Review question

# What if the remainder when we divide by $x-k$ is $k$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8701

## Solution

1. Given that $f(x)=x^3+kx^2-2x+1$ and that when $f(x)$ is divided by $(x-k)$ the remainder is $k$, find the possible values of $k$.

Using the remainder theorem, we know that the remainder when we divide by $(x-k)$ is the same as $f(k)=2k^3-2k+1$.

So we need to find solutions for $2k^3-2k+1=k$, that is $2k^3-3k+1=0$.

It is easy to guess and check that $k=1$ is a solution, so $(k-1)$ is a factor; taking out a factor of $k-1$ gives $2k^3-3k+1=(k-1)(2k^2+2k-1).$ So either $k=1$ or $2k^2+2k-1=0$. Solutions for the latter are, using the quadratic formula, $k=\frac{-2 \pm \sqrt{2^2-4\times 2\times (-1)}}{2\times 2} =\frac{-1 \pm \sqrt{3}}{2}$

So the solutions are $k=1$, $k=\dfrac{-1 \pm \sqrt{3}}{2}$.

1. When the polynomial $p(x)$ is divided by $(x-1)$ the remainder is $5$ and when $p(x)$ is divided by $(x-2)$ the remainder is $7$. Given that $p(x)$ may be written in the form $(x-1)(x-2)q(x)+Ax+B,$ where $q(x)$ is a polynomial and $A$ and $B$ are numbers, find the remainder when $p(x)$ is divided by $(x-1)(x-2)$.

Using the remainder theorem again, the remainder when we divide $p(x)$ by $(x-1)$ is the same as $p(1)=A+B$, so $A+B=5$.

Likewise, the remainder when we divide $p(x)$ by $(x-2)$ is the same as $p(2)=2A+B$, so $2A+B=7$.

These simultaneous equations in $A$ and $B$ give as solutions $A=2$ and $B=3$.

The remainder when $p(x)$ is divided by $(x-1)(x-2)$ is $Ax+B$, since $(x-1)(x-2)$ is a factor of $(x-1)(x-2)q(x)$, and $Ax+B$ has a smaller degree than $(x-1)(x-2)$. So the remainder is $2x+3$.