How do you factor completely x^4 + 5x^2 - 36?

1 Answer
George C.
May 1, 2016

x^4+5x^2-36

=(x^2+9)(x-2)(x+2)

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

Explanation:

Use the difference of squares identity once or twice:

a^2-b^2 = (a-b)(a+b)

We can treat this as a quadratic in x^2 first to find:

x^4+5x^2-36

=(x^2+9)(x^2-4)

=(x^2+9)(x^2-2^2)

=(x^2+9)(x-2)(x+2)

The remaining quadratic factor has no linear factors with Real coefficients, but it can also be treated as a difference of squares using 9 = -(3i)^2 as follows:

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

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

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