# 4.6 The factor and remainder theorems

The remainder theorem states that for any number $a$, the remainder when we divide a polynomial (which can be written as $f(x)$) by $(x-a)$ is equal to $f(a)$. 66 6 We can prove this without too much trouble. Firstly, by the definition of division, we can write the result of any division as a ”quotient” $Q(x)$ and a remainder $R(x)$ as a fraction of $x-a$ - as an equation, that is, that $\frac{f(x)}{(x-a)}=Q(x)+\frac{R(x)}{x-a}$. If we multiply both sides by $x-a$, then we obtain that $f(x)=Q(x)(x-a)+R(x)$. We can then set $x=a$, which gives that $f(a)=R(a)$. Note that $Q(x)(a-a)=0$.

A special case of the remainder theorem is known as the "factor theorem" and it relates the roots of a polynomial to its factors. If $(x-a)$ divides $f(x)$ with no remainder, then by the remainder theorem, this means that $f(a)=0$ and so $a$ is a root as well as a factor. Roots are factors, and factors are roots.

## 4.6.1 An example

Question

: Given that $(x-1)$ is a factor of $P(x)=5x^{3}-9x^{2}+2x+a$, find the value of $a$.

Solution

: We can directly apply the factor theorem here; as $(x-1)$ is a factor of $P(x)$, it must be the case that $P(1)=0$. With this information, we can form an equation where the only variable is $a$.

 $\displaystyle P(1)$ $\displaystyle=0$ (4.65) $\displaystyle 5-9+2+a$ $\displaystyle=0$ (4.66) $\displaystyle-2+a$ $\displaystyle=0$ (4.67) $\displaystyle a$ $\displaystyle=0$ (4.68)