Multiplying Binomials (FOIL)
Binomials can be multiplied very easy by using FOIL.
FOIL method is derived by applying distributive property of multiplication three times.
Indeed, suppose we want to find $$${\left({a}+{b}\right)}{\left({c}+{d}\right)}$$$.
If we treat $$${\left({a}+{b}\right)}$$$ as a single symbol, then, according to distributive property $$${\left({a}+{b}\right)}{\left({c}+{d}\right)}={\left({a}+{b}\right)}{c}+{\left({a}+{b}\right)}{d}$$$.
Finally, apply distributive property two more times: $$${\left({a}+{b}\right)}{c}+{\left({a}+{b}\right)}{d}={a}{c}+{b}{c}+{a}{d}+{b}{d}$$$.
FOIL(First, Outer, Inner, Last)
$$$\left(\color{red}{a}+\color{cyan}{b}\right)\left(\color{green}{c}+\color{blue}{d}\right)=\underbrace{\color{red}{a}\color{green}{c}}_{\text{First}}+\underbrace{\color{red}{a}\color{blue}{d}}_{\text{Outer}}+\underbrace{\color{cyan}{b}\color{green}{c}}_{\text{Inner}}+\underbrace{\color{cyan}{b}\color{blue}{d}}_{\text{Last}}$$$
After applying FOIL, you need to multiply monomials and, if possible, combine like terms.
Warning. FOIL can be applied ONLY to the product of binomials.
It means, that you can't multiply polynomials, using FOIL.
For example, you can't use FOIL for multiplying $$${\left({4}{x}+{5}{y}\right)}{\left({2}{{x}}^{{2}}+{5}{x}-{7}\right)}$$$, because, although first factor is binomial, the second factor is polynomial, not binomial.
Let's go through a couple of examples.
Example 1. Multiply $$${\left({2}{x}+{3}{y}\right)}{\left({5}{x}+{4}{y}\right)}$$$.
Here $$${a}={2}{x}$$$, $$${b}={3}{y}$$$, $$${c}={5}{x}$$$ and $$${d}={4}{y}$$$.
Now, apply FOIL:
$$$(\color{red}{2x}+\color{cyan}{3y})(\color{green}{5x}+\color{blue}{4y})=\underbrace{\color{red}{2x}\cdot\color{green}{5x}}_{\text{First}}+\underbrace{\color{red}{2x}\cdot\color{blue}{4y}}_{\text{Outer}}+\underbrace{\color{cyan}{3y}\cdot\color{green}{5x}}_{\text{Inner}}+\underbrace{\color{cyan}{3y}\cdot\color{blue}{4y}}_{\text{Last}}=$$$
$$$={10}{{x}}^{{2}}+{8}{x}{y}+{15}{y}{x}+{12}{{y}}^{{2}}=$$$ (multiply monomials)
$$$={10}{{x}}^{{2}}+{8}{x}{y}+{\color{red}{{{15}{x}{y}}}}+{12}{{y}}^{{2}}=$$$ (commutative property of multiplication)
$$$={10}{{x}}^{{2}}+{23}{x}{y}+{12}{{y}}^{{2}}$$$ (combine like terms)
Answer: $$${\left({2}{x}+{3}{y}\right)}{\left({5}{x}+{4}{y}\right)}={10}{{x}}^{{2}}+{23}{x}{y}+{12}{{y}}^{{2}}$$$.
Of course, FOIL also applicable if binomials contain minus sign. You just need to handle it carefully.
Example 2. Multiply $$${\left(-{4}{a}+\frac{{1}}{{3}}{b}\right)}{\left(\frac{{1}}{{2}}{b}-{2}{a}\right)}$$$.
Here $$${a}=-{4}{a}$$$, $$${b}=\frac{{1}}{{3}}{b}$$$, $$${c}=\frac{{1}}{{2}}{b}$$$ and $$${d}=-{2}{a}$$$.
