# How do you multiply (x-1)(x+1)(x+2)?

Aug 10, 2018

$p \left(x\right) = {x}^{3} + 2 {x}^{2} - x - 2$

#### Explanation:

Let ,

$p \left(x\right) = \left(x - 1\right) \left(x + 1\right) \left(x + 2\right)$

Multiplying first two factors:

$p \left(x\right) = \left\{x \left(x + 1\right) - 1 \left(x + 1\right)\right\} \left(x + 2\right)$

$p \left(x\right) = \left\{x \cdot x + x \cdot 1 - 1 \cdot x - 1 \cdot 1\right\} \left(x + 2\right)$

$p \left(x\right) = \left\{{x}^{2} + \cancel{x} - \cancel{x} - 1\right\} \left(x + 2\right)$

$p \left(x\right) = \left({x}^{2} - 1\right) \left(x + 2\right)$

Again multiplying two factors :

$p \left(x\right) = {x}^{2} \left(x + 2\right) - 1 \left(x + 2\right)$

$p \left(x\right) = {x}^{2} \cdot x + {x}^{2} \cdot 2 - 1 \cdot x - 1 \cdot 2$

$p \left(x\right) = {x}^{3} + 2 {x}^{2} - x - 2$

Aug 10, 2018

${x}^{3} + 2 {x}^{2} - x - 2$

#### Explanation:

$\text{given } \left(x + a\right) \left(x + b\right) \left(x + c\right)$

$\text{then the expansion is}$

${x}^{3} + \left(a + b + c\right) {x}^{2} + \left(a b + b c + a c\right) x + a b c$

$\text{here } a = - 1 , b = 1 , c = 2$

$= {x}^{3} + \left(- 1 + 1 + 2\right) {x}^{2} + \left(- 1 + 2 - 2\right) x$
$\textcolor{w h i t e}{=} + \left(- 1\right) \left(1\right) \left(2\right)$

$= {x}^{3} + 2 {x}^{2} - x - 2$