Question

A cubic polynomial f(x) has a factor x and f(x) – f(x - 1) = 3x^2 - 5x. By substituting
suitable values of x, show that f(x) has a factor x + 1 but leaves a remainder of
-8 when divided by x + 2. What is the remainder when f(x) is divided by x – 1?
Hence or otherwise, find f(x).​

Answer 1

Step by Step:

We know that f has the factor x. Then f(x)=0. It follows that

$f(0)-f(0-1)=-f(-1)=3(0)^2-5(0) = 0 \implies f(-1) = 0$.

As f(-1)=0, then x+1 is a factor of f.

$f(-1)-f(-1 -1)) = f(-1) -f(-2) = 0 -f(-2) = -f(-2)\\\implies -f(-2)=3(-1)^2-5(-1) = 3+5 = 8\\\implies f(-2) = -8$

Hence, f(x) has a remainder of -8 when divided by x+2, by the remainder thereom (also known as little Bézout's theorem).

Similarily,

$f(1) - f(0) = f(1) = 3-5 = -2$,

Consequently, the remainder when dividing by x-1 is -2 by the remainder theorem.

We now find f(x). We know that

$f(x)=(kx+a)x(x+1)=(kx + a)(x^2+x)$.

As f(x) has remainder -8 when divided by x+2 and remainder -2 when divided by x-1, then we have by the remainder theorem that

$f(-2)=(-2k+a)(4-2)=-4k+2a = -8,$

and

$f(1)=(k+a)(1+1)=2k+2a=-2$.

This gives us a system of linear equations which we can solve:

$2k+2a = -2 \implies k = -a-1$

$-4k+2a = -4(-a-1)+2a = 6a+4 =-8 \implies 6a = -12$.

Then a = -2, which gives k = 1. The function then becomes,

and we're done!