Algebra 2 Quiz: Applying The Remainder Theorem

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Applying The Remainder TheoremQuestion 1 of 20

What is the remainder when dividing $p(x)=2x^4-x^2+5$ by $(x+1)$ ? Use the Remainder Theorem (remainder $=p(a)$ for divisor $(x-a)$ ).

The remainder is

0

The remainder is

8

The remainder is

6

The remainder is

4

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Algebra 2 Quiz

Algebra 2 Quiz: Applying The Remainder Theorem

Practice Applying The Remainder Theorem in Algebra 2 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 Applying The Remainder Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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 remainder when dividing $p(x)=2x^4-x^2+5$ by $(x+1)$ ? Use the Remainder Theorem (remainder $=p(a)$ for divisor $(x-a)$ ).

The remainder is $0$ .
The remainder is $8$ .
The remainder is $6$ . (correct answer)
The remainder is $4$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x + 1) = (x - (-1)), a = -1; p(-1) = 2*(-1)^4 - (-1)^2 + 5 = 21 - 1 + 5 = 2 - 1 + 5 = 6. Choice A correctly evaluates p(-1) as 6. Choice B says 4, which could happen if you mistakenly computed (-1)^4 as -1 instead of 1, leading to 2(-1) -1 +5 = -2 -1 +5 =2, but remember even powers are positive! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 2

If $g(x) = x^4 - 3x^3 + 2x^2 - 5x + 1$ and the remainder when $g(x)$ is divided by $(x - c)$ is $15$ , which equation must be satisfied by $c$ ?

$c^4 - 3c^3 + 2c^2 - 5c + 1 = 15$ (correct answer)
$c^4 - 3c^3 + 2c^2 - 5c - 14 = 0$
$c^4 - 3c^3 + 2c^2 - 5c + 16 = 0$
$c^4 - 3c^3 + 2c^2 - 5c - 1 = 15$

Explanation: By the Remainder Theorem, when

g(x)

is divided by

(x - c)

, the remainder equals

g(c)

. Since the remainder is

15

, we have

g(c) = 15

. Therefore

c^4 - 3c^3 + 2c^2 - 5c + 1 = 15

. Choice B incorrectly sets the expression equal to zero after subtracting 15. Choice C adds 15 instead of setting the original equal to 15. Choice D incorrectly modifies the constant term before setting equal to 15.

Question 3

Consider the polynomial $R(x) = 3x^3 - 7x^2 + 4x - 12$ . If $(x - k)$ is a factor of $R(x)$ , which of the following could be the remainder when $R(x)$ is divided by $(x - 2k)$ ?

Always equal to zero regardless of $k$
Equal to $R(2k)$ which depends on $k$ (correct answer)
Always equal to $-12$ regardless of $k$
Equal to $2R(k)$ which depends on $k$

Explanation: If

(x - k)

is a factor of

R(x)

, then

R(k) = 0

. By the Remainder Theorem, the remainder when

R(x)

is divided by

(x - 2k)

R(2k)

. Since

k

and

2k

are generally different values,

R(2k)

will depend on the specific value of

k

. Choice A is wrong because there's no reason

R(2k)

must equal zero. Choice C incorrectly assumes the remainder is always the constant term. Choice D confuses the relationship between

R(k)

and

R(2k)

Question 4

For the polynomial $T(x) = 2x^4 - 3x^3 + x^2 - 4x + 5$ , suppose that when $T(x)$ is divided by $(x - a)$ , the remainder is $r$ . What is the remainder when $T(2x)$ is divided by $(x - \frac{a}{2})$ ?

