Question

When 12x² + kx + 3 and 3x³ + (3k + 2)x² + 28x +17

Answer 1

Answer:

$\boxed{\sf \ \ \ k=5 \ \ \ }$

Step-by-step explanation:

Hello,

We know that for a real, the polynomial P(x) divided by (x-a) the remainder is P(a)

and as

$3x+2=03x=-2x=\dfrac{-2}{3}$

so we need

$12(\dfrac{-2}{3})^2+k(\dfrac{-2}{3})+3=3(\dfrac{-2}{3})^3+(3k+2)(\dfrac{-2}{3})^2+28\dfrac{-2}{3}+17\\ \ multiply \ by \ 9\\ 12*4-6k+27=-8+4(3k+2)-28*2*3+17*9\\ -6k+48+27=-8+12k+8-168+153\\ -6k+75=12k-15\\ 18k=75+15=90\\ k = \dfrac{90}{18}=\dfrac{15}{3}$

hope this helps