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Special Products

Square a binomial using the Binomial Squares Pattern

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.

Let's start by looking at \({(x+9)}^{2}\).
What does this mean?	\({(x+9)}^{2}\)
It means to multiply \((x+9)\) by itself.	\((x+9)(x+9)\)
Then, using FOIL, we get:	\({x}^{2}+9x+9x+81\)
Combining like terms gives:	\({x}^{2}+18x+81\)

Here's another one:	\({(y-7)}^{2}\)
Multiply \((y-7)\) by itself.	\((y-7)(y-7)\)
Using FOIL, we get:	\({y}^{2}-7y-7y+49\)
And combining like terms:	\({y}^{2}-14y+49\)

And one more:	\({(2x+3)}^{2}\)
Multiply.	\((2x+3)(2x+3)\)
Use FOIL:	\(4{x}^{2}+6x+6x+9\)
Combine like terms.	\(4{x}^{2}+12x+9\)

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.

\[{(a+b)}^{2}=\text{\underline{\hspace{1.5em}}}+\text{\underline{\hspace{1.5em}}}+\text{\underline{\hspace{1.5em}}}\]

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!

\[{(a+b)}^{2}={a}^{2}+\text{\underline{\hspace{1.5em}}}+\text{\underline{\hspace{1.5em}}}\]

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.

\[{(a+b)}^{2}=\text{\underline{\hspace{1.5em}}}+\text{\underline{\hspace{1.5em}}}+{b}^{2}\]\[\begin{array}{l}{(a+b)}^{2}=\text{\underline{\hspace{1.5em}}}+2ab+\text{\underline{\hspace{1.5em}}} \\ {(a-b)}^{2}=\text{\underline{\hspace{1.5em}}}-2ab+\text{\underline{\hspace{1.5em}}}\end{array}\]

square the first term
square the last term
double their product

	\({(10+4)}^{2}\)
Square the first term.	\({10}^{2}+\text{\underline{\hspace{1.5em}}}+\)
Square the last term.	\({10}^{2}+\text{\underline{\hspace{1.5em}}}+{4}^{2}\)
Double their product.	\({10}^{2}+2\cdot 10\cdot 4+{4}^{2}\)
Simplify.	\(100+80+16\)
Simplify.	\(196\)

\[\begin{array}{l}{(10+4)}^{2} \\ {(14)}^{2} \\ 196\end{array}\]

Example

Try it.

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

Solution


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

Example

Try it.

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

Solution


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

Example

Try it.

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

Solution


Use the pattern.
Simplify.

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Multiply Conjugates Using the Product of Conjugates Pattern

We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary.

What do you notice about these pairs of binomials?

\[\begin{array}{lllllll}(x-9)(x+9) & & & \ (y-8)(y+8) & & & \ (2x-5)(2x+5)\end{array}\]

Look at the first term of each binomial in each pair.

Notice the first terms are the same in each pair.

Look at the last terms of each binomial in each pair.

Notice the last terms are the same in each pair.

Notice how each pair has one sum and one difference.

\[\begin{array}{lllllll}(x-9)(x+9) & & & \ (y-8)(y+8) & & & \ (2x-5)(2x+5) \\ {x}^{2}+9x-9x-81 & & & \ {y}^{2}+8y-8y-64 & & & \ 4{x}^{2}+10x-10x-25 \\ {x}^{2}-81 & & & \ {y}^{2}-64 & & & \ 4{x}^{2}-25\end{array}\]\[\begin{array}{l} \\ (a+b)(a-b)={a}^{2}-\text{\underline{\hspace{1.5em}}} \\ \text{To get the}\ \text{first term, square the first term}.\end{array}\]\[\begin{array}{l} \\ (a+b)(a-b)={a}^{2}-{b}^{2} \\ \text{To get the}\ \text{last term, square the last term}.\end{array}\]

	\((10-2)(10+2)\)
It is the product of conjudgates, so the result will be the difference of two squares.	\(\text{\underline{\hspace{1.5em}}}-\text{\underline{\hspace{1.5em}}}\)
Square the first term.	\({10}^{2}-\text{\underline{\hspace{1.5em}}}\)
Square the last term.	\({10}^{2}-{2}^{2}\)
Simplify.	\(100-4\)
Simplify.	\(96\)
What do you get using the order of operations?
	\(\begin{array}{l}(10-2)(10+2) \\ (8)(12) \\ 96\end{array}\)

Example

Try it.

