How do you find all the zeros of $f (x) = x^{4} + - 7 x^{2} - 144$ $f(x) = x^4+ -7x^2 -144$ ?

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Answer 1 · 2016-03-11T02:47:39

$x = \pm 4, \pm 3 i$ $x = +-4, +-3i$

$f (x) - x^{4} - 7 x^{2} - 144$ $f(x) - x^4 - 7x^2 - 144$ (Since $+ - = -$ $"+ -" = -$ )

Let $z = x^{2}$ $z = x^2$
$f (x) = z^{2} - 7 z - 144$ $f(x) = z^2 - 7z - 144$

To find all zeros set $f (x) = 0$ $f(x) = 0$
$z^{2} - 7 z - 144 = 0$ $z^2 - 7z - 144 = 0$

Factorizing:
$(z + 9) (z - 16) = 0$ $(z+9) (z-16) = 0$

Therefore $z = - 9 or 16$ $z = -9 or 16$

But $z = x^{2} \to x = \pm \sqrt{z}$ $z = x^2 -> x = +- sqrt(z)$

With $z = - 9, x = \pm 3 i$ $z = -9, x = +-3i$
And with $z = 16, x = \pm 4$ $z = 16, x = +-4$

How do you find all the zeros of f(x)=x4+−7x2−144f(x) = x^4+ -7x^2 -144?