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

Mar 9, 2016

Zeros of $f \left(x\right)$ are $\left\{- 1 , 1 , i \sqrt{7} , - i \sqrt{7}\right\}$

#### Explanation:

To find all the zeros of $f \left(x\right) = {x}^{4} + 6 {x}^{2} - 7$ means to find the values of $x$ that make $f \left(x\right) = 0$. In other words, it means finding solution of the equation ${x}^{4} + 6 {x}^{2} - 7 = 0$ and for this we should factorize ${x}^{4} + 6 {x}^{2} - 7$.

For this, let us split middle term in to two components $7 {x}^{2}$ and $- {x}^{2}$. Then ${x}^{4} + 6 {x}^{2} - 7 = 0$ becomes

${x}^{4} + 7 {x}^{2} - {x}^{2} - 7 = 0$ i.e. x^2(x^2+7)-1((x^2+7)=0 or

$\left({x}^{2} - 1\right) \left({x}^{2} + 7\right) = 0$. Note that $\left({x}^{2} - 1\right)$ can be further factorized into $\left(x + 1\right) \left(x - 1\right)$. Hence, ${x}^{4} + 6 {x}^{2} - 7 = 0$ can be written as

$\left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 7\right) = 0$ and hence

Rational zeros of $f \left(x\right)$ are $\left\{- 1 , 1\right\}$.

Further if we include complex numbers in domain of $x$, from $\left({x}^{2} + 7\right) = 0$, we get $\left(x - i \sqrt{7}\right) \left(x + i \sqrt{7}\right)$ or $x = \left\{i \sqrt{7} , - i \sqrt{7}\right\}$