Multiply: \((x-8)(x+8).\)

Solution

First, recognize this as a product of conjugates. The binomials have the same first terms, and the same last terms, and one binomial is a sum and the other is a difference.

It fits the pattern.
Square the first term, x.
Square the last term, 8.
The product is a difference of squares.

Example

Try it.

Multiply: \((2x+5)(2x-5).\)

Solution

Are the binomials conjugates?

It is the product of conjugates.
Square the first term, 2x.
Square the last term, 5.
Simplify. The product is a difference of squares.

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Recognize and Use the Appropriate Special Product Pattern

We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences.

Example

Try it.

Choose the appropriate pattern and use it to find the product:

ⓐ \((2x-3)(2x+3)\) ⓑ \({(8x-5)}^{2}\) ⓒ \({(6m+7)}^{2}\) ⓓ \((5x-6)(6x+5)\)

Solution

ⓐ \((2x-3)(2x+3)\) These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern.

This fits the pattern.
Use the pattern.
Simplify.
ⓑ \({(8x-5)}^{2}\) We are asked to square a binomial. It fits the binomial squares pattern.

Use the pattern.
Simplify.
ⓒ \({(6m+7)}^{2}\) Again, we will square a binomial so we use the binomial squares pattern.

Use the pattern.
Simplify.
ⓓ \((5x-6)(6x+5)\) This product does not fit the patterns, so we will use FOIL.

\((5x-6)(6x+5)\)
Use FOIL. \(30{x}^{2}+25x-36x-30\)
Simplify. \(30{x}^{2}-11x-30\)

Key Concepts

Binomial Squares Pattern
- If \(a,b\) are real numbers,
- \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\)
- \({(a-b)}^{2}={a}^{2}-2ab+{b}^{2}\)
- To square a binomial: square the first term, square the last term, double their product.
Product of Conjugates Pattern
- If \(a,b\) are real numbers,
- \((a-b)(a+b)={a}^{2}-{b}^{2}\)
- The product is called a difference of squares.
To multiply conjugates:
- square the first term square the last term write it as a difference of squares

Special Products

Square a Binomial Using the Binomial Squares Pattern

In the following exercises, square each binomial using the Binomial Squares Pattern.

Try it.

\({(w+4)}^{2}\)

Try it.

\({(q+12)}^{2}\)

Solution

\({q}^{2}+24q+144\)

Try it.

\({(y+\frac{1}{4})}^{2}\)

Try it.

\({(x+\frac{2}{3})}^{2}\)

Solution

\({x}^{2}+\frac{4}{3}x+\frac{4}{9}\)

Try it.

\({(b-7)}^{2}\)

Try it.

\({(y-6)}^{2}\)

Solution

\({y}^{2}-12y+36\)

Try it.

\({(m-15)}^{2}\)

Try it.

\({(p-13)}^{2}\)

Solution

\({p}^{2}-26p+169\)

Try it.

\({(3d+1)}^{2}\)

Try it.

\({(4a+10)}^{2}\)

Solution

\(16{a}^{2}+80a+100\)

Try it.

\({(2q+\frac{1}{3})}^{2}\)

Try it.

\({(3z+\frac{1}{5})}^{2}\)

Solution

\(9{z}^{2}+\frac{6}{5}z+\frac{1}{25}\)

Try it.

\({(3x-y)}^{2}\)

Try it.

\({(2y-3z)}^{2}\)

Solution

\(4{y}^{2}-12yz+9{z}^{2}\)

Try it.

\({(\frac{1}{5}x-\frac{1}{7}y)}^{2}\)

Try it.

\({(\frac{1}{8}x-\frac{1}{9}y)}^{2}\)

Solution

\(\frac{1}{64}{x}^{2}-\frac{1}{36}xy+\frac{1}{81}{y}^{2}\)

Try it.

\({(3{x}^{2}+2)}^{2}\)

Try it.

\({(5{u}^{2}+9)}^{2}\)

Solution

\(25{u}^{4}+90{u}^{2}+81\)

Try it.

\({(4{y}^{3}-2)}^{2}\)

Try it.

