Question

Find the remainders when $x^4-3x^3+4x^2-x+4$ is divided by

1) $x+1$

2) $x-\frac{1}{4}$
[3 MARKS]

Answer 1

Concept : 1 Mark
Application : 2 Marks

We know that,

When a polynomial $f\left(x\right)$ is divided by $(x-a)$ the remainder is $f\left(a\right)$

[Remainder theorem]

1) When $x^4-3x^3+4x^2-x+4$ is divided by $x+1$

$\Rightarrow (x+1)=0\Rightarrow x=-1$

$\therefore f(-1)=1+3+4+1+4=13$

2) When $x^4-3x^3+4x^2-x+4$ is divided by $x-\frac{1}{4}$

$\Rightarrow x-\frac{1}{4}=0\Rightarrow x=\frac{1}{4}$

$f\left(\frac{1}{4}\right)=\frac{1}{256}+(-\frac{3}{64})+\frac{1}{4}-\frac{1}{4}+4=\frac{1013}{256}$