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Algebra 1

Course: Algebra 1 > Unit 13

Lesson 3: Special products of binomials
Math
Algebra 1
Quadratics: Multiplying & factoring
Special products of binomials
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Binomial special products review

Google Classroom
A review of the difference of squares pattern (a+b)(a-b)=a^2-b^2, as well as other common patterns encountered while multiplying binomials, such as (a+b)^2=a^2+2ab+b^2.
These types of binomial multiplication problems come up time and time again, so it's good to be familiar with some basic patterns.
The "difference of squares" pattern:
left parenthesis, a, plus, b, right parenthesis, left parenthesis, a, minus, b, right parenthesis, equals, a, squared, minus, b, squared
Two other patterns:
(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2\begin{aligned} &(a+b)^2=a^2+2ab+b^2\\\\ &(a-b)^2=a^2-2ab+b^2 \end{aligned}

Example 1

Expand the expression.
left parenthesis, c, minus, 5, right parenthesis, left parenthesis, c, plus, 5, right parenthesis
The expression fits the difference of squares pattern:
left parenthesis, a, plus, b, right parenthesis, left parenthesis, a, minus, b, right parenthesis, equals, a, squared, minus, b, squared
So our answer is:
left parenthesis, c, minus, 5, right parenthesis, left parenthesis, c, plus, 5, right parenthesis, equals, c, squared, minus, 25
But if you don't recognize the pattern, that's okay too. Just multiply the binomials as normal. Over time, you'll learn to see the pattern.
(c−5)(c+5)=c(c)+c(5)−5(c)−5(5)=c(c)+5c−5c−5(5)=c2−25\begin{aligned} &(\purpleD{c-5})(c+5)\\\\ =&\purpleD{c}(c)+\purpleD{c}(5)\purpleD{-5}(c)\purpleD{-5}(5)\\\\ =&\purpleD{c}(c)+\redD{5c-5c}\purpleD{-5}(5)\\\\ =&c^2-25 \end{aligned}
Notice how the "middle terms" cancel.
Want another example? Check out this video.

Example 2

Expand the expression.
left parenthesis, m, plus, 7, right parenthesis, squared
The expression fits this pattern:
left parenthesis, a, plus, b, right parenthesis, squared, equals, a, squared, plus, 2, a, b, plus, b, squared
So our answer is:
left parenthesis, m, plus, 7, right parenthesis, squared, equals, m, squared, plus, 14, m, plus, 49
But if you don't recognize the pattern, that's okay too. Just multiply the binomials as normal. Over time, you'll learn to see the pattern.
(m+7)2=(m+7)(m+7)=m(m)+m(7)+7(m)+7(7)=m(m)+7m+7m+7(7)=m2+14m+49\begin{aligned} &(m+7)^2\\\\ =&(\blueD{m+7})(m+7)\\\\ =&\blueD{m}(m)+\blueD{m}(7)+\blueD{7}(m)+\blueD{7}(7)\\\\ =&\blueD{m}(m)\greenD{+7m+7m}+\blueD{7}(7)\\\\ =&m^2+14m+49 \end{aligned}
Want another example? Check out this video.

Example 3

Expand this expression.
left parenthesis, 6, w, minus, y, right parenthesis, left parenthesis, 6, w, plus, y, right parenthesis
The expression fits the difference of squares pattern:
left parenthesis, a, plus, b, right parenthesis, left parenthesis, a, minus, b, right parenthesis, equals, a, squared, minus, b, squared
So our answer is:
(6w−y)(6w+y)=(6w)2−y2=36w2−y2\begin{aligned} &(6w-y)(6w+y) \\\\ =&(6w)^2-y^2 \\\\ =&36w^2-y^2 \end{aligned}
But if you don't recognize the pattern, that's okay too. Just multiply the binomials as normal. Over time, you'll learn to see the pattern.
(6w−y)(6w+y)=6w(6w)+6w(y)−y(6w)−y(y)=6w(6w)+6wy−6wy−y(y)=36w2−y2\begin{aligned} &(\purpleD{6w-y})(6w+y)\\\\ =&\purpleD{6w}(6w)+\purpleD{6w}(y)\purpleD{-y}(6w)\purpleD{-y}(y)\\\\ =&\purpleD{6w}(6w)+\redD{6wy-6wy}\purpleD{-y}(y)\\\\ =&36w^2-y^2 \end{aligned}
Notice how the "middle terms" cancel.
Want more practice? Check out this intro exercise and this slightly harder exercise.

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