Remainder theorem 10 pts

The polynomial 2*x^2 + 6*x +3 has the same remainder when divided by x + p or by x - 2q , where p does not equal to -2q. Find the value of p - 2*q.

please help me with the workings

Remainder Theorem:
The remainder of the polynomial f(x) upon division by (x - a) is f(a).

Applying this, to f(x) = 2x^2 + 6x + 3:
The remainder of f(x) / (x - (-p)) equals f(-p) = 2p^2 - 6p + 3.
The remainder of f(x) / (x - 2q) equals f(2q) = 8q^2 + 12q + 3.

Since the reminders are to be equal, we obtain
2p^2 - 6p + 3 = 8q^2 + 12q + 3.

Simplifying this equation:
2p^2 - 6p = 8q^2 + 12q
==> 2(p^2 - 3p) = 2(4q^2 + 6q)
==> p^2 - 3p = 4q^2 + 6q
==> (p^2 - 4q^2) - (3p + 6q) = 0
==> (p - 2q)(p + 2q) - 3(p + 2q) = 0, factoring by grouping
==> ((p - 2q) - 3) (p + 2q) = 0

Since p ≠ -2q, we have p + 2q ≠ 0.
Hence, dividing both sides by p + 2q yields p - 2q - 3 = 0
==> p - 2q = 3.

I hope this helps!

If you long divide f(x)=2x^2+6x+3 by x+p
the remainder is 2p^2-6p+3.
If you long divide f(x) by x-2q the remainder
is 8q^2+12q+3.
Hence you want

2p^2-6p+3=8q^2+12q+3.
Cancelling the "3's" and dividing by 2 and rearranging

p^2-4q^2=3p+6q or

(p+2q)(p-2q)=3(p+2q).

Since p+2q is given as nonzero this implies

p-2q=3.

Hope that helps - i couldn't figure an easy
way to represent the long division. It works
just like long dividing two actual numbers.