Find the value of $k$ such that $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$.

Question

$p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$.

(a) -1
(b) 1
(c) 2
(d) -2

I am getting the value of k: $-17/29$ after equating the remainders.

$p(x)= kx^3 + 4x^2 + 3x - 4/(x – 3)$: remainder= $(30k+32)$
$q(x)= x^3 - 4x + k/(x – 3)$: remainder= $(k+15)$

So, $30k+32=k+15$
=> $k=(-17/29)$

Any help would be appreciated. :)

Answer 1

Your first remainder is wrong since $$\frac{k x^3+4 x^2+3 x-4}{x-3}=k x^2+(3 k+4) x+3 (3 k+5)+\frac{27 k+41}{x-3}$$ I am sure that you can take from here.

Answer 2

By the polynomial remainder theorem, the remainder of $f(x)$ when divided by $x - a$ is equal to $f(a)$. So we have

$$p(3) = q(3)$$ $$27k + 41 = 15 + k$$ $$k = -1$$

Answer 3

Find the value of $k$ if $x^3-4x^2+5x-k$ is completely divisible by $x-4$.