# How do you factor  20x^4+16x^3-5x-4?

Oct 2, 2016

$20 {x}^{4} + 16 {x}^{3} - 5 x - 4 = \left(4 {x}^{3} - 1\right) \left(5 x + 4\right)$

$\textcolor{w h i t e}{20 {x}^{4} + 16 {x}^{3} - 5 x - 4} = \left(\sqrt[3]{4} x - 1\right) \left(2 \sqrt[3]{2} {x}^{2} + \sqrt[3]{4} x + 1\right) \left(5 x + 4\right)$

#### Explanation:

Notice that the ratio of the first and second terms is the same as that of the third and fourth terms. So this quadrinomial will factor by grouping:

$20 {x}^{4} + 16 {x}^{3} - 5 x - 4 = \left(20 {x}^{4} + 16 {x}^{3}\right) - \left(5 x + 4\right)$

$\textcolor{w h i t e}{20 {x}^{4} + 16 {x}^{3} - 5 x - 4} = 4 {x}^{3} \left(5 x + 4\right) - 1 \left(5 x + 4\right)$

$\textcolor{w h i t e}{20 {x}^{4} + 16 {x}^{3} - 5 x - 4} = \left(4 {x}^{3} - 1\right) \left(5 x + 4\right)$

We can factor $4 {x}^{3} - 1$ as a difference of cubes using irrational coefficients:

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

So:

$4 {x}^{3} - 1 = {\left(\sqrt[3]{4} x\right)}^{3} - {1}^{3}$

$\textcolor{w h i t e}{4 {x}^{3} - 1} = \left(\sqrt[3]{4} x - 1\right) \left({\left(\sqrt[3]{4} x\right)}^{2} + \sqrt[3]{4} x + 1\right)$

$\textcolor{w h i t e}{4 {x}^{3} - 1} = \left(\sqrt[3]{4} x - 1\right) \left(\sqrt[3]{16} {x}^{2} + \sqrt[3]{4} x + 1\right)$

$\textcolor{w h i t e}{4 {x}^{3} - 1} = \left(\sqrt[3]{4} x - 1\right) \left(2 \sqrt[3]{2} {x}^{2} + \sqrt[3]{4} x + 1\right)$