# Lesson 15

The Remainder Theorem

### Problem 1

For the polynomial function $$f(x)=x^3-2x^2-5x+6$$, we have $$f(0)=6, f(2)=\text-4, f(\text-2)=0, f(3)=0,f(\text-1)=8, f(1)=0$$. Rewrite $$f(x)$$ as a product of linear factors.

### Solution

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### Problem 2

Select all the polynomials that have $$(x-4)$$ as a factor.

A:

$$x^3-13x-12$$

B:

$$x^3 + 8x^2 + 19x + 12$$

C:

$$x^3+6x + 5x - 12$$

D:

$$x^3-x^2-10x-8$$

E:

$$x^2-4$$

### Solution

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### Problem 3

Write a polynomial function, $$p(x)$$, with degree 3 that has $$p(7) = 0$$.

### Solution

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### Problem 4

Long division was used here to divide the polynomial function $$p(x)=x^3+7x^2-20x-110$$ by $$(x-5)$$ and to divide it by $$(x+5)$$.

$$\displaystyle \require{enclose} \begin{array}{r} x^2+12x+40 \\ x-5 \enclose{longdiv}{x^3+7x^2-20x-110} \\ \underline{-x^3+5x^2} \phantom{-20x-110} \\ 12x^2-20x \phantom{-110}\\ \underline{-12x^2+60x} \phantom{-110} \\ 40x-110 \\ \underline{-40x+200} \\ 90 \end{array}$$

$$\displaystyle \require{enclose} \begin{array}{r} x^2+2x-30 \\ x+5 \enclose{longdiv}{x^3+7x^2-20x-110} \\ \underline{-x^3-5x^2} \phantom{-20x-110} \\ 2x^2-20x \phantom{-110}\\ \underline{-2x^2-10x} \phantom{-110} \\ \text-30x-110 \\ \underline{30x+150} \\ 40 \end{array}$$

1. What is $$p(\text-5)$$?

2. What is $$p(5)$$?

### Solution

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### Problem 5

Which polynomial function has zeros when $$x=5,\frac23,\text-7$$?

A:

$$f(x)=(x+5)(2x+3)(x-7)$$

B:

$$f(x)=(x+5)(3x+2)(x-7)$$

C:

$$f(x)=(x-5)(2x-3)(x+7)$$

D:

$$f(x)=(x-5)(3x-2)(x+7)$$

### Solution

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(From Unit 2, Lesson 5.)

### Problem 6

The polynomial function $$q(x)=3x^4+8x^3-13x^2-22x+24$$ has known factors $$(x+3)$$ and $$(x+2)$$. Rewrite $$q(x)$$ as the product of linear factors.

### Solution

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(From Unit 2, Lesson 12.)

### Problem 7

We know these things about a polynomial function $$f(x)$$: it has degree 3, the leading coefficient is negative, and it has zeros at $$x=\text-5,\text-1, 3$$. Sketch a graph of $$f(x)$$ given this information.