# How do you FOIL with 3 terms?

## (8r^2 + 4r + 6)(3r^2 − 7r + 1) How do I foil when I have 3 terms? Help and steps are much appreciated :)

May 20, 2018

$\left(8 {r}^{2} + 4 r + 6\right) \left(3 {r}^{2} - 7 r + 1\right) = 24 {r}^{4} - 44 {r}^{3} - 2 {r}^{2} - 38 r + 6$

#### Explanation:

FOIL is a mnemonic to help enumerate all individual products of terms when multiplying two binomials. It captures the result of applying the distributive property of multiplication over addition three times:

$\left(a + b\right) \left(c + d\right) = a \left(c + d\right) + b \left(c + d\right)$

$\textcolor{w h i t e}{\left(a + b\right) \left(c + d\right)} = {\overbrace{a c}}^{\text{First"+overbrace(ad)^"Outside"+overbrace(bc)^"Inside"+overbrace(bd)^"Last}}$

FOIL is not applicable to trinomials, but distributivity is.

So we could solve the given problem by:

$\left(8 {r}^{2} + 4 r + 6\right) \left(3 {r}^{2} - 7 r + 1\right)$

$= 8 {r}^{2} \left(3 {r}^{2} - 7 r + 1\right) + 4 r \left(3 {r}^{2} - 7 r + 1\right) + 6 \left(3 {r}^{2} - 7 r + 1\right)$

$= \left(24 {r}^{4} - 56 {r}^{3} + 8 {r}^{2}\right) + \left(12 {r}^{3} - 28 {r}^{2} + 4 r\right) + \left(18 {r}^{2} - 42 r + 6\right)$

$= 24 {r}^{4} - 56 {r}^{3} + 12 {r}^{3} + 8 {r}^{2} - 28 {r}^{2} + 18 {r}^{2} + 4 r - 42 r + 6$

$= 24 {r}^{4} + \left(- 56 + 12\right) {r}^{3} + \left(8 - 28 + 18\right) {r}^{2} + \left(4 - 42\right) r + 6$

$= 24 {r}^{4} - 44 {r}^{3} - 2 {r}^{2} - 38 r + 6$

Alternatively, we can write the coefficients of the $9$ individual products of pairs of terms in a table and sum the reverse diagonals, to find the coefficients of the product like this:

$\underline{\textcolor{w h i t e}{+} \textcolor{w h i t e}{00} \setminus \text{ |} \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} 8 \setminus \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} 4 \setminus \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} 6}$
$\textcolor{w h i t e}{+} \textcolor{w h i t e}{0} 3 \setminus \text{ |} \textcolor{w h i t e}{+} \textcolor{red}{24} \setminus \textcolor{w h i t e}{+} \textcolor{\mathmr{and} a n \ge}{12} \setminus \textcolor{w h i t e}{+} \textcolor{g r e e n}{18}$
$\textcolor{b l a c k}{-} \textcolor{w h i t e}{0} 7 \setminus \text{ |} \textcolor{\mathmr{and} a n \ge}{-} \textcolor{\mathmr{and} a n \ge}{56} \setminus \textcolor{g r e e n}{-} \textcolor{g r e e n}{28} \setminus \textcolor{b l u e}{-} \textcolor{b l u e}{42}$
$\textcolor{w h i t e}{+} \textcolor{w h i t e}{0} 1 \setminus \text{ |} \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} \textcolor{g r e e n}{8} \setminus \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} \textcolor{b l u e}{4} \setminus \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} \textcolor{p u r p \le}{6}$

Hence:

$\left(8 {r}^{2} + 4 r + 6\right) \left(3 {r}^{2} - 7 r + 1\right)$

$= \textcolor{red}{24} {r}^{4} + \left(\textcolor{\mathmr{and} a n \ge}{- 56 + 12}\right) {r}^{3} + \left(\textcolor{g r e e n}{8 - 28 + 18}\right) {r}^{2} + \left(\textcolor{b l u e}{4 - 42}\right) r + \textcolor{p u r p \le}{6}$

$= 24 {r}^{4} - 44 {r}^{3} - 2 {r}^{2} - 38 r + 6$

Alternatively, we could examine the given product of trinomials and think about each power of $r$ in descending order, summing all the ways that a term in the first trinomial multiplied by a term in the second can give rise to that power. With practice, that should allow us to write the answer directly.

$\left(8 {r}^{2} + 4 r + 6\right) \left(3 {r}^{2} - 7 r + 1\right) = 24 {r}^{4} - 44 {r}^{3} - 2 {r}^{2} - 38 r + 6$