# How do you use FOIL to multiply (x-2) (x^2-5x+3)?

Jun 17, 2015

FOIL helps with multiplying two binomials, but is not applicable to multiplying a binomial by a trinomial.

$\left(x - 2\right) \left({x}^{2} - 5 x + 3\right) = {x}^{3} - 7 {x}^{2} + 13 x - 6$

#### Explanation:

$\left(x - 2\right) \left({x}^{2} - 5 x + 3\right)$

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

$= {x}^{3} - 5 {x}^{2} + 3 x - 2 {x}^{2} + 10 x - 6$

$= {x}^{3} - \left(5 {x}^{2} + 2 {x}^{2}\right) + \left(3 x + 10 x\right) - 6$

$= {x}^{2} - \left(5 + 2\right) {x}^{2} + \left(3 + 10\right) x - 6$

$= {x}^{2} - 7 {x}^{2} + 13 x - 6$

Alternatively, do what I do, look at each of the powers of $x$ in descending order in turn and total up the coefficients:

${x}^{3}$ : $1 \cdot 1 = 1$
${x}^{2}$ : $\left(1 \cdot - 5\right) + \left(- 2 \cdot 1\right) = - 5 - 2 = - 7$
$x$ : $\left(1 \cdot 3\right) + \left(- 2 \cdot - 5\right) = 3 + 10 = 13$
$1$ : $- 2 \cdot 3 = - 6$

Putting these together:

${x}^{3} - 7 {x}^{2} + 13 x - 6$