# 8.2: Dividing by (x - c)

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We now restrict our attention to the case where the divisor is $$g(x)=x-c$$ for some real number $$c$$. In this case, the remainder $$r$$ of the division $$f(x)$$ by $$g(x)$$ is a real number. We make the following observations.

## Observation: Remainder

Assume that $$g(x)=x-c$$, and the long division of $$f(x)$$ by $$g(x)$$ has remainder $$r$$, that is,

$\text{Assumption: } f(x)=q(x)\cdot (x-c)+r \nonumber$

When we evaluate both sides in the above equation at $$x=c$$ we see that $$f(c)=q(c)\cdot (c-c)+r = q(c)\cdot 0 + r=r$$. In short:

$\boxed{\text{The remainder when dividing f(x) by (x-c) is } \quad r=f(c)}$

In particular:

$\boxed{f(c)=0 \quad \iff \quad g(x)=x-c \text{ is a factor of } f(x)}$

The above statement $$\PageIndex{1}$$ is called the remainder theorem, and $$\PageIndex{2}$$ is called the factor theorem.

## The Remainder Theorem

The remainder when dividing $$f(x)$$ by $$(x-c)$$ is $$r=f(c)$$

## The Factor Theorem

$$f(c)=0 \quad \iff \quad g(x)=x-c$$ is a factor of $$f(x)$$

## Example $$\PageIndex{1}$$

Find the remainder of dividing $$f(x)=x^2+3x+2$$ by

1. $$x-3$$
2. $$x+4$$
3. $$x+1$$
4. $$x-\dfrac{1}{2}$$

Solution

By Observation, we know that the remainder $$r$$ of the division by $$x-c$$ is $$f(c)$$.

1. Thus, the remainder for part (a), when dividing by $$x-3$$ is $r=f(3)=3^2+3\cdot 3+2=9+9+2=20 \nonumber$
2. For (b), note that $$g(x)=x+4=x-(-4)$$, so that we take $$c=-4$$ for our input giving that $$r= f(-4)=(-4)^2+3\cdot (-4)+2=16-12+2=6$$.

Similarly, the other remainders are:

1. $$r=f(-1)=(-1)^{2}+3 \cdot(-1)+2=1-3+2=0$$
2. $$r=f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{2}+\dfrac{1}{2} \cdot 3+2=\dfrac{1}{4}+\dfrac{3}{2}+2=\dfrac{1+6+8}{4}=\dfrac{15}{4}$$

Note that in part (c), we found a remainder $$0$$, so that $$(x+1)$$ is a factor of $$f(x)$$.

## Example $$\PageIndex{2}$$

Determine whether $$g(x)$$ is a factor of $$f(x)$$.

1. $$f(x)=x^3+2x^2+5x+1, \quad g(x)=x-2$$
2. $$f(x)=x^4+4 x^3+x^2+18, \quad g(x)=x+3$$
3. $$f(x)=x^5+3x^2+7, \quad g(x)=x+1$$

Solution

1. We need to determine whether $$2$$ is a root of $$f(x)=x^3+2x^2+5x+1$$, that is, whether $$f(2)$$ is zero. $f(2)=2^3+2\cdot 2^2+5\cdot 2+1=8+8+10+1=27 \nonumber$Since $$f(2)=27\neq 0$$, we see that $$g(x)=x-2$$ is not a factor of $$f(x)$$.
2. Now, $$g(x)=x+3=x-(-3)$$, so that we calculate: $f(-3)=(-3)^4+4\cdot (-3)^3+(-3)^2-18=81-108+9+18=0 \nonumber$ Since the remainder is zero, we see that $$x+3$$ is a factor of $$x^4+4\cdot x^3+x^2+18$$. Therefore, if we wanted to find the other factor, we could use long division to obtain the quotient.
3. Finally, we have: $f(-1)=(-1)^5+3\cdot (-1)^2+7=-1+3+7=9 \nonumber$ $$g(x)=x+1$$ is not a factor of $$f(x)=x^5+3x^2+7$$.

## Example $$\PageIndex{3}$$

1. Show that $$-2$$ is a root of $$f(x)=x^5-3\cdot x^3+5x^2-12$$, and use this to factor $$f$$.
2. Show that $$5$$ is a root of $$f(x)=x^3-19x-30$$, and use this to factor $$f$$ completely.

Solution

1. First, we calculate that $$-2$$ is a root. $f(-2)=(-2)^5-3\cdot (-2)^3+5\cdot (-2)^2+12=-32+24+20-12=0 \nonumber$ So we can divide $$f(x)$$ by $$g(x)=x-(-2)=x+2$$: So we factored $$f(x)$$ as $$f(x)=(x^4-2x^3+x^2+3x-6)\cdot (x+2)$$.

1. Again we start by calculating $$f(5)$$: $f(5)=5^3-19\cdot 5-30=125-95-30=0 \nonumber$Long division by $$g(x)=x-5$$ gives: Thus, $$x^3-19x-30=(x^2+5x+6)\cdot(x-5)$$. To factor $$f$$ completely, we also factor $$x^2+5x+6$$. $f(x)=(x^2+5x+6)\cdot(x-5)=(x+2)\cdot(x+3)\cdot(x-5) \nonumber$

This page titled 8.2: Dividing by (x - c) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.