Next, Apply FOIL:
$$$\left(\color{red}{(-4a)}+\color{cyan}{\frac{1}{3}b}\right)\left(\color{green}{\frac{1}{2}b}+\color{blue}{(-2a)}\right)=\underbrace{\color{red}{(-4a)}\cdot\color{green}{\left(\frac{1}{2}b\right)}}_{\text{First}}+\underbrace{\color{red}{(-4a)}\cdot\color{blue}{(-2a)}}_{\text{Outer}}+\underbrace{\color{cyan}{\left(\frac{1}{3}b\right)}\cdot\color{green}{\left(\frac{1}{2}b\right)}}_{\text{Inner}}+\underbrace{\color{cyan}{\left(\frac{1}{3}b\right)}\cdot\color{blue}{(-2a)}}_{\text{Last}}=$$$
$$$=-{2}{a}{b}+{8}{{a}}^{{2}}+\frac{{1}}{{6}}{{b}}^{{2}}-\frac{{2}}{{3}}{b}{a}=$$$ (multiply monomials)
$$$=-{2}{a}{b}+{8}{{a}}^{{2}}+\frac{{1}}{{6}}{{b}}^{{2}}{\color{red}{{-\frac{{2}}{{3}}{a}{b}}}}=$$$ (commutative property of multiplication)
$$$={8}{{a}}^{{2}}+\frac{{1}}{{6}}{{b}}^{{2}}-\frac{{8}}{{3}}{a}{b}$$$ (combine like terms)
Answer: $$${\left(-{4}{a}+\frac{{1}}{{3}}{b}\right)}{\left(\frac{{1}}{{2}}{b}-{2}{a}\right)}={8}{{a}}^{{2}}+\frac{{1}}{{6}}{{b}}^{{2}}-\frac{{8}}{{3}}{a}{b}$$$.
Of course, binomials can have more than one variable.
Example 3. Multiply the following: $$${\left({x}{y}+{3}{y}{z}\right)}{\left(-{5}{{x}}^{{2}}{{y}}^{{2}}+{4}{{z}}^{{3}}\right)}$$$.
Here $$${a}={x}{y}$$$, $$${b}={3}{y}{z}$$$, $$${c}=-{5}{{x}}^{{2}}{{y}}^{{2}}$$$ and $$${d}={4}{{z}}^{{3}}$$$.
Next, apply FOIL:
$$$(\color{red}{xy}+\color{cyan}{3yz})\left(\color{green}{\left(-5x^2y^2\right)}+\color{blue}{4z^3}\right)=\underbrace{\color{red}{xy}\cdot\color{green}{\left(-5x^2y^2\right)}}_{\text{First}}+\underbrace{\color{red}{xy}\cdot\color{blue}{4z^3}}_{\text{Outer}}+\underbrace{\color{cyan}{3yz}\cdot\color{green}{\left(-5x^2y^2\right)}}_{\text{Inner}}+\underbrace{\color{cyan}{3yz}\cdot\color{blue}{4z^3}}_{\text{Last}}=$$$
$$$=-{5}{{x}}^{{3}}{{y}}^{{3}}+{4}{x}{y}{{z}}^{{3}}-{15}{{x}}^{{2}}{{y}}^{{3}}{z}+{12}{y}{{z}}^{{4}}$$$ (multiply monomials)
There are no like terms, so we are done.
Answer: $$${\left({x}{y}+{3}{y}{z}\right)}{\left(-{5}{{x}}^{{2}}{{y}}^{{2}}+{4}{{z}}^{{3}}\right)}=-{5}{{x}}^{{3}}{{y}}^{{3}}+{4}{x}{y}{{z}}^{{3}}-{15}{{x}}^{{2}}{{y}}^{{3}}{z}+{12}{y}{{z}}^{{4}}$$$.
Now, it is time to exercise.
Exercise 1. Multiply $$${\left({5}{y}+{3}{z}\right)}{\left({7}{y}+{z}\right)}$$$.
Answer: $$${35}{{y}}^{{2}}+{26}{y}{z}+{3}{{z}}^{{2}}$$$.
Exercise 2. Multiply the following: $$${\left({7}{a}-{14}{b}\right)}{\left(-\frac{{2}}{{7}}{a}-{b}\right)}$$$.
Answer: $$$-{2}{{a}}^{{2}}-{3}{a}{b}+{14}{{b}}^{{2}}$$$.
Exercise 3. Multiply $$${\left(-{3}{x}{{y}}^{{2}}{z}+{5}{z}{y}\right)}{\left({2}{{x}}^{{3}}{{y}}^{{2}}-{{x}}^{{2}}{z}\right)}$$$.
Answer: $$$-{6}{{x}}^{{4}}{{y}}^{{4}}{z}+{3}{{x}}^{{3}}{{y}}^{{2}}{{z}}^{{2}}+{10}{{x}}^{{3}}{{y}}^{{3}}{z}-{5}{{x}}^{{2}}{y}{{z}}^{{2}}$$$.