How do you factor by grouping 15b^2-22bc-5c^2?

May 3, 2015

To solve this using Grouping, we need 4 terms.

We can split the middle term of this expression to get 4 terms

We need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = 15 \cdot - 5 = - 75$
AND
${N}_{1} + {N}_{2} = - 22$

After trying out a few numbers we get ${N}_{1} = - 25$ and ${N}_{2} = 3$

$- 25 \cdot 3 = - 75$, and $- 25 + 3 = - 22$

We write the expression as:

$15 {b}^{2} - 25 b c + 3 b c - 5 {c}^{2}$

We make Groups of two terms:

$= \left(15 {b}^{2} - 25 b c\right) + \left(3 b c - 5 {c}^{2}\right)$

$= 5 b \left(3 b - 5 c\right) + c \left(3 b - 5 c\right)$

$3 b - 5 c$ is a Common Factor to each of the terms

 = color(green)((3b-5c)(5b+c)