# How do you use find the zeroes of f(x)=x^4-4x^2-45?

Aug 13, 2016

$f \left(x\right)$ has zeros: $3 , - 3 , \sqrt{5} i , - \sqrt{5} i$

#### Explanation:

Treat as a quadratic in ${x}^{2}$ then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Note that $9 \cdot 5 = 45$ and $9 - 5 = 4$

Hence:

${x}^{4} - 4 {x}^{2} - 45$

$= \left({x}^{2} - 9\right) \left({x}^{2} + 5\right)$

$= \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {\left(\sqrt{5} i\right)}^{2}\right)$

$= \left(x - 3\right) \left(x + 3\right) \left(x - \sqrt{5} i\right) \left(x + \sqrt{5} i\right)$

Hence zeros:

$3 , - 3 , \sqrt{5} i , - \sqrt{5} i$