Finding the roots of $x^5+x^2-9x+3$

Question

I have to find all the roots of the polynomial $x^5+x^2-9x+3$ over the complex. To start, I used Wolfram to look for a factorization and it is $$x^5+x^2-9x+3=(x^2 + 3) (x^3 - 3 x + 1)$$

I can take it from here employing the quadratic and cubic formula, but how can I get the same answer w/o using any software?

Answer 1

Here's one way to think about it:

$$\begin{align}x^5+x^2-9x + 3 &= x^5-9x+x^2+3\\ &= x(x^4-9)+x^2+3 \\ &= x(x^2-3)(x^2+3)+x^2+3 \\ &= (x^2+3)(x(x^2-3)+1) \\ &= (x^2+3)(x^3-3x+1). \end{align}$$

In words, the trick is to recognize that $x^5-9x$ is "almost" a difference of squares (i.e., up to the factor of $x$ multiplying it). Then everything falls into place.

Answer 2

It can be factorize by adding and subtracting $3x^3$

$$x^5+x^2-9x+3=x^5-3x^3+x^2+3x^3-9x+3=x^2(x^3-3x+1)+3(x^3-3x+1)=(x^2 + 3) (x^3 - 3 x + 1)$$