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2 years ago

9

How do i factor x^4+6x^2-7 completly?

2 answers:

liq [111]2 years ago

8 0

The factoring can be done similarly to a quadratic equation thanks to x^4 being the square value of x^2.

<span>x^4 + 6x^2 - 7
x^4</span><span> - x^2</span> + 7x^2 - 7
(x^4 - x^2) + (<span>7x^2 - 7)
</span>x^2(x^2 - 1) + 7(<span>x^2 - 1)
</span>(x^2 + 7)(x^2 - 1)
<span>(x^2 + 7)(x - 1)(x + 1)
</span>
Factored completely we get: <span>(x^2 + 7)(x - 1)(x + 1)</span>

Nina [5.8K]2 years ago

4 0

Just make it into ax^2+bx+c form
remember that
(x^2)^2=x^4
(x^2)=x^2

rewriete in normal form (for people who are still understnain stuf)

(x^2)^2+6(x^2)-7
now find what 2 numbers multiply to get -7 and add to 6
numbers are -1 and 7
((x^2)-1)((x^2)+7)
(x^2-1)(x^2+7)
diffrence of 2 perfect squaer
a^2-b^2=(a-b)(a+b)
(x)^2-(1)^2=(x-1)(x+1)

totla factored form is
(x-1)(x+1)(x^2+7)

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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