# A calculus of the absurd

#### 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 $$f(a)$$. 1515 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 $$a$$.

\begin{align} & P(1) & = 0 \\ & 5 - 9 + 2 + a & = 0 \\ & -2 + a & = 0 \\ & a & = 0 \end{align}