# 6.4 Special products

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By the end of this section, you will be able to:
• Square a binomial using the Binomial Squares Pattern
• Multiply conjugates using the Product of Conjugates Pattern
• Recognize and use the appropriate special product pattern

Before you get started, take this readiness quiz.

1. Simplify: ${9}^{2}$ ${\left(-9\right)}^{2}$ $\text{−}{9}^{2}.$
If you missed this problem, review [link] .

## Square a binomial using the binomial squares pattern

Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial    twice and using the methods of the last section, there is less work to do if you learn to use a pattern.

$\begin{array}{}\\ \\ \text{Let’s start by looking at}\phantom{\rule{0.2em}{0ex}}{\left(x+9\right)}^{2}.\hfill & & & \\ \text{What does this mean?}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(x+9\right)}^{2}\hfill \\ \text{It means to multiply}\phantom{\rule{0.2em}{0ex}}\left(x+9\right)\phantom{\rule{0.2em}{0ex}}\text{by itself.}\hfill & & & \phantom{\rule{4em}{0ex}}\left(x+9\right)\left(x+9\right)\hfill \\ \text{Then, using FOIL, we get:}\hfill & & & \phantom{\rule{4em}{0ex}}{x}^{2}+9x+9x+81\hfill \\ \text{Combining like terms gives:}\hfill & & & \phantom{\rule{4em}{0ex}}{x}^{2}+18x+81\hfill \\ \\ \\ \\ \text{Here’s another one:}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(y-7\right)}^{2}\hfill \\ \text{Multiply}\phantom{\rule{0.2em}{0ex}}\left(y-7\right)\phantom{\rule{0.2em}{0ex}}\text{by itself.}\hfill & & & \phantom{\rule{4em}{0ex}}\left(y-7\right)\left(y-7\right)\hfill \\ \text{Using FOIL, we get:}\hfill & & & \phantom{\rule{4em}{0ex}}{y}^{2}-7y-7y+49\hfill \\ \text{And combining like terms:}\hfill & & & \phantom{\rule{4em}{0ex}}{y}^{2}-14y+49\hfill \\ \\ \\ \\ \text{And one more:}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(2x+3\right)}^{2}\hfill \\ \text{Multiply.}\hfill & & & \phantom{\rule{4em}{0ex}}\left(2x+3\right)\left(2x+3\right)\hfill \\ \text{Use FOIL:}\hfill & & & \phantom{\rule{4em}{0ex}}4{x}^{2}+6x+6x+9\hfill \\ \text{Combine like terms.}\hfill & & & \phantom{\rule{4em}{0ex}}4{x}^{2}+12x+9\hfill \end{array}$

Look at these results. Do you see any patterns?

What about the number of terms? In each example we squared a binomial and the result was a trinomial    .

${\left(a+b\right)}^{2}=\text{____}+\text{____}+\text{____}$

Now look at the first term in each result. Where did it come from?

The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term!

${\left(a+b\right)}^{2}={a}^{2}+\text{____}+\text{____}$

To get the first term of the product, square the first term .

Where did the last term come from? Look at the examples and find the pattern.

The last term is the product of the last terms, which is the square of the last term.

${\left(a+b\right)}^{2}=\text{____}+\text{____}+{b}^{2}$

To get the last term of the product, square the last term .

Finally, look at the middle term . Notice it came from adding the “outer” and the “inner” terms—which are both the same! So the middle term is double the product of the two terms of the binomial.

$\begin{array}{c}{\left(a+b\right)}^{2}=\text{____}+2ab+\text{____}\hfill \\ {\left(a-b\right)}^{2}=\text{____}-2ab+\text{____}\hfill \end{array}$

To get the middle term of the product, multiply the terms and double their product .

Putting it all together:

## Binomial squares pattern

If $a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b$ are real numbers,

$\begin{array}{}\\ \\ {\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}\hfill \\ {\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}\hfill \end{array}$

To square a binomial:

• square the first term
• square the last term
• double their product

A number example helps verify the pattern.

