# How do you find all the zeros of f(x)=(6x^7+4x^2+6)(x^4+5x-6) with its multiplicities?

Jun 24, 2016

Find zeros of $\left({x}^{4} + 5 x - 6\right)$ algebraically and approximate zeros of $\left(6 {x}^{7} + 4 x + 6\right)$ numerically.

#### Explanation:

$f \left(x\right) = \left(6 {x}^{7} + 4 {x}^{2} + 6\right) \left({x}^{4} + 5 x - 6\right)$

The zeros of $f \left(x\right)$ are the zeros of $\left(6 {x}^{7} + 4 {x}^{2} + 6\right) = 2 \left(3 {x}^{7} + 2 {x}^{2} + 3\right)$ and the zeros of $\left({x}^{4} + 5 x - 6\right)$

$\textcolor{w h i t e}{}$
Zeros of $\boldsymbol{{x}^{4} + 5 x - 6}$

Notice that the sum of the coefficients is $0$. That is: $1 + 5 - 6 = 0$.

Hence $x = \textcolor{b l u e}{1}$ is a zero and $\left(x - 1\right)$ a factor.

${x}^{4} + 5 x - 6 = \left(x - 1\right) \left({x}^{3} + {x}^{2} + x + 6\right)$

By the rational root theorem, the possible rational zeros of ${x}^{3} + {x}^{2} + x + 6$ are: $\pm 1 , \pm 2 , \pm 3$. Note that $\textcolor{b l u e}{- 2}$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{3} + {x}^{2} + x + 6 = \left(x + 2\right) \left({x}^{2} - x + 3\right)$

$x = \frac{1 \pm \sqrt{- 11}}{2} = \textcolor{b l u e}{\frac{1}{2} \pm \frac{\sqrt{11}}{2} i}$

$\textcolor{w h i t e}{}$
Zeros of $\boldsymbol{6 {x}^{7} + 4 {x}^{2} + 6 = 2 \left(3 {x}^{7} + 2 {x}^{2} + 3\right)}$

This septic has no simple factorisation and no rational zeros.

About the best you can do is use a numerical algorithm to find approximations for the $1$ Real and $6$ non-Real Complex zeros.

The Durand-Kerner method is straightforward to code and gives all of the approximate zeros at once.

$x \approx - 1.08646$

$x \approx - 0.539746 \pm 0.832444 i$

$x \approx 0.15543 \pm 0.896241 i$

$x \approx 0.927548 \pm 0.519443 i$