How do you factor #y= x^4 + 5x^2 - 6# ?

2 Answers
Dec 5, 2015

Answer:

#y=(x^2+6)(x+1)(x-1)#

Explanation:

Given #y=x^4+5x^2-6#

We notice that all terms containing the variable #x# have #x^2# as a factor,
so to simplify things initially we will replace #x^2# with #z#

#color(white)("XXX")y=z^2+5z-6#
which can easily be factored as:
#color(white)("XXX")y=(z+6)(z-1)#

Restoring #x^2# back in place of #z#
#color(white)("XXX")y=(x^2+6)(x^2-1)#

The first factor, #(x^2+6)#, has no obvious sub-factors
but the second, #(x^2-1)#, is the difference of squares with sub-factors #(x+1)(x-1)#

giving
#color(white)("XXX")y=(x^2+6)(x+1)(x-1)#

Dec 5, 2015

Answer:

#y = (x^2 + 6)(x+1)(x-1)#

Explanation:

Perhaps it will help if I rewrite this expression for #y# in a slightly different way:

#y = (x^2)^2 + 5(x^2) - 6#

Let #u = x^2#. Then we have

#y = u^2 + 5u - 6#

This is something we can factor pretty easily.

#y = (u + 6)(u - 1)#

Just substitute back to get #y# in terms of #x#:

#y = (x^2 + 6)(x^2 - 1)#

And now, the last step is to factor the second bit which is a difference of two squares:

#y = (x^2 + 6)(x - 1)(x+1)#