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(10+4\right)}^{2}\hfill \\ \text{Square the first term.}\hfill & & & \phantom{\rule{4em}{0ex}}{10}^{2}+\text{___}+\text{___}\hfill \\ \text{Square the last term.}\hfill & & & \phantom{\rule{4em}{0ex}}{10}^{2}+\text{___}+{4}^{2}\hfill \\ \text{Double their product.}\hfill & & & \phantom{\rule{4em}{0ex}}{10}^{2}+2·10·4+{4}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}100+80+16\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}196\hfill \end{array}$

To multiply ${\left(10+4\right)}^{2}$ usually you’d follow the Order of Operations.

$\begin{array}{c}\hfill {\left(10+4\right)}^{2}\hfill \\ \hfill {\left(14\right)}^{2}\hfill \\ \hfill 196\hfill \end{array}$

The pattern works!

Multiply: ${\left(x+5\right)}^{2}.$

## Solution

 Square the first term. Square the last term. Double the product. Simplify.

Multiply: ${\left(x+9\right)}^{2}.$

${x}^{2}+18x+81$

Multiply: ${\left(y+11\right)}^{2}.$

${y}^{2}+22y+121$

Multiply: ${\left(y-3\right)}^{2}.$

## Solution

 Square the first term. Square the last term. Double the product. Simplify.

Multiply: ${\left(x-9\right)}^{2}.$

${x}^{2}-18x+81$

Multiply: ${\left(p-13\right)}^{2}.$

${p}^{2}-26p+169$

Multiply: ${\left(4x+6\right)}^{2}.$

## Solution

 Use the pattern. Simplify.

Multiply: ${\left(6x+3\right)}^{2}.$

$36{x}^{2}+36x+9$

Multiply: ${\left(4x+9\right)}^{2}.$

$16{x}^{2}+72x+81$

Multiply: ${\left(2x-3y\right)}^{2}.$

## Solution

 Use the pattern. Simplify.

Multiply: ${\left(2c-d\right)}^{2}.$

$4{c}^{2}-4cd+{d}^{2}$

Multiply: ${\left(4x-5y\right)}^{2}.$

$16{x}^{2}-40xy+25{y}^{2}$

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snigdha
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hello, I have algebra phobia. Subtracting negative numbers always seem to get me confused.
what do you need help in?
Felix
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look at the numbers if they have different signs, it's like subtracting....but you keep the sign of the largest number...
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for example.... -19 + 7.... different signs...subtract.... 12 keep the sign of the "largest" number 19 is bigger than 7.... 19 has the negative sign... Therefore, -12 is your answer...
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—12
Thanks Felix.l also get confused with signs.
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Thank you for this
Shatey
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think about it like you lost $19 (-19), then found$7(+7). Totally you lost just $12 (-12) Annushka I used to struggle a lot with negative numbers and math in general what I typically do is look at it in terms of money I have -$5 in my account I then take out 5 more dollars how much do I have in my account well-\$10 ... I also for a long time would draw it out on a number line to visualize it
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practicing with smaller numbers to understand then working with larger numbers helps too and the song/rhyme same sign add and keep opposite signs subtract keep the sign of the bigger # then you'll be exact
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john is 16. wanjiru is 11.
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1y x 15y
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I believe it's x^2-3x+2
NerdNamedGerg
because the X's multiply by the -2 and the -1 and than combine like terms
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find the equation whose roots are -1 and 4
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Gee
find the equation whose roots are -2 and -1
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Gee
yeah
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there's a chatting option in the app wow
Nana
That's cool cool
Nana
Nice to meet you all
Nana
you too.
Joan
😃
Nana
Hey you all there are several Free Apps that can really help you to better solve type Equations.
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Debra, which apps specifically. ..?
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am having a course in elementary algebra ,any recommendations ?
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Samuel Addai, me too at ucc elementary algebra as part of my core subjects in science
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me too as part of my core subjects in R M E
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at ABETIFI COLLEGE OF EDUCATION
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ok great. Good to know.
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Thks ranu
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the previous equation should be 3x = 1/6 x=1/18
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vida
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a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya
a trader gains 20 rupees loses 42 rupees and then gains 10 rupees Express algebraically the result of his three transactions
vinaya