Binomials can be multipliedFOIL.
FOIL method is derived by applying distributive property of multiplication three times.
Indeed, suppose we want to find ( a + b ) ( c + d ) {\left({a}+{b}\right)}{\left({c}+{d}\right)} ( a + b ) ( c + d )
If we treat ( a + b ) {\left({a}+{b}\right)} ( a + b ) ( a + b ) ( c + d ) = ( a + b ) c + ( a + b ) d {\left({a}+{b}\right)}{\left({c}+{d}\right)}={\left({a}+{b}\right)}{c}+{\left({a}+{b}\right)}{d} ( a + b ) ( c + d ) = ( a + b ) c + ( a + b ) d
Finally, apply distributive property two more times: ( a + b ) c + ( a + b ) d = a c + b c + a d + b d {\left({a}+{b}\right)}{c}+{\left({a}+{b}\right)}{d}={a}{c}+{b}{c}+{a}{d}+{b}{d} ( a + b ) c + ( a + b ) d = a c + b c + a d + b d
F O I L (First , Outer , Inner , Last )
( a + b ) ( c + d ) = a c ⏟ First + a d ⏟ Outer + b c ⏟ Inner + b d ⏟ 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}} ( a + b ) ( c + d ) = First a c + Outer a d + Inner b c + Last b d
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 ( 4 x + 5 y ) ( 2 x 2 + 5 x − 7 ) {\left({4}{x}+{5}{y}\right)}{\left({2}{{x}}^{{2}}+{5}{x}-{7}\right)} ( 4 x + 5 y ) ( 2 x 2 + 5 x − 7 )
Let's go through a couple of examples.
Example 1 ( 2 x + 3 y ) ( 5 x + 4 y ) {\left({2}{x}+{3}{y}\right)}{\left({5}{x}+{4}{y}\right)} ( 2 x + 3 y ) ( 5 x + 4 y )
Here a = 2 x {a}={2}{x} a = 2 x b = 3 y {b}={3}{y} b = 3 y c = 5 x {c}={5}{x} c = 5 x d = 4 y {d}={4}{y} d = 4 y
Now, apply FOIL:
( 2 x + 3 y ) ( 5 x + 4 y ) = 2 x ⋅ 5 x ⏟ First + 2 x ⋅ 4 y ⏟ Outer + 3 y ⋅ 5 x ⏟ Inner + 3 y ⋅ 4 y ⏟ Last = (\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}}= ( 2 x + 3 y ) ( 5 x + 4 y ) = First 2 x ⋅ 5 x + Outer 2 x ⋅ 4 y + Inner 3 y ⋅ 5 x + Last 3 y ⋅ 4 y =
= 10 x 2 + 8 x y + 15 y x + 12 y 2 = ={10}{{x}}^{{2}}+{8}{x}{y}+{15}{y}{x}+{12}{{y}}^{{2}}= = 10 x 2 + 8 x y + 15 y x + 12 y 2 =
= 10 x 2 + 8 x y + 15 x y + 12 y 2 = ={10}{{x}}^{{2}}+{8}{x}{y}+{\color{red}{{{15}{x}{y}}}}+{12}{{y}}^{{2}}= = 10 x 2 + 8 x y + 15 x y + 12 y 2 = commutative property of multiplication )
= 10 x 2 + 23 x y + 12 y 2 ={10}{{x}}^{{2}}+{23}{x}{y}+{12}{{y}}^{{2}} = 10 x 2 + 23 x y + 12 y 2
Answer : ( 2 x + 3 y ) ( 5 x + 4 y ) = 10 x 2 + 23 x y + 12 y 2 {\left({2}{x}+{3}{y}\right)}{\left({5}{x}+{4}{y}\right)}={10}{{x}}^{{2}}+{23}{x}{y}+{12}{{y}}^{{2}} ( 2 x + 3 y ) ( 5 x + 4 y ) = 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 ( − 4 a + 1 3 b ) ( 1 2 b − 2 a ) {\left(-{4}{a}+\frac{{1}}{{3}}{b}\right)}{\left(\frac{{1}}{{2}}{b}-{2}{a}\right)} ( − 4 a + 3 1 b ) ( 2 1 b − 2 a )
Here a = − 4 a {a}=-{4}{a} a = − 4 a b = 1 3 b {b}=\frac{{1}}{{3}}{b} b = 3 1 b c = 1 2 b {c}=\frac{{1}}{{2}}{b} c = 2 1 b d = − 2 a {d}=-{2}{a} d = − 2 a
Next, Apply FOIL:
( ( − 4 a ) + 1 3 b ) ( 1 2 b + ( − 2 a ) ) = ( − 4 a ) ⋅ ( 1 2 b ) ⏟ First + ( − 4 a ) ⋅ ( − 2 a ) ⏟ Outer + ( 1 3 b ) ⋅ ( 1 2 b ) ⏟ Inner + ( 1 3 b ) ⋅ ( − 2 a ) ⏟ Last = \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}}= ( ( − 4 a ) + 3 1 b ) ( 2 1 b + ( − 2 a ) ) = First ( − 4 a ) ⋅ ( 2 1 b ) + Outer ( − 4 a ) ⋅ ( − 2 a ) + Inner ( 3 1 b ) ⋅ ( 2 1 b ) + Last ( 3 1 b ) ⋅ ( − 2 a ) =
= − 2 a b + 8 a 2 + 1 6 b 2 − 2 3 b a = =-{2}{a}{b}+{8}{{a}}^{{2}}+\frac{{1}}{{6}}{{b}}^{{2}}-\frac{{2}}{{3}}{b}{a}= = − 2 a b + 8 a 2 + 6 1 b 2 − 3 2 b a =
= − 2 a b + 8 a 2 + 1 6 b 2 − 2 3 a b = =-{2}{a}{b}+{8}{{a}}^{{2}}+\frac{{1}}{{6}}{{b}}^{{2}}{\color{red}{{-\frac{{2}}{{3}}{a}{b}}}}= = − 2 a b + 8 a 2 + 6 1 b 2 − 3 2 a b =
= 8 a 2 + 1 6 b 2 − 8 3 a b ={8}{{a}}^{{2}}+\frac{{1}}{{6}}{{b}}^{{2}}-\frac{{8}}{{3}}{a}{b} = 8 a 2 + 6 1 b 2 − 3 8 a b
Answer : ( − 4 a + 1 3 b ) ( 1 2 b − 2 a ) = 8 a 2 + 1 6 b 2 − 8 3 a b {\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} ( − 4 a + 3 1 b ) ( 2 1 b − 2 a ) = 8 a 2 + 6 1 b 2 − 3 8 a b
Of course, binomials can have more than one variable.
