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

May 4, 2016

$4 {x}^{4} + 11 {x}^{2} + 9 = \left(2 {x}^{2} - x + 3\right) \left(2 {x}^{2} + x + 3\right)$

#### Explanation:

Given:

$4 {x}^{4} + 11 {x}^{2} + 9$

First note that as a 'quadratic in ${x}^{2}$', this has a negative discriminant:

$\Delta = {11}^{2} - \left(4 \cdot 4 \cdot 9\right) = 121 - 144 = - 23$

So treated as a quadratic in ${x}^{2}$ we would be looking at Complex zeros to take the square root of.

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

Note that:

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

Let $a = \sqrt{2} x$ and $b = \sqrt{3}$

Then:

${a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4} = 4 {x}^{4} + 6 \left(2 - {k}^{2}\right) {x}^{2} + 9$

To make $6 \left(2 - {k}^{2}\right)$ match the coefficient $11$ in the original quartic, let $k = \frac{1}{\sqrt{6}}$.

Then:

$6 \left(2 - {k}^{2}\right) = 12 - 6 {\left(\frac{1}{\sqrt{6}}\right)}^{2} = 12 - 1 = 11$

and:

$k a b = \frac{1}{\sqrt{6}} \cdot \sqrt{2} x \cdot \sqrt{3} = x$

So we find:

$4 {x}^{4} + 11 {x}^{2} + 9$

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

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

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

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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. ${x}^{5} + 4 x + 2$), the Real coefficients of those linear and quadratic factors may not be expressible in simple algebraic form, but it is always possible to calculate numerical approximations.