Answer:
[tex]\boxed{\sf \ \ \ k=5 \ \ \ }[/tex]
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
[tex]3x+2=0<=>3x=-2<=>x=\dfrac{-2}{3}[/tex]
so we need
[tex]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}[/tex]
hope this helps
