# How do you foil (9x + 7)(6x + 4)?

Aug 10, 2015

$54 {x}^{2} + 78 x + 28$

#### Explanation:

FOIL stands for: First (times First), Outside (times Outside), Inside (times Inside), Last (times Last).

In the problem to expand $\left(9 x + 7\right) \left(6 x + 4\right)$, First-times-First is $9 x \cdot 6 x = 54 {x}^{2}$, Outside-times-Outside is $9 x \cdot 4 = 36 x$, Inside-times-Inside is $7 \cdot 6 x = 42 x$, and Last-times-Last is $7 \cdot 4 = 28$.

Combining these gives $\left(9 x + 7\right) \left(6 x + 4\right) = 54 {x}^{2} + 36 x + 42 x + 28$. The $36 x$ and $42 x$ are "like terms" and combine to give $36 x + 42 x = 78 x$. Therefore, the answer is:

$\left(9 x + 7\right) \left(6 x + 4\right) = 54 {x}^{2} + 78 x + 28$

The reason this works is the distributive property: $a \cdot \left(b + c\right) = a \cdot b + a \cdot c$, and the commutative property: $a \cdot b = b \cdot a$

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

$= a \cdot c + b \cdot c + a \cdot d + b \cdot d$

$= a \cdot c + a \cdot d + b \cdot c + b \cdot d$ (which is what you'd get by using "FOIL" on the original product)