Factorization of Polynomial without Polynomial Division

Question

How do I factorize $2p^3 -7p^2+8p+6$ without polynomial division? I know, that the solution is $(2p + 1)(p^2 - 4p + 6)$.

Answer 1

Hint: $$2p^3 -7p^2+8p+6=2p^3+p^2-8p^2-4p+12p+6$$

Answer 2

A real cubic polynomial has at least one real root. Fetch!

Classically, try some easy values first. So what if $p=0$? $P(0) = 6$. And $p=1$? $P(1) = 9$. Unlucky. So $p=-1$? $P(-1) = -11$. So there is (a least) a root between $-1$ and $0$.

Keep it simple. Try dichotomy. $P(-1/2) = 0$? Bingo! From now try to find $(p+1/2)$ terms. You are quite safe now. You can collect $2$-ratios (8 and 4, 6 and 3) between some $x^p$ and $x^{p-1}$ coefficients.

First, $8p+6=(2\times 4)p+(4) +2 = 8(p+1/2)+2$. Then $2p^3−7p^2=(2\times 1)p^3+(1)p^2−8p^2 = (p+1/2)(2p^2)−8p^2$.

So you already have: $$2p^3−7p^2+8p+6 = (p+1/2)(2p^2)−8p^2+ 8(p+1/2)+2 = (p+1/2)(2p^2+8)-8p^2+2 \,. $$

As you know that $-1/2$ is a root, you recognize that $-8p^2+2=-8(p+1/2)(p-1/2)$. Hence you have: $$(p+1/2)(2p^2+8)-8p^2+2 = (p+1/2)(2p^2+8-8(p-1/2)) = (p+1/2)(2p^2-8p+12)\,.$$

You are almost done (by a factor of $2$).