# How do you solve 10b^2 = 27b - 18 by factoring?

Aug 21, 2015

The solutions are:
color(blue)(b=6/5

 color(blue)(b=3/2

#### Explanation:

$10 {b}^{2} - 27 b + 18 = 0$

We can Split the Middle Term of this expression to factorise it and thereby find solutions.

In this technique we need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = a \cdot c = 10 \cdot 18 = 180$
and
${N}_{1} + {N}_{2} = b = - 27$

After trying out a few numbers we get ${N}_{1} = - 12$ and ${N}_{2} = - 15$

$\left(- 12\right) \cdot \left(- 15\right) = 180$, and $\left(- 12\right) + \left(- 15\right) = - 27$

$10 {b}^{2} - 27 b + 18 = 10 {b}^{2} - 15 b - 12 b + 18$

$10 {b}^{2} - 15 b - 12 b + 18 = 0$

$5 b \left(2 b - 3\right) - 6 \left(2 b - 3\right) = 0$

$\left(5 b - 6\right) \left(2 b - 3\right) = 0$ is the factorised form of the expression.

Now we equate the factors to zero.

5b-6=0, color(blue)(b=6/5

2b-3=0, color(blue)(b=3/2