Verified answer

u^4 - 3u^2 - 4

= u^4 + u^2 - 4u^2 - 4

= u^2(u^2 + 1) - 4(u^2 + 1)

= (u^2 + 1)(u^2 - 4)

= (u^2 + 1)(u + 2)(u - 2) Ans.

u⁴−3u²−4

=u⁴−4u² +u²−4

= u²(u²−4) +(u²−4)

=(u²−4) (u²+1)

=(u+2)(u−2) (u²+1)

(u^2)^2 - 3(u^2) - 4

You can use quadratic formula on this, using u^2 as your "x".

u^2 = [3 +or- sqrt(9 + 16)] / 2

u^2 = [3 +or- 5] / 2

u^2 = -1 or 4

u = √(-1) or √(4)

Since square roots of negatives aren't possible, u must equal √(4) = 2.

(it also equals i, if you want to do complex/imaginary numbers.)

(u^2 - 4)(u^2 + 1)

v = u²

u^4 - 3u^2 - 4 = v² - 3v - 4 = (v - 4)(v + 1)

ASSUMING the original equation was equal to 0

(v - 4)(v + 1) = 0

For that to be true, v = 4 or -1

Therefore

u² = 4

u = ±√4 = ±2

u² = -1

u = ±√(-1) = ±i

u = 2, -2, i or -i

***************

Oops, just now saw you are trying to factor

jumping back to

(v - 4)(v + 1)

(u² - 4)(u² + 1) = (u + 2)(u - 2)(u + i)(u - 1) using difference of two squares twice

are you trying to solve for u?

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