Example 3 ( x y + 3 y z ) ( − 5 x 2 y 2 + 4 z 3 ) {\left({x}{y}+{3}{y}{z}\right)}{\left(-{5}{{x}}^{{2}}{{y}}^{{2}}+{4}{{z}}^{{3}}\right)} ( x y + 3 y z ) ( − 5 x 2 y 2 + 4 z 3 )
Here a = x y {a}={x}{y} a = x y b = 3 y z {b}={3}{y}{z} b = 3 y z c = − 5 x 2 y 2 {c}=-{5}{{x}}^{{2}}{{y}}^{{2}} c = − 5 x 2 y 2 d = 4 z 3 {d}={4}{{z}}^{{3}} d = 4 z 3
Next, apply FOIL:
( x y + 3 y z ) ( ( − 5 x 2 y 2 ) + 4 z 3 ) = x y ⋅ ( − 5 x 2 y 2 ) ⏟ First + x y ⋅ 4 z 3 ⏟ Outer + 3 y z ⋅ ( − 5 x 2 y 2 ) ⏟ Inner + 3 y z ⋅ 4 z 3 ⏟ Last = (\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}}= ( x y + 3 yz ) ( ( − 5 x 2 y 2 ) + 4 z 3 ) = First x y ⋅ ( − 5 x 2 y 2 ) + Outer x y ⋅ 4 z 3 + Inner 3 yz ⋅ ( − 5 x 2 y 2 ) + Last 3 yz ⋅ 4 z 3 =
= − 5 x 3 y 3 + 4 x y z 3 − 15 x 2 y 3 z + 12 y z 4 =-{5}{{x}}^{{3}}{{y}}^{{3}}+{4}{x}{y}{{z}}^{{3}}-{15}{{x}}^{{2}}{{y}}^{{3}}{z}+{12}{y}{{z}}^{{4}} = − 5 x 3 y 3 + 4 x y z 3 − 15 x 2 y 3 z + 12 y z 4
There are no like terms, so we are done.
Answer : ( x y + 3 y z ) ( − 5 x 2 y 2 + 4 z 3 ) = − 5 x 3 y 3 + 4 x y z 3 − 15 x 2 y 3 z + 12 y z 4 {\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}} ( x y + 3 y z ) ( − 5 x 2 y 2 + 4 z 3 ) = − 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 .( 5 y + 3 z ) ( 7 y + z ) {\left({5}{y}+{3}{z}\right)}{\left({7}{y}+{z}\right)} ( 5 y + 3 z ) ( 7 y + z )
Answer : 35 y 2 + 26 y z + 3 z 2 {35}{{y}}^{{2}}+{26}{y}{z}+{3}{{z}}^{{2}} 35 y 2 + 26 y z + 3 z 2
Exercise 2 .( 7 a − 14 b ) ( − 2 7 a − b ) {\left({7}{a}-{14}{b}\right)}{\left(-\frac{{2}}{{7}}{a}-{b}\right)} ( 7 a − 14 b ) ( − 7 2 a − b )
Answer : − 2 a 2 − 3 a b + 14 b 2 -{2}{{a}}^{{2}}-{3}{a}{b}+{14}{{b}}^{{2}} − 2 a 2 − 3 a b + 14 b 2
Exercise 3 .( − 3 x y 2 z + 5 z y ) ( 2 x 3 y 2 − x 2 z ) {\left(-{3}{x}{{y}}^{{2}}{z}+{5}{z}{y}\right)}{\left({2}{{x}}^{{3}}{{y}}^{{2}}-{{x}}^{{2}}{z}\right)} ( − 3 x y 2 z + 5 z y ) ( 2 x 3 y 2 − x 2 z )
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 -{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}} − 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
