Rewriting $f(x) = 2x^3+7x^2-14x-40$ as the product of three linear factors

Question

I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics" and I am stuck on factorizing a particular polynomial to a product of three linear factors. Through the trial-and-error method, I've gotten that the factors are $(x+2)$ and $(x+4)$ (although, it is my understanding that only quadratic polynomials are candidate for looking for multiple factors in this way, in case the factor isn't already given).

I've been banging my head against this, and gotten a different result every time, and the best I could come up with is: $$(x+1)(x+4)(2x+5).$$

It just seems that every time I try it out, a different result comes out.

Can anybody help me out or give me some pointers? Generally I have this down pretty well, but this particular one is doing my head in.

Answer 1

It's actually $$2x^3+7x^2-14x-40=(2x\color{red}-5)(x+\color{red}2)(x+4)$$

Once you've identified that $(x+2)$ and $(x+4)$ are factors, use the polynomial long division to obtain the last factor.

Answer 2

Observe that $$2x^3+7x^2-14x-40 = (x+4) (2x^2-x-10).$$

Can you proceed now?

Answer 3

A general way to find such a factorization for a polynomial $p(x)$ of third degree is:

1) Try to find "by hand" a root of the polynomial: let's call it $a$.

2) Since $a$ is a root, (x-a) divides your polynomial. Hence you can write it as $p(x)=(x-a)(ax^2+bx+c)$. Now all you have to do is finding the roots of the second degree polynomial we just found.