4.6 The factor and remainder theorems

The remainder theorem states that for any number aa, the remainder when we divide a polynomial (which can be written as f(x)f(x)) by (xa)(x-a) is equal to f(a)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)Q(x) and a remainder R(x)R(x) as a fraction of xax-a - as an equation, that is, that f(x)(xa)=Q(x)+R(x)xa\frac{f(x)}{(x-a)}=Q(x)+\frac{R(x)}{x-a}. If we multiply both sides by xax-a, then we obtain that f(x)=Q(x)(xa)+R(x)f(x)=Q(x)(x-a)+R(x). We can then set x=ax=a, which gives that f(a)=R(a)f(a)=R(a). Note that Q(x)(aa)=0Q(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 (xa)(x-a) divides f(x)f(x) with no remainder, then by the remainder theorem, this means that f(a)=0f(a)=0 and so aa is a root as well as a factor. Roots are factors, and factors are roots.

4.6.1 An example


: Given that (x1)(x-1) is a factor of P(x)=5x39x2+2x+aP(x)=5x^{3}-9x^{2}+2x+a, find the value of aa.


: We can directly apply the factor theorem here; as (x1)(x-1) is a factor of P(x)P(x), it must be the case that P(1)=0P(1)=0. With this information, we can form an equation where the only variable is aa.

P(1)\displaystyle P(1) =0\displaystyle=0 (4.65)
59+2+a\displaystyle 5-9+2+a =0\displaystyle=0 (4.66)
2+a\displaystyle-2+a =0\displaystyle=0 (4.67)
a\displaystyle a =0\displaystyle=0 (4.68)