# How do you factor completely 56a^2b^3 - 35ab?

Nov 10, 2017

See a solution process below:

#### Explanation:

We can factor each term as:

$56 {a}^{2} {b}^{3} = 2 \times 2 \times 2 \times 7 \times a \times a \times b \times b \times b$

$35 a b = 5 \times 7 \times a \times b$

The common factors are in read:

$56 {a}^{2} {b}^{3} = 2 \times 2 \times 2 \times \textcolor{red}{7} \times \textcolor{red}{a} \times a \times \textcolor{red}{b} \times b \times b$

$35 a b = 5 \times \textcolor{red}{7} \times \textcolor{red}{a} \times \textcolor{red}{b}$

Therefore the common factor is:

$\textcolor{red}{7} \times \textcolor{red}{a} \times \textcolor{red}{b} = \textcolor{red}{7 a b}$

We can now write the expression as the product of the common factor and the remaining factors for each term:

$\textcolor{red}{7 a b} \left(\left[2 \times 2 \times 2 \times a \times b \times b\right] - 5\right) \implies$

$\textcolor{red}{7 a b} \left(8 a {b}^{2} - 5\right)$