# How do you factor a^4+4b^4 ?

Jan 20, 2017

${a}^{4} + 4 {b}^{4} = \left({a}^{2} - 2 a b + 2 {b}^{2}\right) \left({a}^{2} + 2 a b + 2 {b}^{2}\right)$

#### Explanation:

This polynomial factors nicely into two quadratic polynomials with integer coefficients:

${a}^{4} + 4 {b}^{4} = \left({a}^{2} - 2 a b + 2 {b}^{2}\right) \left({a}^{2} + 2 a b + 2 {b}^{2}\right)$

These quadratic factors have no simpler linear factors with Real coefficients. To see that, you can check their discriminants:

${\Delta}_{{a}^{2} - 2 a b + 2 {b}^{2}} = {\left(- 2\right)}^{2} - 4 \left(1\right) \left(2\right) = 4 - 8 = - 4$

${\Delta}_{{a}^{2} + 2 a b + 2 {b}^{2}} = {2}^{2} - 4 \left(1\right) \left(2\right) = 4 - 8 = - 4$

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Footnotes

${a}^{4} + 4 {b}^{4}$ is an example of a homogeneous polynomial in two variables, $a$ and $b$. That is, all of the terms are of the same degree, namely $4$ in this example.

Factoring homogeneous polynomials is very similar to factoring a corresponding polynomial in one variable.

In our example, we could let $t = \frac{a}{b}$ and look at how to factor:

${t}^{4} + 4 = \left({t}^{2} - 2 t + 2\right) \left({t}^{2} + 2 t + 2\right)$

Then we could multiply this through by ${b}^{4}$ to get our original factorisation.

Similarly, when we look at ${a}^{2} - 2 a b + 2 {b}^{2}$, the corresponding polynomial in $\frac{a}{b}$ formed by dividing it by ${b}^{2}$ can be expressed as:

${t}^{2} - 2 t + 2$

When you have a quadratic of the form $A {t}^{2} + B t + C$ in one variable you are probably familiar with finding the discriminant $\Delta = {B}^{2} - 4 A C$. The sign of the discriminant $\Delta$ allows us to determine whether the quadratic in $t$ has Real zeros and therefore linear factors with Real coefficients.

Faced with $A {a}^{2} + B a b + C {b}^{2}$ the same discriminant tells us whether this factors into linear factors with Real coefficients.