Find the remainder when $x^{4} + 4 x^{3} - 5 x^{2} - 6 x + 7$ is divided by $x + 3$ .

Open in App

Solution

Let $f (x) = x^{4} + 4 x^{3} - 5 x^{2} - 6 x + 7$ .
Let $R$ be the remainder obtained when $f (x)$ is divided by $x + 3$ .

Equate $x + 3$ to $0$ and put the value of $x$ in $f (x)$ .

$x + 3$ =0
$⟹ x = - 3$

$R = (- 3)^{4} + 4 (- 3)^{3} - 5 (- 3)^{2} - 6 (- 3) + 7$
$= 81 - 108 - 45 + 18 + 7$
$= - 47$

Thus required remainder is $- 47$

Suggest Corrections

Similar questions

p (x) = x^{4} + 4 x^{3} + 7 x^{2} - 4

x + 1

x^{4} + 4 x^{3} + 2

x + 1

x^{3} + 5 x^{2} + a x - 7

x - 3

47

a

4 x^{3} + 5 x - 10

x - 3

4 x^{3} - 3 x^{2} + 2 x - 4

x + 3

Join BYJU'S Learning Program