$4r$
$2r$
$\frac{r}{2}$
$r$ (correct answer)

Explanation: When you encounter polynomial remainder problems involving transformations, think about how the Remainder Theorem connects division and function evaluation. By the Remainder Theorem, when

T(x)

is divided by

(x - a)

, the remainder equals

T(a) = r

. Now let's analyze what happens with

T(2x)

divided by

(x - \frac{a}{2})

. Using the Remainder Theorem again, this remainder equals

T(2x)

evaluated at

x = \frac{a}{2}

. Substituting:

T(2 \cdot \frac{a}{2}) = T(a) = r

. So the remainder is still

r

, making (D) correct. Let's see why the other answers represent common misconceptions. (A) $4r$ likely comes from thinking the coefficient 2 in

T(2x)

gets squared when applied to the remainder. (B) $2r$ probably results from assuming the transformation

x \to 2x

directly scales the remainder by the same factor of 2. **(C)

\frac{r}{2}

might come from incorrectly thinking that since we're dividing by

(x - \frac{a}{2})

instead of

(x - a)

, the remainder gets scaled by

\frac{1}{2}

. The key insight is that transformations of the variable don't automatically scale the remainder in obvious ways. Instead, you must carefully apply the Remainder Theorem to the transformed situation. Remember: when working with polynomial transformations and remainders, always go back to the fundamental relationship that remainder equals the polynomial evaluated at the root of the divisor.

Question 5

The polynomial $P(x) = x^3 + ax^2 + bx + c$ satisfies $P(2) = 0$ and $P(-1) = 6$ . If $Q(x) = P(x) - 3$ , what is the remainder when $Q(x)$ is divided by $(x + 1)$ ?

$-3$
$6$
$9$
$3$ (correct answer)

Explanation: When you see polynomial questions involving remainders and given conditions, think about the Remainder Theorem: when a polynomial

P(x)

is divided by

(x - a)

, the remainder equals

P(a)

. Since

Q(x) = P(x) - 3

, we need to find

Q(-1)

to get the remainder when

Q(x)

is divided by

(x + 1)

(which is

(x - (-1))

). We know

P(-1) = 6

from the given information. Therefore:

Q(-1) = P(-1) - 3 = 6 - 3 = 3

By the Remainder Theorem, the remainder is

3

. Let's examine why the other answers are wrong: Answer A (

-3

): This would be correct if you mistakenly calculated

3 - 6 = -3

instead of

6 - 3 = 3

, reversing the subtraction order. Answer B (

6

): This trap answer uses

P(-1)

directly, forgetting that we need

Q(-1)

, not

P(-1)

. Remember that

Q(x) = P(x) - 3

, so you must subtract

3

. Answer C (

9

): This incorrect result comes from adding instead of subtracting:

P(-1) + 3 = 6 + 3 = 9

. The definition

Q(x) = P(x) - 3

requires subtraction. Study tip: When working with transformed polynomials like

Q(x) = P(x) + k

, always apply the transformation to your known values. If you know

P(a)

, then

Q(a) = P(a) + k

. Don't skip this step—it's where most errors occur on these problems.

Question 6

Given that $W(x) = x^5 - 2x^4 + 3x^3 - x^2 + 4x - 6$ and the remainder when $W(x)$ is divided by $(x^2 - 4)$ is $ax + b$ , what is the value of $a + b$ ?

$W(2) + W(-2) - 4$
$\frac{W(2) + W(-2)}{2}$
$\frac{W(2) - W(-2)}{4} + W(2)$ (correct answer)
$\frac{W(2) - W(-2)}{2} + \frac{W(2) + W(-2)}{2}$

Explanation: Since

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

, the remainder when dividing by this quadratic is linear:

R(x) = ax + b

. By the Remainder Theorem,

R(2) = W(2)

and

R(-2) = W(-2)

. So

2a + b = W(2)

and

-2a + b = W(-2)

. Solving:

4a = W(2) - W(-2)

, so

a = \frac{W(2) - W(-2)}{4}

. And

b = W(2) - 2a = W(2) - \frac{W(2) - W(-2)}{2}

. Therefore

a + b = \frac{W(2) - W(-2)}{4} + W(2) - \frac{W(2) - W(-2)}{2} = \frac{W(2) - W(-2)}{4} + W(2)

. The other choices give incorrect combinations of these values.

Question 7

Determine if $x=2$ is a zero of $p(x)=x^3-6x^2+11x-6$ . (Use: $x=a$ is a zero iff $p(a)=0$ .)

