How do you factor the expression #x^4-13x^2+36#?

2 Answers
Feb 8, 2016

Answer:

#x^4-13x^2+36=(x-3)(x+3)(x-2)(x+2)#

Explanation:

We will use the difference of squares identity which can be written:

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

First, notice that this quartic is a quadratic in #x^2#. Note also that #36 = 9xx4# and #13 = 9+4#, so we find:

#x^4-13x^2+36#

#=(x^2)^2-13(x^2)+36#

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

#=(x^2-3^2)(x^2-2^2)#

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

Feb 8, 2016

Answer:

Use an alternative method to find:

#x^4-13x^2+36 = (x-3)(x-2)(x+3)(x+2)#

Explanation:

Another method of factoring #x^4-13x^2+36# goes as follows:

Note that #x^4 = (x^2)^2# and #36 = 6^2# are both perfect squares.

Note also that there are no terms in #x^3# or #x#.

Consider:

#(x^2-kx+6)(x^2+kx+6)=x^4+(12-k^2)x^2+36#

The opposite middle terms #-kx# and #+kx# cause the #x^3# and #x# terms to cancel out.

If #(12-k^2)x^2 = -13x^2# then we find #k^2=25# so #k=+-5# and our original quartic factors as:

#x^4-13x^2+36 = (x^2-5x+6)(x^2+5x+6)#

In our example, each of the remaining quadratic factors is reducible to linear factors with Real coefficients:

#(x^2-5x+6)(x^2+5x+6) = (x-3)(x-2)(x+3)(x+2)#

Note however, that in general we may often be able to factor a quartic with no #x^3# or #x# terms into two quadratics in this way regardless of whether the resulting quadratics have Real roots.

For example:

#x^4-x^2+1 = (x^2-sqrt(3)x+1)(x^2+sqrt(3)x+1)#