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

##### 2 Answers

#### 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^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)#

#### 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

Note that

Note also that there are no terms in

Consider:

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

The opposite middle terms

If

#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

For example:

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