# How do you simplify  (x-3)(2x-5)-(x+1)(x-6)?

Oct 21, 2017

${x}^{2} - 6 x + 21$

#### Explanation:

You can expend the expression.

$\left(x - 3\right) \left(2 x - 5\right) - \left(x + 1\right) \left(x - 6\right)$

$= \left(x\right) \left(2 x\right) + \left(x\right) \left(- 5\right) - 3 \left(2 x\right) - 3 \left(- 5\right) - \left[\left(x\right) \left(x\right) + \left(x\right) \left(- 6\right) + 1 \left(x\right) + 1 \left(- 6\right)\right]$

$= 2 {x}^{2} - 5 x - 6 x + 15 - \left({x}^{2} - 6 x + x - 6\right)$
$= 2 {x}^{2} - 11 x + 15 - {x}^{2} + 5 x + 6$
$= {x}^{2} - 6 x + 21$

Oct 21, 2017

$\left(x - 3\right) \left(2 x - 5\right) - \left(x + 1\right) \left(x - 6\right) = {x}^{2} - 6 x + 21$

#### Explanation:

(x-3)(2x-5) - (x+1)(x-6)

Use the distributive law to expand both equations:
(a+b)*(c+d)= (ac + ad + bc + bd)

=$\left(\left(x \cdot 2 x\right) + \left(x \cdot - 5\right) + \left(- 3 \cdot 2 x\right) + \left(- 3 \cdot - 5\right)\right) - \left(\left(x \cdot x\right) + \left(x \cdot - 6\right) + \left(1 \cdot x\right) + \left(1 \cdot - 6\right)\right)$
=$\left(2 {x}^{2} - 5 x - 6 x + 15\right) - \left({x}^{2} - 6 x + x - 6\right)$
=$\left(2 {x}^{2} - 11 x + 15\right) - \left({x}^{2} - 5 x - 6\right)$
=$2 {x}^{2} - 11 x + 15 - {x}^{2} + 5 x + 6$
=${x}^{2} - 6 x + 21$

Therefore:
$\left(x - 3\right) \left(2 x - 5\right) - \left(x + 1\right) \left(x - 6\right) = {x}^{2} - 6 x + 21$