Answer:
[tex]c=8[/tex]
Step-by-step explanation:
When we plug in the solution [tex](x+2)[/tex], we will substitute [tex]x=-2[/tex], as [tex]-2+2=0[/tex].
To be divisible, a remainder of zero must be yielded.
Let's substitute [tex]x=-2[/tex] to see what we need to make [tex]c[/tex]:
[tex]p(-2)=4\cdot(-2)^3+c\cdot (-2)^2+(-2)+2,\\p(-2)=-32+4c[/tex]
Remember that we want a remainder of zero:
[tex]-32+4c=0,\\4c=32,\\c=\boxed{8}[/tex]
