11.2.4: Multiplying Special Cases
- Page ID
- 67622
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- Square a binomial.
- Multiply the sum and difference of the same two terms.
Introduction
While the distributive property can be used for all polynomial multiplication, some products with binomials can be found using short cuts. These methods are sometimes called special products.
Square of a Binomial Sum
Multiplying a number by itself is often called squaring. You can represent this multiplication as a square. The number being squared is the length of the sides of the square and the product is represented by the area of that square. Consider a square whose side length is described by the binomial \(\ x+3\):
The area of this square is \(\ (x+3)(x+3) \text { or }(x+3)^{2}\).
The following shows the same square, but separates the variable and constant terms:
From this area model, you can see that the area can be described by the sum of the red, green and yellow pieces. That is, the area is \(\ x^{2}+3 x+3 x+9=x^{2}+6 x+9\).
So \(\ (x+3)^{2}=x^{2}+3 x+3 x+9=x^{2}+6 x+9\).
You can also find the square of a sum of two terms by using the FOIL method.
Square the binomial. \(\ (x+5)^{2}\).
Solution
| \(\ (x+5)(x+5)\) | Multiply the binomial by itself. |
| \(\ \begin{aligned} \text { First: } x \cdot x &=x^{2} \\ \text { Outer: } x \cdot 5 &=5 x \\ \text { Inner: } 5 \cdot x &=5 x \\ \text { Last: } 5 \cdot 5 &=25 \end{aligned}\) |
|
| \(\ x^{2}+5 x+5 x+25\) | Add the terms. |
| \(\ x^{2}+10 x+25\) | Combine like terms. |
\(\ (x+5)^{2}=x^{2}+10 x+25\)
Notice that the Outer and Inner terms are the same.
So far, we have two examples of squaring a sum of two terms, one from constructing an area model and one from algebraic calculations:
\(\ \begin{array}{l}
(x+3)^{2}=x^{2}+6 x+9 \\
(x+5)^{2}=x^{2}+10 x+25
\end{array}\)
In both cases, the first term is squared to get the first term of the product. The two terms are multiplied and doubled for the middle term of the product, and the last term is squared for the last term of the product.
This pattern will hold true for the square of the sum of any two terms: Square the first term, add twice the product of the first and the last term, add the last term squared.
To square a binomial, do the following:
- Square the first term.
- Add the product of the two middle terms, times two.
- Add the square of the last term.
This process is illustrated by the following case:
\(\ (a+b)^{2}=a^{2}+2 a b+b^{2}\)
Square the binomial. \(\ (2 x+6)^{2}\)
Solution
| \(\ (2 x)^{2}=4 x^{2}\) | Square the first term. |
| \(\ (2 x)(6)(2)=24 x\) | Multiply the two terms together and double the product. |
| \(\ 6^{2}=36\) | Square the last term. |
| \(\ 4 x^{2}+24 x+36\) | Combine the terms. |
\(\ (2 x+6)^{2}=4 x^{2}+24 x+36\)
You can check this answer using the distributive property or FOIL.
Find the product: \(\ (2 s+9)^{2}\)
- \(\ 4 s^{2}+81\)
- \(\ 4 s+18\)
- \(\ 4 s^{2}+36 s+81\)
- \(\ 4 s^{2}+18 s+18\)
- Answer
-
- Incorrect. You squared the first term, and you squared the last term. But you forgot the middle term, which is 2 times the product of 2s and 9, which is 36s. The correct answer is \(\ 4 s^{2}+36 s+81\).
- Incorrect. You must square \(\ (2 s+9)\), not double it. \(\ (a+b)^{2}=a^{2}+2 a b+b^{2}\); in this example, \(\ a=2 s\) and \(\ b=9\), so \(\ (2 s+9)^{2}=4 s^{2}+36 s+81\).
- Correct. \(\ (a+b)^{2}=a^{2}+2 a b+b^{2}\); in this example, \(\ a=2 s\) and \(\ b=9\) so \(\ (2 s+9)^{2}=4 s^{2}+2(18 s)+81=4 s^{2}+36 s+81\).
- Incorrect. The terms \(\ 18 s\) and 18 are incorrect. \(\ (a+b)^{2}=a^{2}+2 a b+b^{2}\); in this example, \(\ a=2 s\) and \(\ b=9\), so \(\ (2 s+9)^{2}=4 s^{2}+36 s+81\).
Square of a Binomial Difference
Is there also a pattern when you square the difference between two terms? Yes! Since subtraction can be expressed as adding the opposite, a similar pattern does occur.
Consider the square of the binomial \(\ (x-7)\). You can use FOIL.
Square the binomial. \(\ (x-7)^{2}\)
Solution
| \(\ (x-7)(x-7)\) | Rewrite as multiplication. |
| \(\ \begin{array}{c} \text { First: } x \cdot x=x^{2} \\ \text { Outer: } x \cdot-7=-7 x \\ \text { Inner: }-7 \cdot x=-7 x \\ \text { Last: }-7 \cdot-7=49 \end{array}\) |
|
| \(\ x^{2}+(-7 x)+(-7 x)+49\) | Add the terms. |
| \(\ x^{2}-14 x+49\) | Combine the like terms. |
\(\ (x-7)^{2}=x^{2}-14 x+49\)
Notice that the pattern is similar to when you square the sum of a binomial.