Yes, because $p(2)=0$ . (correct answer)
No, because $p(2)=1$ .
Yes, because $p(-2)=0$ .
No, because $p(-2)=0$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. To check if x=2 is a zero, evaluate p(2)=2^3 -62^2 +112 -6=8-24+22-6; step by step, 8-24=-16, -16+22=6, 6-6=0, so yes. Choice A correctly determines yes because p(2)=0. A distractor like B might come from stopping at 6 before subtracting 6. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 8

Use the Remainder Theorem to find the remainder when $p(x)=3x^3-x^2-7x+2$ is divided by $(x+2)$ .

The remainder is $-6$ .
The remainder is $-12$ . (correct answer)
The remainder is $12$ .
The remainder is $0$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x + 2) = (x - (-2)), a = -2; p(-2) = 3*(-2)^3 - (-2)^2 - 7*(-2) + 2 = 3*(-8) - 4 + 14 + 2 = -24 - 4 + 14 + 2, which is (-24 - 4) + (14 + 2) = -28 + 16 = -12. Choice A correctly evaluates p(-2) as -12. Choice B says 12, possibly from changing all signs incorrectly or forgetting the negative in (-2)^3, but use parentheses to track signs properly! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 9

If the remainder when $p(x)=x^4+mx^2-2x+1$ is divided by $(x-2)$ is $9$ , what is $m$ ? (Use: remainder $=p(2)$ .)

$m=0$
$m=1$
$m=2$
$m=-1$ (correct answer)

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x-2), a=2, p(2)=2^4 +m2^2 -22 +1=16+4m-4+1=13+4m; set to 9: 13+4m=9, 4m=-4, m=-1. Choice A correctly finds m=-1. A distractor like C (1) might come from setting 16+4m-4+1=9 as 13+4m=9 but solving 4m=-4 incorrectly. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 10

Is $(x+2)$ a factor of $p(x)=x^3+3x^2-4x-12$ ? Use the Remainder Theorem / Factor Theorem (i.e., $(x-a)$ is a factor iff $p(a)=0$ ).

Yes, because $p(-2)=0$ . (correct answer)
No, because $p(-2)=8$ .
Yes, because $p(2)=0$ .
No, because the remainder is $0$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. For (x + 2) = (x - (-2)), a = -2; p(-2) = (-2)^3 + 3*(-2)^2 - 4*(-2) - 12 = -8 + 3*4 + 8 - 12 = -8 + 12 + 8 - 12, which is (-8 - 12) + (12 + 8) = -20 + 20 = 0. Choice A correctly determines that yes, it is a factor because p(-2) = 0. Choice B says no because p(-2) = 8, which might come from forgetting to include the -12 or mishandling signs, but actually it's 0—keep track of all terms! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 11

Use the Remainder Theorem to find the remainder when $p(x)=3x^3-x^2-7x+2$ is divided by $(x+2)$ .

The remainder is $-12$ . (correct answer)
The remainder is $12$ .
The remainder is $-6$ .
The remainder is $0$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For (x + 2) = (x - (-2)), a = -2; p(-2) = 3*(-2)^3 - (-2)^2 - 7*(-2) + 2 = 3*(-8) - 4 + 14 + 2 = -24 - 4 + 14 + 2, which is (-24 - 4) + (14 + 2) = -28 + 16 = -12. Choice A correctly evaluates p(-2) as -12. Choice B says 12, possibly from changing all signs incorrectly or forgetting the negative in (-2)^3, but use parentheses to track signs properly! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 12

Verify the claim using the Remainder Theorem: When dividing $p(x)=x^4-3x^2+2x+1$ by $(x-1)$ , the remainder is $1$ .

