How do you find all the zeros of $f (x) = x^{4} - 5 x^{2} - 36$ $f(x)=x^4-5x^2-36$ ?

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How do you find all the zeros of $f (x) = x^{4} - 5 x^{2} - 36$ $f(x)=x^4-5x^2-36$ ?

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Answer 1 · 2016-08-14T13:27:12

$f (x)$ $f(x)$ has zeros $\pm 3$ $+-3$ and $\pm 2 i$ $+-2i$

Explanation:

Note that $9 \cdot 4 = 36$ $9*4 = 36$ and $9 - 4 = 5$ $9-4 = 5$ .

So we find:

$x^{4} - 5 x^{2} - 36$ $x^4-5x^2-36$

$= (x^{2} - 9) (x^{2} + 4)$ $=(x^2-9)(x^2+4)$

$= (x^{2} - 3^{2}) (x^{2} - {(2 i)}^{2})$ $=(x^2-3^2)(x^2-(2i)^2)$

$= (x - 3) (x + 3) (x - 2 i) (x + 2 i)$ $=(x-3)(x+3)(x-2i)(x+2i)$

Hence zeros: $\pm 3$ $+-3$ and $\pm 2 i$ $+-2i$

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How do you find all the zeros of $f (x) = x^{4} - 5 x^{2} - 36$ $f(x)=x^4-5x^2-36$ ?

1 Answer

Explanation:

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How do you find all the zeros of f(x)=x4−5x2−36f(x)=x^4-5x^2-36?

1 Answer

Explanation:

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How do you find all the zeros of $f (x) = x^{4} - 5 x^{2} - 36$ $f(x)=x^4-5x^2-36$ ?