# How do you find all the zeros of f(x) = x⁴ - 10x² + 24?

Jul 20, 2018

$x = \pm 2 \text{ }$ and $\text{ } x = \pm \sqrt{6}$

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} - 10 {x}^{2} + 24$

Note that this quartic contains only terms of even degree, so we can start to factor it as a quadratic in ${x}^{2}$.

Note also that $10 = 4 + 6$ and $24 = 4 \cdot 6$

Hence we find:

${x}^{4} - 10 {x}^{2} + 24 = \left({x}^{2} - 4\right) \left({x}^{2} - 6\right)$

$\textcolor{w h i t e}{{x}^{4} - 10 {x}^{2} + 24} = \left({x}^{2} - {2}^{2}\right) \left({x}^{2} - {\left(\sqrt{6}\right)}^{2}\right)$

$\textcolor{w h i t e}{{x}^{4} - 10 {x}^{2} + 24} = \left(x - 2\right) \left(x + 2\right) \left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right)$

Hence zeros:

$x = \pm 2 \text{ }$ and $\text{ } x = \pm \sqrt{6}$

graph{x^4-10x^2+24 [-5.067, 4.933, -6, 32]}

$x = \setminus \pm 2 , \setminus \pm \setminus \sqrt{6}$

#### Explanation:

Given function:

$f \left(x\right) = {x}^{4} - 10 {x}^{2} + 24$

The zeroes of above bi-quadratic polynomial is given by setting $f \left(x\right) = 0$ as follows

${x}^{4} - 10 {x}^{2} + 24 = 0$

${x}^{4} - 6 {x}^{2} - 4 {x}^{2} + 24 = 0$

${x}^{2} \left({x}^{2} - 6\right) - 4 \left({x}^{2} - 6\right) = 0$

$\left({x}^{2} - 6\right) \left({x}^{2} - 4\right) = 0$

${x}^{2} - 6 = 0 \setminus \setminus \mathmr{and} \setminus \setminus {x}^{2} - 4 = 0$

${x}^{2} = 6 \setminus \setminus \mathmr{and} \setminus \setminus {x}^{2} = 4$

$x = \setminus \pm \setminus \sqrt{6} \setminus \setminus \mathmr{and} \setminus \setminus x = \setminus \pm 2$

$x = \setminus \pm 2 , \setminus \pm \setminus \sqrt{6}$