros. The choice that gives

2

and

-3

multiplicity

1

ignores that the squares in the factors double those multiplicities. Your study tip: always factor completely, including complex factors when necessary, and count the exponent on each factor as the multiplicity. Watch for sign flips when reading zeros from

(x - r)^k

Question 8

Let $R(x)=\frac{(x-2)^2(x+3)}{(x-2)(x+3)^2}$ . Which statement correctly describes the $x$ -intercepts and the graph of $R$ ?

There is no $x$ -intercept; the graph has a hole at $x=2$ and a vertical asymptote at $x=-3$ . (correct answer)
The $x$ -intercept is $x=2$ ; the graph has a vertical asymptote at $x=-3$ .
The $x$ -intercept is $x=2$ ; the graph has a hole at $x=-3$ .
The $x$ -intercept is $x=-3$ ; the graph has a vertical asymptote at $x=2$ .

Explanation: When you see a rational function, factor everything and compare the numerator and denominator before simplifying. Here,

R(x)=\frac{(x-2)^2(x+3)}{(x-2)(x+3)^2}

simplifies to

\frac{x-2}{x+3}

, but only where

x\neq 2

and

x\neq -3

. The factor

(x-2)

cancels completely, so

x=2

is a hole: the function is undefined there, even though the simplified form would be

0

at that point. Because the graph has no point at

x=2

, it does not cross the

x

-axis there—so there is no

x

-intercept. The remaining denominator factor

(x+3)

is zero at

x=-3

, so that is a vertical asymptote: the graph increases or decreases without bound near it. Thus the correct description is: no

x

-intercept, a hole at

x=2

, anda vertical asymptote at

x=-3

. The wrong choices each come froma simplifying mistake. Saying "the

x

-intercept is

x=2

" confuses the value of the simplified expression with an actual point on the original graph; the hole removes that point. Saying"the graph has a hole at

x=-3

" mistakes the vertical asymptote for a hole—the

(x+3)

factor does not cancel completely, so it remains in the denominator. Saying"the

x

-intercept is

x=-3

and the vertical asymptote is

x=2

" reverses both roles:

x=-3

makes the denominator zero, so it cannot be an intercept, and

x=2

is cancelled, so it cannot be a vertical asymptote. Remember: after simplifying a rational function, any canceled factor creates a hole, whileany factor left in the denominator creates a vertical asymptote. An

x

-intercept must come from a numerator zero that is actually in the domain of the original function.

Question 9

A polynomial of least possible degree has real coefficients, touches the $x$ -axis at $x=-1$ , crosses at $x=2$ , and satisfies $p(0)=6$ . Which polynomial could be $p(x)$ ?

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

Explanation: When you see "touches" versus "crosses" at an x-intercept, think about multiplicity: touching means the factor has even multiplicity, crossing means odd multiplicity. Here, touching at

x=-1

requires

(x+1)

squared (or any even power), and crossing at

x=2

requires

(x-2)

to the first power (or odd). For the least possible degree, use the smallest exponents:

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

. Now use

p(0)=6

: plugging in

0

gives

a(1)^2(-2)=-2a=6

, so

a=-3

. Thus

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

. The choice

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

reverses the multiplicities—it touches at 2 and crosses at -1, which is the opposite of what you need. The choice

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

has the correct shape but the wrong leading coefficient:

p(0)=3(1)^2(-2)=-6

, not 6. And

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

has zeros at

1

and

-2

, not at

-1

and

2

. Only the correct polynomial satisfies both the multiplicity conditions and the value at

0

. A quick strategy: first translate the graph behavior into factors with the right exponents, then use the given point to solve for the constant. Always check that the constant's sign matches the y-intercept you need.

Question 10

A 6th-degree polynomial $p$ has real coefficients. It has $-1$ as a zero of multiplicity 2, and it also has $2+i$ and $-3i$ as zeros, each of multiplicity 1. Which of the following is the complete list of zeros of $p$ with their multiplicities?

$-1$ , multiplicity 2; $2+i$ and $2-i$ , each multiplicity 1; $-3i$ and $3i$ , each multiplicity 1 (correct answer)
$-1$ , multiplicity 2; $2+i$ , multiplicity 2; $-3i$ , multiplicity 2
$-1$ , multiplicity 2; $2+i$ , multiplicity 1; $-3i$ , multiplicity 1
$-1$ , multiplicity 2; $-2+i$ and $-2-i$ , each multiplicity 1; $-3i$ and $3i$ , each multiplicity 1

Explanation: Whenever you see a polynomial with real coefficients and a complex zero, the central rule is: complex zeros must appear in conjugate pairs. That means if