\({(8{p}^{3}-3)}^{2}\)

Solution

\(64{p}^{6}-48{p}^{3}+9\)

Multiply Conjugates Using the Product of Conjugates Pattern

In the following exercises, multiply each pair of conjugates using the Product of Conjugates Pattern.

Try it.

\((m-7)(m+7)\)

Try it.

\((c-5)(c+5)\)

Solution

\({c}^{2}-25\)

Try it.

\((x+\frac{3}{4})(x-\frac{3}{4})\)

Try it.

\((b+\frac{6}{7})(b-\frac{6}{7})\)

Solution

\({b}^{2}-\frac{36}{49}\)

Try it.

\((5k+6)(5k-6)\)

Try it.

\((8j+4)(8j-4)\)

Solution

\(64{j}^{2}-16\)

Try it.

\((11k+4)(11k-4)\)

Try it.

\((9c+5)(9c-5)\)

Solution

\(81{c}^{2}-25\)

Try it.

\((11-b)(11+b)\)

Try it.

\((13-q)(13+q)\)

Solution

\(169-{q}^{2}\)

Try it.

\((5-3x)(5+3x)\)

Try it.

\((4-6y)(4+6y)\)

Solution

\(16-36{y}^{2}\)

Try it.

\((9c-2d)(9c+2d)\)

Try it.

\((7w+10x)(7w-10x)\)

Solution

\(49{w}^{2}-100{x}^{2}\)

Try it.

\((m+\frac{2}{3}n)(m-\frac{2}{3}n)\)

Try it.

\((p+\frac{4}{5}q)(p-\frac{4}{5}q)\)

Solution

\({p}^{2}-\frac{16}{25}{q}^{2}\)

Try it.

\((ab-4)(ab+4)\)

Try it.

\((xy-9)(xy+9)\)

Solution

\({x}^{2}{y}^{2}-81\)

Try it.

\((uv-\frac{3}{5})(uv+\frac{3}{5})\)

Try it.

\((rs-\frac{2}{7})(rs+\frac{2}{7})\)

Solution

\({r}^{2}{s}^{2}-\frac{4}{49}\)

Try it.

\((2{x}^{2}-3{y}^{4})(2{x}^{2}+3{y}^{4})\)

Try it.

\((6{m}^{3}-4{n}^{5})(6{m}^{3}+4{n}^{5})\)

Solution

\(36{m}^{6}-16{n}^{10}\)

Try it.

\((12{p}^{3}-11{q}^{2})(12{p}^{3}+11{q}^{2})\)

Try it.

\((15{m}^{2}-8{n}^{4})(15{m}^{2}+8{n}^{4})\)

Solution

\(225{m}^{4}-64{n}^{8}\)

Recognize and Use the Appropriate Special Product Pattern

In the following exercises, find each product.

Try it.

Solution

Try it.

ⓐ \({({a}^{5}-7b)}^{2}\) ⓑ \(({x}^{2}+8y)(8x-{y}^{2})\) ⓒ \(({r}^{6}+{s}^{6})({r}^{6}-{s}^{6})\) ⓓ \({({y}^{4}+2z)}^{2}\)

Try it.

ⓐ \(({x}^{5}+{y}^{5})({x}^{5}-{y}^{5})\) ⓑ \({({m}^{3}-8n)}^{2}\) ⓒ \({(9p+8q)}^{2}\) ⓓ \(({r}^{2}-{s}^{3})({r}^{3}+{s}^{2})\)

Solution

ⓐ \({x}^{10}-{y}^{10}\) ⓑ \({m}^{6}-16{m}^{3}n+64{n}^{2}\) ⓒ \(81{p}^{2}+144pq+64{q}^{2}\) ⓓ \({r}^{5}+{r}^{2}{s}^{2}-{r}^{3}{s}^{3}-{s}^{5}\)

Condensed: the full section is in OpenStax Elementary Algebra 2e.

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Kuphunzitsa (40)

Kuyesa zonse pa pepala koyamba. Kusonyeza yankho kuti aone; kutsimikizika amene angatsegulidwe mu wothamanga kwa aliyense gawo.

