Question

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

(i) x - 1 (ii) x + 2 (iii) $x+\frac{1}{2}$

Answer 1

Let p(x) = $4x^3-3x^2+2x-4$

(i) When p(x) is divided by (x - 1), then by remainder theorem, the required remainder will be p(1).
$p\left(1\right)=4(1{)}^3-3(1{)}^2+2\left(1\right)-4$
= 4 × 1 - 3 × 1 + 2 × 1 - 4
= 4 - 3 + 2 - 4 = -1
(ii) When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p (-2).
$p(-2)=4(-2{)}^3-3(-2{)}^2+2(-2)-4$
= 4 × (-8) - 3 × 4 - 4 - 4
= - 32 - 12 - 8 = - 52
(iii) When p(x) is divided by, $x+\frac{1}{2}$ then by remainder theorem, the required remainder will be.
$p(-\frac{1}{2})=4{(-\frac{1}{2})}^3-3{(-\frac{1}{2})}^2+2(-\frac{1}{2})-4$

$=4\times (-\frac{1}{8})-3\times \frac{1}{4}-2\times \frac{1}{2}-4$

= $-\frac{1}{2}-\frac{3}{4}-1-4=\frac{1}{2}-\frac{3}{4}=5$

$=\frac{-2-3-20}{4}=\frac{-25}{4}$