Math 3 Quiz: Building Polynomials From Zeros

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Building Polynomials From ZerosQuestion 1 of 8

A polynomial $f(x)$ of degree $4$ has zeros at $x = -3$ , $x = 1$ , and $x = 5$ , where the zero at $x = 1$ has multiplicity $2$ . If $f(x)$ has a $y$ -intercept of $30$ , what is the leading coefficient?

-2

-1

1

2

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Math 3 Quiz

Math 3 Quiz: Building Polynomials From Zeros

Practice Building Polynomials From Zeros in Math 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 Building Polynomials From Zeros, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 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

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

Explanation: Since

f(x)

has degree 4 with zeros at

x = -3

x = 1

(multiplicity 2), and

x = 5

, we can write

f(x) = a(x + 3)(x - 1)^2(x - 5)

where

a

is the leading coefficient. The

y

-intercept is

f(0) = 30

. Substituting:

f(0) = a(0 + 3)(0 - 1)^2(0 - 5) = a(3)(1)(-5) = -15a = 30

. Solving for

a

a = 30/(-15) = -2

. Therefore, the leading coefficient is

-2

Question 2

A polynomial function has zeros at $x = -1$ (multiplicity 3) and $x = 2$ (multiplicity 2). If the leading coefficient is negative and the degree is 5, which statement about the end behavior is correct?

As $x \to -\infty$ , $f(x) \to +\infty$ and as $x \to +\infty$ , $f(x) \to +\infty$
As $x \to -\infty$ , $f(x) \to +\infty$ and as $x \to +\infty$ , $f(x) \to -\infty$ (correct answer)
As $x \to -\infty$ , $f(x) \to -\infty$ and as $x \to +\infty$ , $f(x) \to +\infty$
As $x \to -\infty$ , $f(x) \to -\infty$ and as $x \to +\infty$ , $f(x) \to -\infty$

Explanation: The polynomial has the form

f(x) = a(x+1)^3(x-2)^2

where

a < 0

and the degree is

3 + 2 = 5

(odd). For polynomials with odd degree and negative leading coefficient, the end behavior is: as

x \to -\infty

f(x) \to +\infty

and as

x \to +\infty

f(x) \to -\infty

. This is because the leading term is

ax^5

with

a < 0

. Choice A describes even degree with positive leading coefficient. Choice C describes odd degree with positive leading coefficient. Choice D describes even degree with negative leading coefficient.

Question 3

The polynomial $f(x) = 2x^4 - 8x^3 - 2x^2 + 32x - 24$ can be written in factored form as $f(x) = 2(x-a)(x-b)(x-c)(x-d)$ where $a, b, c, d$ are the zeros. If one of the zeros is $x = 3$ , what is the sum of all four zeros?

$2$
$4$ (correct answer)
$6$
$8$

Explanation: For a monic polynomial

x^4 + px^3 + qx^2 + rx + s

, the sum of the zeros equals

-p

by Vieta's formulas. First, we factor out the leading coefficient:

f(x) = 2x^4 - 8x^3 - 2x^2 + 32x - 24 = 2(x^4 - 4x^3 - x^2 + 16x - 12)

. For the monic polynomial

g(x) = x^4 - 4x^3 - x^2 + 16x - 12

, the sum of zeros is

-(-4) = 4

. The fact that one zero is

x = 3

confirms our polynomial but doesn't change the sum calculation. Choice A would be the sum if the coefficient of

x^3

were

-2

. Choice C would be the sum if we incorrectly used

-(-6)

. Choice D would be the sum if we incorrectly used the coefficient

-8

directly without factoring out the

2

Question 4

If the polynomial $g(x) = 2x^3 - 12x^2 + 22x - 12$ can be written as $g(x) = 2(x - a)(x - b)(x - c)$ where $a \leq b \leq c$ , what is the value of $b^2$ ?

$1$
$4$ (correct answer)
$9$
$16$

Explanation: We have

g(x) = 2x^3 - 12x^2 + 22x - 12 = 2(x^3 - 6x^2 + 11x - 6)

. Let

h(x) = x^3 - 6x^2 + 11x - 6

, so

g(x) = 2h(x)

. We need to find the zeros of

h(x)

. By the rational root theorem, possible rational roots are

\pm 1, \pm 2, \pm 3, \pm 6

. Testing

x = 1

h(1) = 1 - 6 + 11 - 6 = 0

✓. So

(x - 1)

is a factor. Using polynomial division:

h(x) = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)

. Therefore,

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

. The zeros are

a = 1

b = 2

, and

c = 3

(since

1 \leq 2 \leq 3

). Thus,

b^2 = 2^2 = 4

Question 5

Consider the polynomial $h(x) = (x - a)(x - b)(x - c)$ where $a < b < c$ . If $h(x)$ has a local maximum at $x = 2$ and a local minimum at $x = 6$ , what is $a + b + c$ ?