True, because $p(1)=1$ . (correct answer)
False, because $p(1)=0$ so the remainder is $0$ .
False, because $p(-1)=1$ so the remainder is $1$ .
True, because $p(-1)=0$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! To verify, evaluate p(1): 1^4 = 1, -31^2 = -3, +21 = +2, +1; combine: 1 - 3 = -2, -2 + 2 = 0, 0 + 1 = 1. Choice A correctly verifies it's true because p(1)=1 matches the claimed remainder. Choice B incorrectly claims false with p(1)=0, perhaps from miscalculating 1 -3 +2 +1 as 0 by ignoring the last +1. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 13

If the remainder when $p(x)=x^3-2x^2+kx-5$ is divided by $(x-1)$ is $3$ , find $k$ . (Use $\text{remainder} = p(1)$ .)

$k=7$
$k=9$ (correct answer)
$k=11$
$k=3$

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works:

p(x) = (x - a) \cdot q(x) + r

, where r is the remainder. Substituting x = a:

p(a) = (a - a) \cdot q(a) + r = 0 + r = r

. So the remainder r equals p(a)—brilliant! Here, p(1) = 3 and we set it equal to 3:

1^3 - 2(1)^2 + k \cdot 1 - 5 = 1 - 2 + k - 5 = -6 + k = 3

, so

k = 9

. Choice B correctly finds k=9. A distractor like k=7 in choice A might come from misadding -6 + k =3 as k=-3+6=3 or similar arithmetic error. Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember:

(x + 3) = (x - (-3))

, so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 14

Is $(x+2)$ a factor of $p(x)=x^3+3x^2-4x-12$ ? Use the Remainder Theorem / Factor Theorem (i.e., $(x-a)$ is a factor iff $p(a)=0$ ).

Yes, because $p(-2)=0$ . (correct answer)
No, because the remainder is $0$ .
Yes, because $p(2)=0$ .
No, because $p(-2)=8$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. For (x + 2) = (x - (-2)), a = -2; p(-2) = (-2)^3 + 3*(-2)^2 - 4*(-2) - 12 = -8 + 3*4 + 8 - 12 = -8 + 12 + 8 - 12, which is (-8 - 12) + (12 + 8) = -20 + 20 = 0. Choice A correctly determines that yes, it is a factor because p(-2) = 0. Choice B says no because p(-2) = 8, which might come from forgetting to include the -12 or mishandling signs, but actually it's 0—keep track of all terms! Sign safety for negatives: when testing (x + 2) = (x - (-2)), you're evaluating at a = -2. Substitute carefully: if p(x) = x³ - 3x + 5, then p(-2) = (-2)³ - 3(-2) + 5 = -8 + 6 + 5 = 3. Use parentheses around negative values to avoid sign errors! This is where most mistakes happen with the Remainder Theorem.

Question 15

Determine if $x=-1$ is a zero of $p(x)=x^3+2x^2-x-2.$ (Use the fact that $x=a$ is a zero iff $p(a)=0$ .)

Yes; $p(-1)=0$ . (correct answer)
No; $p(-1)=2$ .
Yes; $p(1)=0$ .
No; $p(1)=-2$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! A special case is the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. This means the remainder is zero, so the division is exact with no remainder. We can test potential factors by just evaluating the polynomial—if p(a) = 0, we've found a factor! This beats trial-and-error factoring when testing specific values. To check if x = -1 is a zero of p(x) = x³ + 2x² - x - 2, we evaluate p(-1): p(-1) = (-1)³ + 2(-1)² - (-1) - 2 = -1 + 2(1) + 1 - 2 = -1 + 2 + 1 - 2 = 0. Since p(-1) = 0, x = -1 is indeed a zero! Choice A correctly states that x = -1 is a zero because p(-1) = 0. Using the Remainder Theorem: (1) A zero of p(x) is a value a where p(a) = 0, (2) This also means (x - a) is a factor of p(x), (3) Calculate carefully with parentheses for negative values, (4) If p(a) = 0, you've found both a zero and a factor! This connection between zeros and factors is fundamental in algebra.

Question 16

If $p(x) = 2x^3 - 5x^2 + 3x - 7$ and $p(2) = -5$ , what can be concluded about the polynomial $q(x) = p(x) + 5$ ?

