Question

(8x^4-2x^2+6x-7) / (2x+1) = 4x^3+px^2-qx+3 find p,q and remainder

Answer 1

Answer:

p=-2

q=0

Remainder is -10

Step-by-step explanation:

Given:

$\frac{8x^4-2x^2+6x-7}{2x+1}=4x^3+px^2-qx+3$

$\implies\,8x^4-2x^2+6x-7=(4x^3+px^2-qx+3)(2x+1)$

Equating coefficient of x² and x on both sides, we get

$p-2q=-2.........(1)$ and

$-q+6=6$

$\implies\bf\,q=0$

put q=0 in (1) we get

$\bf\,p=-2$

Instead of using long division method, I have applied remainder theorem to find the remainder  

Remainder theorem:

$\boxed{\bf\text{The remainer when P(x) is divided by (ax+b) is }P(-b/a)}$

$\text{The remainder when }8x^4-2x^2+6x-7\text{ is divided by }(2x+1)\text{ is}$

$8(\frac{-1}{2})^4-2(\frac{-1}{2})^2+6(\frac{-1}{2})-7$

$=8(\frac{1}{16})-2(\frac{1}{4})-6(\frac{1}{2})-7$

$=\frac{1}{2}-\frac{1}{2}-3-7$

$=-10$