Question

In expansion of (2x+1)(x^2+px+4), the coefficient of x is twice the coefficient of x^2.

Answer 1

You're probably supposed to find the value of p and not x, which is impossible to do.

Expanding gives

$(2x+1)(x^2+px+4)=2x^3+(1+2p)x^2+(8+p)x+4$

We're told that

$8+p=2(1+2p)$

which we can use to solve for p:

$8+p=2+4p$

$6=3p$

$\boxed{p=2}$