# How do you find all the zeros of (3x^6+x^2-4)(x^3+6x+7)?

Jul 16, 2016

This product has zeros:

• $1$

• $- 1$ (with multiplicity $2$)

• $\pm \left(\sqrt{\frac{4 \sqrt{3} - 3}{12}}\right) \pm \left(\sqrt{\frac{4 \sqrt{3} + 3}{12}}\right) i$

• $\frac{1}{2} \pm \frac{3 \sqrt{3}}{2} i$

#### Explanation:

The zeros of $\left(3 {x}^{6} + {x}^{2} - 4\right) \left({x}^{3} + 6 x + 7\right)$ are all the values of $x$ which are zeros of $\left(3 {x}^{6} + {x}^{2} - 4\right)$ or $\left({x}^{3} + 6 x + 7\right)$.

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Zeros of $\boldsymbol{3 {x}^{6} + {x}^{2} - 4}$

Note that all of the degrees are even and the sum of the coefficients is $0$.

Hence this sextic has zeros $\pm 1$ and quadratic factor:

$\left(x - 1\right) \left(x + 1\right) = {x}^{2} - 1$

We find:

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

Treating the remaining quartic factor as a quadratic in ${x}^{2}$ and using the quadratic formula, we find:

${x}^{2} = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(3\right) \left(4\right)}}{2 \cdot 3}$

$= \frac{- 3 \pm \sqrt{9 - 48}}{6}$

$= \frac{- 3 \pm \sqrt{- 39}}{6}$

$= - \frac{1}{2} \pm \frac{\sqrt{39}}{6} i$

The square roots of $a \pm b i$ are:

$\pm \left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) \pm \left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i$

So, putting $a = - \frac{1}{2}$ and $b = \frac{\sqrt{39}}{6}$ we find:

${a}^{2} + {b}^{2} = {\left(- \frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{39}}{6}\right)}^{2} = \frac{1}{4} + \frac{39}{36} = \frac{48}{36} = \frac{4}{3}$

So:

$\sqrt{{a}^{2} + {b}^{2}} = \sqrt{\frac{4}{3}} = \sqrt{\frac{4}{9} \cdot 3} = \frac{2 \sqrt{3}}{3}$

Hence:

$x = \pm \left(\sqrt{\frac{\frac{2 \sqrt{3}}{3} - \frac{1}{2}}{2}}\right) \pm \left(\sqrt{\frac{\frac{2 \sqrt{3}}{3} + \frac{1}{2}}{2}}\right) i$

$= \pm \left(\sqrt{\frac{4 \sqrt{3} - 3}{12}}\right) \pm \left(\sqrt{\frac{4 \sqrt{3} + 3}{12}}\right) i$

$\textcolor{w h i t e}{}$
Zeros of $\boldsymbol{{x}^{3} + 6 x + 7}$

Notice that $- 1$ is a zero, so $\left(x + 1\right)$ is a factor and we find:

${x}^{3} + 6 x + 7 = \left(x + 1\right) \left({x}^{2} - x + 7\right)$

The remaining quadratic factor has zeros given by the quadratic formula:

$x = \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(7\right)}}{2 \cdot 1}$

$= \frac{1 \pm \sqrt{- 27}}{2}$

$= \frac{1}{2} \pm \frac{3 \sqrt{3}}{2} i$