# 4.6 Special binomial products

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Operations with algebraic expressions and numerical evaluations are introduced in this chapter. Coefficients are described rather than merely defined. Special binomial products have both literal and symbolic explanations and since they occur so frequently in mathematics, we have been careful to help the student remember them. In each example problem, the student is "talked" through the symbolic form.Objectives of this module: be able to expand (a + b)^2, (a - b)^2, and (a + b)(a - b).

## Overview

• Expanding ${\left(a+b\right)}^{2}$ and ${\left(a-b\right)}^{2}$
• Expanding $\left(a+b\right)\left(a-b\right)$

Three binomial products occur so frequently in algebra that we designate them as special binomial products . We have seen them before (Sections [link] and [link] ), but we will study them again because of their importance as time saving devices and in solving equations (which we will study in a later chapter).

These special products can be shown as the squares of a binomial

${\left(a+b\right)}^{2}$      and      ${\left(a-b\right)}^{2}$

and as the sum and difference of two terms .

$\left(a+b\right)\left(a-b\right)$

There are two simple rules that allow us to easily expand (multiply out) these binomials. They are well worth memorizing, as they will save a lot of time in the future.

## Squaring a binomial

To square a binomial: $*$

1. Square the first term.
2. Take the product of the two terms and double it.
3. Square the last term.
4. Add the three results together.

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

## Sum and difference of two terms

To expand the sum and difference of two terms: $†$

1. Square the first term and square the second term.
2. Subtract the square of the second term from the square of the first term.

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

$*$ See problems 56 and 57 at the end of this section.
$†$ See problem 58.

## Sample set a

Note that ${\left(x+4\right)}^{2}\ne {x}^{2}+{4}^{2}$ . The $8x$ term is missing!

Notice that the sign of the last term in this expression is “ $+$ .” This will always happen since the last term results from a number being squared . Any nonzero number times itself is always positive.

The sign of the second term in the trinomial will always be the sign that occurs inside the parentheses.

$\begin{array}{ll}\left(x+6\right)\left(x-6\right)\hfill & \text{Square}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{first}\text{\hspace{0.17em}}\text{term:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}.\hfill \\ \hfill & \text{Subtract}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{second}\text{\hspace{0.17em}}\text{term}\text{\hspace{0.17em}}\left(36\right)\text{\hspace{0.17em}}\text{from}\hfill \\ \hfill & \text{the}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{first}\text{\hspace{0.17em}}\text{term:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}-36.\hfill \\ \left(x+6\right)\left(x-6\right)={x}^{2}-36\hfill & \hfill \end{array}$

$\begin{array}{ll}\left(4a-12\right)\left(4a+12\right)\hfill & \text{Square}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{first}\text{\hspace{0.17em}}\text{term:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}16{a}^{2}.\hfill \\ \hfill & \text{Subtract}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{second}\text{\hspace{0.17em}}\text{term}\text{\hspace{0.17em}}\left(144\right)\text{\hspace{0.17em}}\text{from}\hfill \\ \hfill & \text{the}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{first}\text{\hspace{0.17em}}\text{term:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}16{a}^{2}-144.\hfill \\ \left(4a-12\right)\left(4a+12\right)=16{a}^{2}-144\hfill & \hfill \end{array}$

$\begin{array}{ll}\left(6x+8y\right)\left(6x-8y\right)\hfill & \text{Square}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{first}\text{\hspace{0.17em}}\text{term:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}36{x}^{2}.\hfill \\ \hfill & \text{Subtract}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{second}\text{\hspace{0.17em}}\text{term}\text{\hspace{0.17em}}\left(64{y}^{2}\right)\text{\hspace{0.17em}}\text{from}\hfill \\ \hfill & \text{the}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{first}\text{\hspace{0.17em}}\text{term:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}36{x}^{2}-64{y}^{2}.\hfill \\ \left(6x+8y\right)\left(6x-8y\right)=36{x}^{2}-64{y}^{2}\hfill & \hfill \end{array}$

## Practice set a

Find the following products.

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

${x}^{2}+10x+25$

${\left(x+7\right)}^{2}$

${x}^{2}+14x+49$

${\left(y-6\right)}^{2}$

${y}^{2}-12y+36$

${\left(3a+b\right)}^{2}$

$9{a}^{2}+6ab+{b}^{2}$

${\left(9m-n\right)}^{2}$

$81{m}^{2}-18mn+{n}^{2}$

${\left(10x-2y\right)}^{2}$

$100{x}^{2}-40xy+4{y}^{2}$

${\left(12a-7b\right)}^{2}$

$144{a}^{2}-168ab+49{b}^{2}$

${\left(5h-15k\right)}^{2}$

$25{h}^{2}-150hk+225{k}^{2}$

## Exercises

For the following problems, find the products.

