# How do you factor completely 16x^3y^4+8x^2y^3?

Jun 2, 2016

$8 {x}^{2} {y}^{3} \left(2 x y + 1\right)$

#### Explanation:

$16 {x}^{3} {y}^{4} + 8 {x}^{2} {y}^{3}$

Let us write the factors of the terms given to us:

$16 {x}^{3} {y}^{4} = 2 \times 2 \times 2 \times 2 \times x \times x \times x \times y \times y \times y \times y$

$8 {x}^{2} {y}^{3} = 2 \times 2 \times 2 \times x \times x \times y \times y \times y$

Now, let us mark the common factors between the two terms:

$16 {x}^{3} {y}^{4} = \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{2} \times 2 \times \textcolor{red}{x} \times \textcolor{red}{x} \times x \times \textcolor{red}{y} \times \textcolor{red}{y} \times \textcolor{red}{y} \times y$

$8 {x}^{2} {y}^{3} = \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{x} \times \textcolor{red}{x} \times \textcolor{red}{y} \times \textcolor{red}{y} \times \textcolor{red}{y}$

Write the common factors and put the remaining factors in brackets, with the appropriate sign:

$= \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{x} \times \textcolor{red}{x} \times \textcolor{red}{y} \times \textcolor{red}{y} \times \textcolor{red}{y} \left(\textcolor{b l u e}{2} \times \textcolor{b l u e}{x} \times \textcolor{b l u e}{y} + \textcolor{g r e e n}{1}\right)$

$= \textcolor{red}{8 {x}^{2} {y}^{3}} \left(\textcolor{b l u e}{2 x y} + \textcolor{g r e e n}{1}\right)$

Therefore, the factor of $16 {x}^{3} {y}^{4} + 8 {x}^{2} {y}^{3} = 8 {x}^{2} {y}^{3} \left(2 x y + 1\right)$