# How do you factor 4a^4 + 6a^3b^2 + 2a^2b^3?

Aug 28, 2016

$4 {a}^{4} + 6 {a}^{3} {b}^{2} + 2 {a}^{2} {b}^{3} = 2 {a}^{2} \left(2 {a}^{2} + 3 a {b}^{2} + {b}^{3}\right)$

$4 {a}^{4} b + 6 {a}^{3} {b}^{2} + 2 {a}^{2} {b}^{3} = 2 {a}^{2} b \left(2 a + b\right) \left(a + b\right)$

$4 {a}^{4} + 6 {a}^{3} {b}^{2} + 2 {a}^{2} {b}^{4} = 2 {a}^{2} \left(2 a + {b}^{2}\right) \left(a + {b}^{2}\right)$

#### Explanation:

If the expression is correct as given then we find:

$4 {a}^{4} + 6 {a}^{3} {b}^{2} + 2 {a}^{2} {b}^{3} = 2 {a}^{2} \left(2 {a}^{2} + 3 a {b}^{2} + {b}^{3}\right)$

with no further simplification.

If the first term was supposed to be $4 {a}^{4} b$ then we find:

$4 {a}^{4} b + 6 {a}^{3} {b}^{2} + 2 {a}^{2} {b}^{3}$

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

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

If instead the last term was supposed to be $2 {a}^{2} {b}^{4}$ then we find:

$4 {a}^{4} + 6 {a}^{3} {b}^{2} + 2 {a}^{2} {b}^{4}$

$= 2 {a}^{2} \left(2 {a}^{2} + 3 a {b}^{2} + {b}^{4}\right)$

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