How do you factor #x^5-5x^3-36x# completely?

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Answer 1 · 2015-02-05T17:17:16

We need to start out by factoring out any common factors. We notice that we can factor out x.

#x(x^4-5x^2-36)#

Note that #x^4=(x^2)^2#

#x((x^2)^2-5x^2-36)#

Let #r=x^2#

#x(r^2-5r-36)#

Now factor #(r^2-5r-36)#

#x(r-9)(r+4)#

Replace #r# with #x^2#

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

We should recognize #(x^2-9)# as a difference of squares.

Leaving me with

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