# How do you factor #4x^4 + 11 x^2 + 9#?

##### 1 Answer

#### Explanation:

Given:

#4x^4+11x^2+9#

First note that as a 'quadratic in

#Delta = 11^2-(4*4*9) = 121 - 144 = -23#

So treated as a quadratic in

We could do this, but I would like to show you another way...

Note that:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

Let

Then:

#a^4+(2-k^2)a^2b^2+b^4=4x^4+6(2-k^2)x^2+9#

To make

Then:

#6(2-k^2) = 12-6(1/sqrt(6))^2 = 12 - 1 = 11#

and:

#kab = 1/sqrt(6)*sqrt(2)x*sqrt(3) = x#

So we find:

#4x^4+11x^2+9#

#=a^4+(2-k^2)a^2b^2+b^4#

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

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

**Footnote**

Any polynomial in one variable of even degree with Real coefficients is expressible as the product of quadratic factors with Real coefficients.

Any polynomial in one variable of odd degree with Real coefficients is expressible as the product of one linear factor and zero or more quadratic factors with Real coefficients.

In both these cases, further factorisation with linear factors with Real coefficients may or may not be possible.

Quadratic factors which do not factorise further with Real coefficients correspond to Complex conjugate pairs of zeros.

For more complicated polynomials (e.g.