# How to factor this polynomial?

I was trying to factor this polynomial:

$$x^3 + x^2 - 16x + 20$$

There are four options in this question:

1. (a) It could be factored in the following form: $$(x^2 + b)(x+c)$$;
2. (b) It could be factored in the following form: $$(x+b)(x+c)(x+d)$$, assuming that $$b \neq c \neq d$$
3. (c) It could not be factored.
4. (d) It could be factored in the following form: $$(x+b)^2 (x+c)$$

Here's how I've tried to do it: I've tried to factor by grouping the x, therefore I've obtained: $$x(x^2 + x - 16) + 20$$. Now, I've put the $$x$$ and the $$20$$ together: $$(x+20)(x^2 + x - 16)$$. Then, I've tried to factor the second term: $$(x+20)(x-16)(x+1)$$. So, the answer would be "b", according to this algorithm.

I've completed the test (it's a simulation for the admission test I'm going to do), I submit the answers, and I've noticed that this question isn't correct.

• Doesn't it factor as $(x-2)^2(x+5)$? Nov 25, 2020 at 7:19
• You cannot factor out the $(x+20)$ as you have done. There is no common factor of $(x+20)$ between $x^2 + x -16$ and $20$. Hint on what to do: Guess at a root of the polynomial and perform synthetic division.
– Möb
Nov 25, 2020 at 7:19
• Theorem: every polynomial to be factorized in a homework question has small, integer roots :) Nov 25, 2020 at 15:42

As @Fernis pointed out in the comments,

You cannot factor out the $$(x+20)$$ as you have done. There is no common factor of $$(x+20)$$ between $$x^2+x−16$$ and $$20$$.

Using the Rational Root Theorem, you can know that the possible rational roots are $$\pm 1, \pm2, \pm4, \pm5, \pm10, \pm20$$.

Through inspection and polynomial/synthetic division, you can get $$(x-2)^2(x-5)$$, as @saulspatz said. Therefore, (d) is your answer.

The easiest to try is (d), because it says that the polynomial has a double root. We will look for a root of the derivative and check if it cancels the polynomial.

$$3x^2+2x-16=0\iff x=2\text{ or }x=-\dfrac83.$$

Now $$p(2)=0$$, bingo !

• Good solution, but not sure if OP understands calculus Nov 25, 2020 at 20:22

This one is easy to factor. Let's see how.

Start by plugging in $$x=0,1,-1,2$$ and so on.

You will find by inspection that $$x=2$$ is a zero of the polynomial. Therefore,$$(x-2)$$ is its factor.

Now factor the polynomial in such a way that $$(x-2)$$ gets common.

$$x^3+x^2-16x+20$$ $$=x^3-2x^2+3x^2-6x-10x+20$$ $$=x^2(x-2)+3x(x-2)-10(x-2)$$ $$=(x-2)(x^2+3x-10)$$ $$=(x-2)(x^2+5x-2x-10)$$ $$=(x-2)[x(x+5)-2(x+5)]$$ $$=(x-2)^2(x+5)$$