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

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Answer 1 · 2016-02-25T10:37:54

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

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

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

Solve by factoring because it is factorable!

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

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

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

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

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

How do you find all the zeros of P(x)=x4−26x2+25P(x) = x^4 − 26x^2 + 25 ?