If the quotient obtained on dividing $(8 x^{4} - 2 x^{2} + 6 x - 7)$ by $(2 x + 1)$ is $(4 x^{3} + p x^{2} - q x + 3)$ then find $p, q$ and also the remainder.

Open in App

Solution

The division of $(8 x^{4} - 2 x^{2} + 6 x - 7)$ and $(2 x + 1)$ is shown above:

Therefore, from the division, we get that the quotient is $(\frac{1}{2} (8 x^{3} - 4 x^{2} + 6))$ that is $(4 x^{3} - 2 x^{2} + 3)$ and the remainder is $- 10$ .

But it is also given that the quotient is $(4 x^{3} + p x^{2} - q x + 3)$ , thus on comparing the coefficients of both the quotients $(4 x^{3} - 2 x^{2} + 3)$ and $(4 x^{3} + p x^{2} - q x + 3)$ , we get

$p = - 2$ and
$q = 0$

Hence, $p = - 2$ , $q = 0$ and the remainder is $- 10$ .

Suggest Corrections

Similar questions

8 x^{4} - 2 x^{2} + 6 x - 7

2 x + 1

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

q - p

(4 x^{3} - 2 x^{2} + 6 x + 7) \div (3 + 2 x)

(8 x^{4} - 2 x^{2} + 6 x - 5) \div (4 x + 1)

(8 x^{4} - 2 x^{2} + 6 x - 5) \div (4 x + 1)

g (x)

p (x) = 4 x^{3} + 2 x^{2} - 10 x + 2

g (x)

(2 x^{2} + 4 x + 1)

5

Join BYJU'S Learning Program