# How do you multiply (x-1)(x-2)(x-3)?

Jun 11, 2018

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

#### Explanation:

It is helpful to know that:

$\left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right)$

$= {x}^{3} - \left(\alpha + \beta + \gamma\right) {x}^{2} + \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) x - \alpha \beta \gamma$

In this identity the expressions forming the coefficients of ${x}^{k}$ are (apart from the alternating signs), the elementary symmetric polynomials in $\alpha , \beta , \gamma$, namely:

• $\alpha + \beta + \gamma$

• $\alpha \beta + \beta \gamma + \gamma \alpha$

• $\alpha \beta \gamma$

With $\alpha = 1$, $\beta = 2$ and $\gamma = 3$, we find:

$\alpha + \beta + \gamma = 1 + 2 + 3 = 6$

$\alpha \beta + \beta \gamma + \gamma \alpha = \left(1 \cdot 2\right) + \left(2 \cdot 3\right) + \left(3 \cdot 1\right) = 2 + 6 + 3 = 11$

$\alpha \beta \gamma = 1 \cdot 2 \cdot 3 = 6$

So:

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