# How do you find all the zeros of x^5-2x^4-4x^3-10x^2-21x-12?

May 24, 2016

${\left(x + 1\right)}^{2} \left(x - 4\right) \left(x + i \sqrt{3}\right) \left(x + i \sqrt{3}\right)$

#### Explanation:

The polynomial has constant coefficient $- 12$ which is the product of its roots.

Trying the sequence of $- 12$ factors given by

$\left\{\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12\right\}$

we get

$p \left(- 1\right) = p \left(4\right) = 0$

then at last we have found two roots ( keep in mind that some of this roots can be multiple). So we can write

$p \left(x\right) = {x}^{5} - 2 {x}^{4} - 4 {x}^{3} - 10 {x}^{2} - 21 x - 12$

and also

$p \left(x\right) - \left(x + 1\right) \left(x - 4\right) \left({x}^{3} + a {x}^{2} + b x + c\right) = 0$ for all $x$

Taking the coefficients equal to zero we have

$\left(\begin{matrix}12 - 4 c = 0 \\ 21 - 4 b - 3 c = 0 \\ 10 - 4 a - 3 b + c = 0 \\ - 3 a + b = 0 \\ - 1 + a = 0\end{matrix}\right)$.

Solving for $a , b , c$ we get

$\left(a = 1 , b = 3 , c = 3\right)$

Focusing now in the roots of

${x}^{3} + {x}^{2} + 3 x + 3 = 0$

and having in mind the ideas about the composition of the constant term we find that $x = - 1$ is a root so we can write

${x}^{3} + {x}^{2} + 3 x + 3 = \left(x + 1\right) \left({x}^{2} + d x + e\right)$

proceeding as before we find $d = 0 , e = 3$. Putting all together we have

$p \left(x\right) = {\left(x + 1\right)}^{2} \left(x - 4\right) \left(x + i \sqrt{3}\right) \left(x + i \sqrt{3}\right)$
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