# How do you find all the zeros of P(x) = x^4 − 26x^2 + 25 ?

$x = \pm 1$ and $x = \pm 5$

#### Explanation:

to find the zeros of $P \left(x\right) = {x}^{4} - 26 {x}^{2} + 25$, set $P \left(x\right) = 0$

it becomes ${x}^{4} - 26 {x}^{2} + 25 = 0$

Solve by factoring because it is factorable!

${x}^{4} - 26 {x}^{2} + 25 = 0$

$\left({x}^{2} - 1\right) \left({x}^{2} - 25\right) = 0$
set both factors to zero

$\left({x}^{2} - 1\right) = 0$ and $\left({x}^{2} - 25\right) = 0$
solve for x:

$x = - 1$ and $x = + 1$, and $x = - 5$ and $x = + 5$

God bless you...I hope the explanation helps...