Question

Find the factors of $x^4+9x^2+81$

Answer 1

In the given expression $x^4+9x^2+81$, add and subtract $9x^2$ to make it a perfect square as shown below:

$x^4+9x^2+81=(x^4+9x^2+81+9x^2)-9x^2=(x^4+18x^2+81)-9x^2=\left[\right(x^2{)}^2+(9{)}^2+2(x^2\left)\right(3\left)\right]-9x^2$

$=(x^2+9{)}^2-9x^2$

(using the identity $(a+b{)}^2=a^2+b^2+2ab$

We also know the identity $a^2-b^2=(a+b)(a-b)$, therefore,

Using the above identity, the expression $(x^2+9{)}^2-9x^2$can be factorised as follows:

$(x^2+9{)}^2-9x^2=(x^2+9{)}^2-(3x{)}^2=(x^2+9+3x\left)\right(x^2+9-3x)$

Hence, $x^4+9x^2+81=(x^2+9+3x)(x^2+9-3x)$