$8$
$10$
$12$ (correct answer)
$16$

Explanation: For the cubic polynomial

h(x) = (x - a)(x - b)(x - c)

, we have

h'(x) = (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b)

. The critical points occur where

h'(x) = 0

. Given that there's a local maximum at

x = 2

and a local minimum at

x = 6

, these are the two critical points of the cubic. For a cubic with positive leading coefficient, the critical points occur in the order: local maximum, then local minimum as

x

increases. So

h'(2) = 0

and

h'(6) = 0

. We can write

h'(x) = 3x^2 - 2(a + b + c)x + (ab + bc + ca)

. Since the critical points are

x = 2

and

x = 6

, we have

h'(x) = 3(x - 2)(x - 6) = 3(x^2 - 8x + 12) = 3x^2 - 24x + 36

. Comparing coefficients:

-2(a + b + c) = -24

, so

a + b + c = 12

Question 6

A quartic polynomial $P(x)$ has zeros at $x = -2, 1, 3, 5$ and satisfies $P(0) = 30$ . What is $P(2)$ ?

$0$
$12$
$-12$ (correct answer)
$24$

Explanation: Since

P(x)

has zeros at

x = -2, 1, 3, 5

, we can write

P(x) = a(x+2)(x-1)(x-3)(x-5)

for some constant

a

. Using

P(0) = 30

P(0) = a(0+2)(0-1)(0-3)(0-5) = a(2)(-1)(-3)(-5) = a(2)(-15) = -30a

. Since

P(0) = 30

, we have

-30a = 30

, so

a = -1

. Therefore,

P(x) = -(x+2)(x-1)(x-3)(x-5)

. Now we calculate

P(2) = -(2+2)(2-1)(2-3)(2-5) = -(4)(1)(-1)(-3) = -(4)(3) = -12

. Choice A is incorrect because

x = 2

is not a zero. Choice B gives the positive value. Choice D is twice the correct magnitude.

Question 7

The polynomial $g(x) = 3x^3 - 12x^2 - 15x + 60$ has three real zeros. If one zero is $x = 4$ , what is the product of the other two zeros?

$-5$ (correct answer)
$5$
$-3$
$3$

Explanation: First, factor out the leading coefficient:

g(x) = 3(x^3 - 4x^2 - 5x + 20)

. Let

h(x) = x^3 - 4x^2 - 5x + 20

. Since

x = 4

is a zero, we can factor out

(x-4)

h(x) = (x-4)(x^2 + bx + c)

for some

b

and

c

. Using polynomial division or substitution:

h(x) = (x-4)(x^2 - 5) = (x-4)(x^2 + 0x - 5)

. Wait, let me verify:

(x-4)(x^2 - 5) = x^3 - 5x - 4x^2 + 20 = x^3 - 4x^2 - 5x + 20

✓. So the zeros of

h(x)

are

x = 4

and the zeros of

x^2 - 5 = 0

, which are

x = \pm\sqrt{5}

. The product of the other two zeros is

\sqrt{5} \cdot (-\sqrt{5}) = -5

. Choice B would be the result if both zeros were positive. Choice C would be the result if we incorrectly factored as

x^2 - 3

. Choice D would be the product

\sqrt{3} \cdot \sqrt{3} = 3

from

x^2 - 3

Question 8

If $g(x) = (x-2)^2(x+3)(x-1)$ and $h(x) = k \cdot g(x)$ where $k$ is a nonzero constant, which statement is always true regardless of the value of $k$ ?

$h(x)$ has the same $y$ -intercept as $g(x)$
$h(x)$ has the same maximum value as $g(x)$
$h(x)$ has the same end behavior as $g(x)$
$h(x)$ has the same zeros as $g(x)$ (correct answer)

Explanation: When you see a function multiplied by a constant, think about how that transformation affects the key features of the original function. Here,

h(x) = k \cdot g(x)

represents a vertical scaling of

g(x)

. The zeros of a function occur where the function equals zero. For

h(x) = k \cdot g(x)

, we need

k \cdot g(x) = 0

. Since

k

is nonzero, this happens exactly when

g(x) = 0

. The zeros of

g(x)

occur when

(x-2)^2(x+3)(x-1) = 0

, giving us

x = 2

(with multiplicity 2),

x = -3

, and

x = 1

. These same values make

h(x) = 0

, so choice D is correct. Choice A is wrong because the

y

-intercept changes. When

x = 0

g(0) = (0-2)^2(0+3)(0-1) = 4 \cdot 3 \cdot (-1) = -12

, but

h(0) = k \cdot (-12) = -12k

. Unless

k = 1

, the

y

-intercepts differ. Choice B is incorrect because multiplying by

k

scales all

y

-values, including maximum values. If

k = 2

, every maximum doubles; if

k = -1

, maxima become minima. Choice C is wrong because the end behavior can reverse. Since

g(x)

has degree 4 with positive leading coefficient,

g(x) \to +\infty

x \to \pm\infty

. But if

k < 0

, then

h(x) \to -\infty

x \to \pm\infty

. Remember: Multiplying a function by a constant preserves zeros but changes almost everything else. The zeros are the "anchor points" that remain fixed under vertical scaling.