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

Kusonyeza yankho

Square the first term.
Square the last term.
Double the product.
Simplify.
Multiply: \({(x+9)}^{2}.\)

Kusonyeza yankho

\({x}^{2}+18x+81\)
Multiply: \({(y+11)}^{2}.\)

Kusonyeza yankho

\({y}^{2}+22y+121\)
Multiply: \({(y-3)}^{2}.\)

Kusonyeza yankho

Square the first term.
Square the last term.
Double the product.
Simplify.
Multiply: \({(x-9)}^{2}.\)

Kusonyeza yankho

\({x}^{2}-18x+81\)
Multiply: \({(p-13)}^{2}.\)

Kusonyeza yankho

\({p}^{2}-26p+169\)
Multiply: \({(4x+6)}^{2}.\)

Kusonyeza yankho

Use the pattern.
Simplify.
Multiply: \({(6x+3)}^{2}.\)

Kusonyeza yankho

\(36{x}^{2}+36x+9\)
Multiply: \({(4x+9)}^{2}.\)

Kusonyeza yankho

\(16{x}^{2}+72x+81\)
Multiply: \({(2x-3y)}^{2}.\)

Kusonyeza yankho

Use the pattern.
Simplify.
Multiply: \({(2c-d)}^{2}.\)

Kusonyeza yankho

\(4{c}^{2}-4cd+{d}^{2}\)
Multiply: \({(4x-5y)}^{2}.\)

Kusonyeza yankho

\(16{x}^{2}-40xy+25{y}^{2}\)
Multiply: \({(4{u}^{3}+1)}^{2}.\)

Kusonyeza yankho

Use the pattern.
Simplify.
Multiply: \({(2{x}^{2}+1)}^{2}.\)

Kusonyeza yankho

\(4{x}^{4}+4{x}^{2}+1\)
Multiply: \({(3{y}^{3}+2)}^{2}.\)

Kusonyeza yankho

\(9{y}^{6}+12{y}^{3}+4\)
Multiply: \((x-8)(x+8).\)

Kusonyeza yankho

First, recognize this as a product of conjugates. The binomials have the same first terms, and the same last terms, and one binomial is a sum and the other is a difference.
It fits the pattern.
Square the first term, x.
Square the last term, 8.
The product is a difference of squares.
Multiply: \((x-5)(x+5).\)

Kusonyeza yankho

\({x}^{2}-25\)
Multiply: \((w-3)(w+3).\)

Kusonyeza yankho

\({w}^{2}-9\)
Multiply: \((2x+5)(2x-5).\)

Kusonyeza yankho

Are the binomials conjugates?
It is the product of conjugates.
Square the first term, 2x.
Square the last term, 5.
Simplify. The product is a difference of squares.
Multiply: \((6x+5)(6x-5).\)

Kusonyeza yankho

\(36{x}^{2}-25\)
Multiply: \((2x+7)(2x-7).\)

Kusonyeza yankho

\(4{x}^{2}-49\)
Find the product: \((3+5x)(3-5x).\)

Kusonyeza yankho

It is the product of conjugates.
Use the pattern.
Simplify.
Multiply: \((7+4x)(7-4x).\)

Kusonyeza yankho

\(49-16{x}^{2}\)
Multiply: \((9-2y)(9+2y).\)

Kusonyeza yankho

\(81-4{y}^{2}\)
Find the product: \((5m-9n)(5m+9n).\)

Kusonyeza yankho

This fits the pattern.
Use the pattern.
Simplify.
Find the product: \((4p-7q)(4p+7q).\)

Kusonyeza yankho

\(16{p}^{2}-49{q}^{2}\)
Find the product: \((3x-y)(3x+y).\)

Kusonyeza yankho

\(9{x}^{2}-{y}^{2}\)
Find the product: \((cd-8)(cd+8).\)

Kusonyeza yankho

This fits the pattern.
Use the pattern.
Simplify.
Find the product: \((xy-6)(xy+6).\)

Kusonyeza yankho

\({x}^{2}{y}^{2}-36\)
Find the product: \((ab-9)(ab+9).\)

Kusonyeza yankho

\({a}^{2}{b}^{2}-81\)
Find the product: \((6{u}^{2}-11{v}^{5})(6{u}^{2}+11{v}^{5}).\)

Kusonyeza yankho

This fits the pattern.
Use the pattern.
Simplify.
Find the product: \((3{x}^{2}-4{y}^{3})(3{x}^{2}+4{y}^{3}).\)

Kusonyeza yankho

\(9{x}^{4}-16{y}^{6}\)
Find the product: \((2{m}^{2}-5{n}^{3})(2{m}^{2}+5{n}^{3}).\)

Kusonyeza yankho

\(4{m}^{4}-25{n}^{6}\)
Choose the appropriate pattern and use it to find the product:
ⓐ \((2x-3)(2x+3)\) ⓑ \({(8x-5)}^{2}\) ⓒ \({(6m+7)}^{2}\) ⓓ \((5x-6)(6x+5)\)
Kusonyeza yankho
1. ⓐ \((2x-3)(2x+3)\) These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern.
  
