# How do you factor 4m^4 - 37m^2 + 9?

Nov 13, 2016

$\left(4 {m}^{2} - 1\right) \left({m}^{2} - 9\right)$

#### Explanation:

Factoring this polynomial gives:

$\left(4 {m}^{2} - 1\right) \left({m}^{2} - 9\right)$

Nov 18, 2016

$4 {m}^{4} - 37 {m}^{2} + 9 = \left(2 m - 1\right) \left(2 m + 1\right) \left(m - 3\right) \left(m + 3\right)$

#### Explanation:

The difference of squares identity can be written:

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

We will use this a couple of times, but first note that:

$4 {m}^{4} - 37 {m}^{2} + 9 = 4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9$

So we can treat this quartic as a quadratic in ${m}^{2}$

Factor using an AC method:

Look for a pair of factors of $A C = 4 \cdot 9 = 36$ with sum $B = 37$

The pair $36 , 1$ works.

Use this pair to split the middle term and factor by grouping:

$4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9 = 4 {\left({m}^{2}\right)}^{2} - 36 \left({m}^{2}\right) - \left({m}^{2}\right) + 9$

$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = \left(4 {\left({m}^{2}\right)}^{2} - 36 \left({m}^{2}\right)\right) - \left(\left({m}^{2}\right) - 9\right)$

$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = 4 {m}^{2} \left({m}^{2} - 9\right) - 1 \left({m}^{2} - 9\right)$

$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = \left(4 {m}^{2} - 1\right) \left({m}^{2} - 9\right)$

$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = \left({\left(2 m\right)}^{2} - {1}^{2}\right) \left({m}^{2} - {3}^{2}\right)$

$\textcolor{w h i t e}{4 {\left({m}^{2}\right)}^{2} - 37 \left({m}^{2}\right) + 9} = \left(2 m - 1\right) \left(2 m + 1\right) \left(m - 3\right) \left(m + 3\right)$