To square a binomial difference, do the following:
- Square the first term
- Subtract the product of the two middle terms, times two
- Add the square of the last term.
\(\ (a-b)^{2}=a^{2}-2 a b+b^{2}\)
Square the binomial. \(\ (4 s-3)^{2}\)
Solution
| \(\ (4 s)^{2}=16 s^{2}\) | Square the first term, including the coefficient. |
| \(\ (4 s)(-3)(2)=-24 s\) | Multiply the two terms together and double the product. |
| \(\ (-3)^{2}=9\) | Square the last term. |
| \(\ 16 s^{2}-24 s+9\) | Add the terms. |
\(\ (4 s-3)^{2}=16 s^{2}-24 s+9\)
Square the binomial. \(\ (2 r-9)^{2}\)
- \(\ 4 r^{2}-81\)
- \(\ 4 r-18\)
- \(\ 4 r^{2}-36 r+81\)
- \(\ 4 r^{2}+18 r-18\)
- Answer
-
- Incorrect. You squared the first term correctly, and you squared the last term correctly, but you forgot that there is a middle term, which is \(\ (2 r)(-9)(2)=-36 r\). The correct answer is \(\ 4 r^{2}-36 r+81\).
- Incorrect. The exponent 2 means the binomial is to be squared, not doubled. The correct answer is \(\ 4 r^{2}-36 r+81\).
- Correct. \(\ (a-b)^{2}=a^{2}-2 a b+b^{2}\), and in this case \(\ a=2 r\) and \(\ b=9\). So \(\ (2 r-9)^{2}=4 r^{2}-36 r+81\).
- Incorrect. The terms \(\ 18 r\) and -18 are incorrect. \(\ (a-b)^{2}=a^{2}-2 a b+b^{2}\), and in this case \(\ a=2 r\) and \(\ b=9\). So \(\ (2 r-9)^{2}=4 r^{2}-36 r+81\).
Product of a Sum and a Difference
There is a third special product to consider among binomials: the product of the sum of two terms and the difference of the same two terms. In this case, there is also a pattern. Here is an example:
Multiply the binomials. \(\ (x+8)(x-8)\)
Solution
| \(\ \begin{array}{r} \text { First: } x \cdot x=x^{2} \\ \text { Outer: } x \cdot-8=-8 x \\ \text { Inner: } 8 \cdot x=+8 x \\ \text { Last: } 8 \cdot-8=-64 \end{array}\) |
|
| \(\ x^{2}-8 x+8 x-64\) | Add the terms. |
\(\ (x+8)(x-8)=x^{2}-64\)
Note that the answer to this binomial product is a binomial itself: the difference of two perfect squares. There is no middle term in this case. Why does this happen? The two terms are opposites and therefore add to zero.
The product of the sum of two terms \(\ (a+b)\) and the difference of the same terms \(\ (a-b)\) is the difference of the squares of the two terms.
\(\ (a+b)(a-b)=(a-b)(a+b)=a^{2}-b^{2}\)
Multiply the binomials. \(\ (2 n-5)(2 n+5)\)
Solution
| \(\ (2 n)^{2}=4 n^{2}\) | Square the first term, including the coefficient. |
| \(\ (5)^{2}=25\) | Square the last term. |
| \(\ 4 n^{2}-25\) | Take the difference. |
\(\ (2 n-5)(2 n+5)=4 n^{2}-25\)
Find the product: \(\ (2 r-9)(2 r+9)\)
- \(\ 4 r^{2}-81\)
- \(\ 4 r-18\)
- \(\ 4 r^{2}-36 r+81\)
- \(\ 4 r^{2}-36 r-81\)
- Answer
-
- Correct. \(\ (a-b)(a+b)=a^{2}-b^{2}\), and in this case \(\ a=2 r\) and \(\ b=9\). So \(\ (2 r-9)(2 r+9)=4 r^{2}-81\).
- Incorrect. Each term must be squared, not doubled. \(\ (2 r-9)(2 r+9)=4 r^{2}-81\).
- Incorrect. This is not the square of a difference. One binomial is a difference but the other is a sum. \(\ (a-b)(a+b)=a^{2}-b^{2}\), and in this case \(\ a=2 r\) and \(\ b=9\). So \(\ (2 r-9)(2 r+9)=4 r^{2}-81\).
- Incorrect. \(\ (a-b)(a+b)=a^{2}-b^{2}\), and in this case \(\ a=2 r\) and \(\ b=9\). So \(\ (2 r-9)(2 r+9)=4 r^{2}-81\).
Summary
Some products of multiplying binomials follow a predictable pattern that makes it easy to multiply them. These are known as special products. There are three special products of binomials that each follow a specific formula: squaring the sum of a binomial, squaring the difference of a binomial, and the product of a sum and a difference.