${\left(x+3\right)}^{2}$

${x}^{2}+6x+9$

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

${\left(x+8\right)}^{2}$

${x}^{2}+16x+64$

${\left(x+6\right)}^{2}$

${\left(y+9\right)}^{2}$

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

${\left(y+1\right)}^{2}$

${\left(a-4\right)}^{2}$

${a}^{2}-8a+16$

${\left(a-6\right)}^{2}$

${\left(a-7\right)}^{2}$

${a}^{2}-14a+49$

${\left(b+10\right)}^{2}$

${\left(b+15\right)}^{2}$

${b}^{2}+30b+225$

${\left(a-10\right)}^{2}$

${\left(x-12\right)}^{2}$

${x}^{2}-24x+144$

${\left(x+20\right)}^{2}$

${\left(y-20\right)}^{2}$

${y}^{2}-40y+400$

${\left(3x+5\right)}^{2}$

${\left(4x+2\right)}^{2}$

$16{x}^{2}+16x+4$

${\left(6x-2\right)}^{2}$

${\left(7x-2\right)}^{2}$

$49{x}^{2}-28x+4$

${\left(5a-6\right)}^{2}$

${\left(3a-9\right)}^{2}$

$9{a}^{2}-54a+81$

${\left(3w-2z\right)}^{2}$

${\left(5a-3b\right)}^{2}$

$25{a}^{2}-30ab+9{b}^{2}$

${\left(6t-7s\right)}^{2}$

${\left(2h-8k\right)}^{2}$

$4{h}^{2}-32hk+64{k}^{2}$

${\left(a+\frac{1}{2}\right)}^{2}$

${\left(a+\frac{1}{3}\right)}^{2}$

${a}^{2}+\frac{2}{3}a+\frac{1}{9}$

${\left(x+\frac{3}{4}\right)}^{2}$

${\left(x+\frac{2}{5}\right)}^{2}$

${x}^{2}+\frac{4}{5}x+\frac{4}{25}$

${\left(x-\frac{2}{3}\right)}^{2}$

${\left(y-\frac{5}{6}\right)}^{2}$

${y}^{2}-\frac{5}{3}y+\frac{25}{36}$

${\left(y+\frac{2}{3}\right)}^{2}$

${\left(x+1.3\right)}^{2}$

${x}^{2}+2.6x+1.69$

${\left(x+5.2\right)}^{2}$

${\left(a+0.5\right)}^{2}$

${a}^{2}+a+0.25$

${\left(a+0.08\right)}^{2}$

${\left(x-3.1\right)}^{2}$

${x}^{2}-6.2x+9.61$

${\left(y-7.2\right)}^{2}$

${\left(b-0.04\right)}^{2}$

${b}^{2}-0.08b+0.0016$

${\left(f-1.006\right)}^{2}$

$\left(x+5\right)\left(x-5\right)$

${x}^{2}-25$

$\left(x+6\right)\left(x-6\right)$

$\left(x+1\right)\left(x-1\right)$

${x}^{2}-1$

$\left(t-1\right)\left(t+1\right)$

$\left(f+9\right)\left(f-9\right)$

${f}^{2}-81$

$\left(y-7\right)\left(y+7\right)$

$\left(2y+3\right)\left(2y-3\right)$

$4{y}^{2}-9$

$\left(5x+6\right)\left(5x-6\right)$

$\left(2a-7b\right)\left(2a+7b\right)$

$4{a}^{2}-49{b}^{2}$

$\left(7x+3t\right)\left(7x-3t\right)$

$\left(5h-2k\right)\left(5h+2k\right)$

$25{h}^{2}-4{k}^{2}$

$\left(x+\frac{1}{3}\right)\left(x-\frac{1}{3}\right)$

$\left(a+\frac{2}{9}\right)\left(a-\frac{2}{9}\right)$

${a}^{2}-\frac{4}{81}$

$\left(x+\frac{7}{3}\right)\left(x-\frac{7}{3}\right)$

$\left(2b+\frac{6}{7}\right)\left(2b-\frac{6}{7}\right)$

$4{b}^{2}-\frac{36}{49}$

Expand ${\left(a+b\right)}^{2}$ to prove it is equal to ${a}^{2}+2ab+{b}^{2}$ .

Expand ${\left(a-b\right)}^{2}$ to prove it is equal to ${a}^{2}-2ab+{b}^{2}$ .

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

Expand $\left(a+b\right)\left(a-b\right)$ to prove it is equal to ${a}^{2}-{b}^{2}$ .

Fill in the missing label in the equation below.

first term squared

Label the parts of the equation below.

Label the parts of the equation below.

(a) Square the first term.
(b) Square the second term and subtract it from the first term.

## Exercises for review

( [link] ) Simplify ${\left({x}^{3}{y}^{0}{z}^{4}\right)}^{5}$ .

( [link] ) Find the value of ${10}^{-1}\cdot {2}^{-3}$ .

$\frac{1}{80}$

( [link] ) Find the product. $\left(x+6\right)\left(x-7\right)$ .

( [link] ) Find the product. $\left(5m-3\right)\left(2m+3\right)$ .

$10{m}^{2}+9m-9$

( [link] ) Find the product. $\left(a+4\right)\left({a}^{2}-2a+3\right)$ .

#### Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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a perfect square v²+2v+_
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or infinite solutions?
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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J, combine like terms 7x-4y
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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