How do you factor #3x^5-2x^2+1#?

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Answer 1 · 2018-01-17T21:46:42

This quintic has no factorisation expressible in terms of #n#th roots.

Given:

#3x^5-2x^2+1#

In common with most quintics, this has no zeros expressible in terms of #n#th roots - i.e. square roots, cube roots, etc.

It has one real zero and four non-real complex zeros occurring as two complex conjugate pairs.

In theory, it has a factorisation with real coefficients of the form:

#3x^5-2x^2+1 = 3(x+a)(x^2+bx+c)(x^2+dx+e)#

In practice, we can only find numerical approximations to the coefficients.

Using real and complex coefficients, it has a factorisation of the form:

#3x^5-2x^2+1 = 3(x-a)(x-b)(x-c)(x-d)(x-e)#

where:

#a ~~ -0.610535#

#b ~~ 0.711015 + 0.257119i#

#c ~~ 0.711015 - 0.257119i#

#d ~~ -0.405747 + 0.889067i#

#e ~~ -0.405747 - 0.889067i#