$(x - 2)$ is a factor of $q(x)$ (correct answer)
$(x + 2)$ is a factor of $q(x)$
$q(x)$ has remainder $-5$ when divided by $(x - 2)$
$q(x)$ has remainder $5$ when divided by $(x + 2)$

Explanation: Since

p(2) = -5

, we have

q(2) = p(2) + 5 = -5 + 5 = 0

. By the Remainder Theorem, since

q(2) = 0

, the factor

(x - 2)

divides

q(x)

evenly. Choice B is incorrect because it uses

(x + 2)

instead of

(x - 2)

. Choice C is wrong because

q(x)

has remainder

0

, not

-5

, when divided by

(x - 2)

. Choice D is incorrect because it involves division by

(x + 2)

and the wrong remainder value.

Question 17

A polynomial $f(x)$ has the property that $f(-3) = 8$ and $f(1) = 0$ . When $f(x)$ is divided by $x^2 + 2x - 3$ , what is the remainder?

$4x + 12$
$2x + 2$
$-2x + 2$ (correct answer)
$3x + 9$

Explanation: Since

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

, when

f(x)

is divided by this quadratic, the remainder must be linear:

R(x) = ax + b

. By the Remainder Theorem,

R(-3) = f(-3) = 8

and

R(1) = f(1) = 0

. So

-3a + b = 8

and

a + b = 0

. Solving:

b = -a

and

-3a + (-a) = 8

, so

-4a = 8

and

a = -2

b = 2

. Therefore

R(x) = -2x + 2

. The other choices don't satisfy both conditions simultaneously.

Question 18

A cubic polynomial $f(x)$ has leading coefficient $1$ and satisfies $f(1) = f(3) = f(5) = 0$ . When $f(x)$ is divided by $(x - 7)$ , the remainder is:

$24$
$48$ (correct answer)
$72$
$0$

Explanation: Since

f(x)

is cubic with leading coefficient 1 and has zeros at

x = 1, 3, 5

, we have

f(x) = (x-1)(x-3)(x-5)

. By the Remainder Theorem, the remainder when divided by

(x-7)

f(7) = (7-1)(7-3)(7-5) = 6 \cdot 4 \cdot 2 = 48

Question 19

Given that $(x - 4)$ is a factor of $h(x) = x^3 - 6x^2 + kx + 8$ , which statement about $h(x + 4)$ is necessarily true?

$h(x + 4)$ has $(x - 4)$ as a factor
$h(x + 4)$ has $x$ as a factor (correct answer)
$h(x + 4)$ has $(x + 4)$ as a factor
$h(x + 4)$ has $(x - 8)$ as a factor

Explanation: Since

(x - 4)

is a factor of

h(x)

, we know

h(4) = 0

by the Remainder Theorem. For the transformed function

h(x + 4)

, we evaluate at

x = 0

h(0 + 4) = h(4) = 0

. Therefore, by the Remainder Theorem,

x

is a factor of

h(x + 4)

. Choice A is incorrect because the horizontal shift changes the factor. Choice C incorrectly suggests

(x + 4)

is a factor. Choice D incorrectly identifies

(x - 8)

as the factor.

Question 20

Use the Remainder Theorem to find the remainder when $p(x)=x^3+4x^2-x-12$ is divided by $(x+3).$

The remainder is $0$ . (correct answer)
The remainder is $-6$ .
The remainder is $6$ .
The remainder is $-24$ .

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! For p(x) = x³ + 4x² - x - 12 divided by (x + 3) = (x - (-3)), we need p(-3): p(-3) = (-3)³ + 4(-3)² - (-3) - 12 = -27 + 4(9) + 3 - 12 = -27 + 36 + 3 - 12 = 0. Choice A correctly identifies the remainder as 0, which also means (x + 3) is a factor of p(x)! Sign safety for negatives: when testing (x + 3) = (x - (-3)), you're evaluating at a = -3. Substitute carefully using parentheses: (-3)³ = -27 and (-3)² = 9. The zero remainder reveals that this polynomial is exactly divisible by (x + 3)!