How do you factor #x^4 - 13x^2 + 4#?

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Answer 1 · 2017-05-25T20:02:47

#x^4-13x^2+4 = (x-3/2-sqrt(17)/2)(x-3/2+sqrt(17)/2)(x+3/2-sqrt(17)/2)(x+3/2+sqrt(17)/2)#

This quartic can be factored as a "quadratic in #x^2#" then taking square roots, but the resulting form of the answer is not immediately simple.

Alternatively, consider the following:

#(x^2-kx+2)(x^2+kx+2) = x^4+(4-k^2)x^2+4#

So putting #k=sqrt(17)#, we find:

#(x^2-sqrt(17)x+2)(x^2+sqrt(17)x+2) = x^4-13x^2+4#

Using the quadratic formula, the zeros of #x^2+-sqrt(17)x+2# are:

#(+-sqrt(17)+-sqrt(17-4(1)(2)))/(2*1) = +-sqrt(17)/2+-sqrt(9)/2 = +-sqrt(17)/2+-3/2#

So we find:

#x^4-13x^2+4 = (x-3/2-sqrt(17)/2)(x-3/2+sqrt(17)/2)(x+3/2-sqrt(17)/2)(x+3/2+sqrt(17)/2)#

Alternatively again, consider the following:

#(x^2-hx-2)(x^2+hx-2) = x^4-(4+h^2)x^2+4#

So putting #h=3#, we find:

#(x^2-3x-2)(x^2+3x-2) = x^4-13x^2+4#

Using the quadratic formula, the zeros of #x^2+-3x-2# are:

#(+-3+-sqrt(9-4(1)(-2)))/(2*1) = +-3/2+-sqrt(17)/2#

Hence the same linear factorisation as before.