2+i

is a zero, then

2-i

must also be a zero; if

-3i

is a zero, then

3i

must also be a zero. The given zero

-1

has multiplicity 2, so adding these conjugates gives

2+1+1+1+1=6

total multiplicity, exactly matching a 6th-degree polynomial. The correct list is therefore

-1

, multiplicity 2;

2+i

and

2-i

, each multiplicity 1;

-3i

and

3i

, each multiplicity 1. This satisfies both the conjugate-pair rule and the degree count. The choice saying

-1

, multiplicity 2;

2+i

, multiplicity 2;

-3i

, multiplicity 2 misses the conjugates entirely and assigns incorrect multiplicities. The choice listing only

-1

2+i

, and

-3i

with multiplicity 1 ignores the conjugate pairs and would produce degree 4, not 6. The choice with

-2+i

and

-2-i

has the right conjugate structure but the wrong real part — it changes the given zero

2+i

into

-2+i

, which is not a zero of this polynomial. For any real-coefficient polynomial, as soon as you see one nonreal zero, write its conjugate next to it. Then check that the total multiplicities add up to the degree — that double-check will catch most mistakes.

Question 11

Which leading-coefficient-1 polynomial of least degree with real coefficients has $4$ as a zero of multiplicity 1 and $2-3i$ as a zero?

$(x-4)(x^2-4x-5)$
$(x-4)(x^2+4x+13)$
$(x-4)(x-2+3i)$
$(x-4)(x^2-4x+13)$ (correct answer)

Explanation: Whenever you see a polynomial with real coefficients and a complex zero, the Complex Conjugate Root Theorem is your compass: if

a+bi

is a zero, then

a-bi

must also be a zero. Since

2-3i

is given as a zero,

2+3i

is forced to be one as well. The quadratic factor containing these two zeros is

(x - (2-3i))(x - (2+3i))

. Expanding gives

(x-2+3i)(x-2-3i) = (x-2)^2 - (3i)^2 = x^2 -4x +4 - (-9) = x^2 -4x +13

. Multiplying by the linear factor for the zero of multiplicity 1 at 4 yields

(x-4)(x^2-4x+13)

. This is the least-degree polynomial (degree 3) with real coefficients and leading coefficient 1. Now, check the traps.

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

has a quadratic that factors to

(x-5)(x+1)

, giving real zeros, not complex ones.

(x-4)(x^2+4x+13)

has the wrong sign on the

x

-term; its zeros are

-2 \pm 3i

, not

2 \pm 3i

. The option

(x-4)(x-2+3i)

only includes one complex factor—it forgets the conjugate, so its expanded form would have complex coefficients, violating the requirement for real coefficients. For your exam, remember the shortcut:

(x - (a+bi))(x - (a-bi)) = x^2 - 2ax + (a^2 + b^2)

. This instantly gives you the correct quadratic, so you can avoid sign errors and missing conjugates. Always verify the middle term's sign and that complex roots come in pairs.

Question 12

Let $f(x)=(x-3)^4(x+2)^2$ and $g(x)=[f(x)]^3$ . Which statement gives all zeros of $g$ with their multiplicities?

$3$ , multiplicity 12; $2$ , multiplicity 6
$3$ , multiplicity 4; $-2$ , multiplicity 2
$3$ , multiplicity 12; $-2$ , multiplicity 6 (correct answer)
$3$ , multiplicity 4; $2$ , multiplicity 2

Explanation: Whenever you see zeros and multiplicities, look at the factored form: each factor

(x-a)^k

contributes a zero at

x=a

with multiplicity

k

. Here,

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

, so

f

has zero

x=3

with multiplicity

4

, and

x=-2

with multiplicity

2

. Then

g(x)=[f(x)]^3

cubes the entire polynomial; cubing doesn't change where the zeros are, but it does multiply every exponent by

3

. Thus,

x=3

gets multiplicity

4\cdot =12

,and

x=-2

gets multiplicity

2\cdot3=6

. So the correct statement is

3

, multiplicity

12

;

-2

, multiplicity

6

. The choice saying

3

, multiplicity

4

;

-2

, multiplicity

2

gives the zeros of

f

, not

g

, because it forgets to multiply by the exponent

3

. The choice saying

3

, multiplicity

12

;

2

, multiplicity

6

catches the cubing but has the sign wrong: solving

x+2=0

gives

x=-2

, not

x=2

. The choice saying

3

, multiplicity

4

;

2

, multiplicity

2

has both mistakes: it gives

f

's multiplicities and also flips the sign of the second zero. When you raise a factored polynomial to a power,take each original zero and multiply its multiplicity by that power. Also, always solve each factor equal to zero carefully — the sign is easy to reverse.