# How do you use FOIL to multiply (x-2)^2 (x+17)?

Jun 23, 2015

FOIL will only get you part of the way, but...

${\left(x - 2\right)}^{2} \left(x + 17\right)$

$= \left({x}^{2} - 4 x + 4\right) \left(x + 17\right)$

$= {x}^{3} + 13 {x}^{2} - 64 x + 68$

#### Explanation:

FOIL will only get you part of the way since once you multiply two of the binomials together you will be left with multiplying a binomial by a trinomial - which is not covered by FOIL.

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

$= F + O + I + L$ (First, Outside, Inside, Last)

$= \left(x \cdot x\right) + \left(x \cdot - 2\right) + \left(- 2 \cdot x\right) + \left(- 2 \cdot - 2\right)$

$= {x}^{2} - 2 x - 2 x + 4$

$= {x}^{2} - 4 x + 4$

Then using distributivity...

$\left({x}^{2} - 4 x + 4\right) \left(x + 17\right)$

$= \left({x}^{2} - 4 x + 4\right) x + \left({x}^{2} - 4 x + 4\right) 17$

$= {x}^{3} - 4 {x}^{2} + 4 x + 17 {x}^{2} - 68 x + 68$

$= {x}^{2} + \left(17 - 4\right) {x}^{2} - \left(68 - 4\right) x + 68$

$= {x}^{2} + 13 {x}^{2} - 64 x + 68$