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

Jun 3, 2015

This is where FOIL is not quite enough - It's meant for multiplying two binomials together, but here the second expression is a trinomial.

Instead you can break the problem into separate terms using the distributive law, multiply out, then recombine the terms...

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

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

$= \left(2 {x}^{3} + 10 {x}^{2} - 6 x\right) - \left(3 {x}^{2} + 15 x - 9\right)$

$= 2 {x}^{3} + 10 {x}^{2} - 6 x - 3 {x}^{2} - 15 x + 9$

$= 2 {x}^{3} + \left(10 - 3\right) {x}^{2} - \left(6 + 15\right) x + 9$

$= 2 {x}^{3} + 7 {x}^{2} - 21 x + 9$

Jun 3, 2015

Alternatively, working with each of the powers of $x$ from ${x}^{3}$ down to ${x}^{0}$ (i.e. the constant term), match the terms in the first and second bracketed expressions which will multiply to give that power of $x$ and add them together, thus:

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

${x}^{3} : 2 x \times {x}^{2} = 2 {x}^{3}$

${x}^{2} : \left(2 x \times 5 x\right) + \left(- 3 \times {x}^{2}\right) = 10 {x}^{2} - 3 {x}^{2} = 7 {x}^{2}$

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

${x}^{0} : - 3 \times - 3 = 9$

Added, give $2 {x}^{3} + 7 {x}^{2} - 21 x + 9$