  This fits the pattern.
  Use the pattern.
  Simplify.
2. ⓑ \({(8x-5)}^{2}\) We are asked to square a binomial. It fits the binomial squares pattern.
  
  Use the pattern.
  Simplify.
3. ⓒ \({(6m+7)}^{2}\) Again, we will square a binomial so we use the binomial squares pattern.
  
  Use the pattern.
  Simplify.
4. ⓓ \((5x-6)(6x+5)\) This product does not fit the patterns, so we will use FOIL.
  
  \((5x-6)(6x+5)\)
  Use FOIL. \(30{x}^{2}+25x-36x-30\)
  Simplify. \(30{x}^{2}-11x-30\)
Choose the appropriate pattern and use it to find the product:
ⓐ \((9b-2)(2b+9)\) ⓑ \({(9p-4)}^{2}\) ⓒ \({(7y+1)}^{2}\) ⓓ \((4r-3)(4r+3)\)

Kusonyeza yankho

ⓐ FOIL; \(18{b}^{2}+77b-18\) ⓑ Binomial Squares; \(81{p}^{2}-72p+16\) ⓒ Binomial Squares; \(49{y}^{2}+14y+1\) ⓓ Product of Conjugates; \(16{r}^{2}-9\)
Choose the appropriate pattern and use it to find the product:
ⓐ \({(6x+7)}^{2}\) ⓑ \((3x-4)(3x+4)\) ⓒ \((2x-5)(5x-2)\) ⓓ \({(6n-1)}^{2}\)

Kusonyeza yankho

ⓐ Binomial Squares; \(36{x}^{2}+84x+49\) ⓑ Product of Conjugates; \(9{x}^{2}-16\) ⓒ FOIL; \(10{x}^{2}-29x+10\) ⓓ Binomial Squares; \(36{n}^{2}-12n+1\)
\({(w+4)}^{2}\)
\({(q+12)}^{2}\)

Kusonyeza yankho

\({q}^{2}+24q+144\)
\({(y+\frac{1}{4})}^{2}\)

Kusunga ntchito yanu yokha

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Zithunzi

Pitani pa dzina lililonse lachifaniziro kuti mudziwe tanthauzo lake, chithunzi chake, ndi zimene limatanthauza.

Momwe mungagwiritsire ntchito: Special Products

Square a binomial using the Binomial Squares Pattern
Multiply conjugates using the Product of Conjugates Pattern
Recognize and use the appropriate special product pattern
square the first term
square the last term
double their product
If
To square a binomial: square the first term, square the last term, double their product.

Mafunso omwe anthu amafunsa

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Zigawo za m'nkhaniyi ndi zochokera OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Kuphatikizapo ndi kufotokozanso pano; zolakwika ndi zathu. Chifukwa chakuti malemba ake ali ndi ufulu wogwiritsa ntchito popanda cholinga chogulitsa, maphunzirowa ndi aulere, sagulitsa chilichonse, ndipo amagawidwa pansi pa ufulu wogwiritsa ntchito umenewo.

Zambiri pa Algebra

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Binomial Squares	Product of Conjugates
\({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\)	\((a-b)(a+b)={a}^{2}-{b}^{2}\)
\({(a-b)}^{2}={a}^{2}-2ab+{b}^{2}\)
- Squaring a binomial	- Multiplying conjugates
- Product is a trinomial	- Product is a binomial
- Inner and outer terms with FOIL are the same.	- Inner and outer terms with FOIL are opposites.
- Middle term is double the product of the terms.	- There is no middle term.

	\((5x-6)(6x+5)\)
Use FOIL.	\(30{x}^{2}+25x-36x-30\)
Simplify.	\(30{x}^{2}-11x-30\)