Example 7.30.
- \(\displaystyle (w^5)^2 = w^5 \cdot w^5 = w^{10}\)
- \(\displaystyle (4x^3)^2 = 4x^3 \cdot 4x^3 = 4 \cdot 4 \cdot x^3 \cdot x^3 = 16x^6\)
Section 7.4 Special Products and Factors
Subsection Squares of Monomials
A few special binomial products occur so frequently that it is useful to recognize their forms. This will enable you to write their factored forms directly, without trial and error. To prepare for these special products, we first consider the squares of monomials.
Study the squares of monomials in Example 7.30. Do you see a quick way to find the product?
Look Closer.
In Example 7.30a, we doubled the exponent and kept the same base. In Example 7.30b, we squared the numerical coefficient and doubled the exponent.
Reading Questions Reading Questions
1.
Why do we double the exponent when we square a power?
Answer.
Because when we multiply powers with the same base, we add the exponents.
Example 7.31.
Find a monomial whose square is \(36t^8\text{.}\)
Solution.
When we square a power, we double the exponent, so \(t^8\) is the square of \(t^4\text{.}\) Because 36 is the square of 6, the monomial we want is \(6t^4\text{.}\) To check our result, we square \(6t^4\) to see that \((6t^4)^2=36t^8\text{.}\)
Subsection Squares of Binomials
You can use the distributive law to verify each of the following special products.
\begin{align*}
(a+b)^2 \amp =(a+b)(a+b) \amp \amp \amp (a-b)^2 \amp =(a-b)(a-b)\\
\amp = a^2+ab+ab+b^2 \amp \amp \amp \amp =a^2-ab-ab+b^2\\
\amp = a^2+2ab+b^2 \amp \amp \amp \amp =a^2-2ab+b^2
\end{align*}
Squares of Binomials.
- \(\displaystyle \blert{(a+b)^2=a^2+2ab+b^2}\)
- \(\displaystyle \blert{(a-b)^2=a^2-2ab+b^2}\)
Reading Questions Reading Questions
2.
Explain why it is NOT true that \((a+b)^2=a^2+b^2\text{.}\)
Answer.
Because we must square \((a+b)\) using FOIL.
We can use these results as formulas to compute the square of any binomial.
Example 7.32.
Expand \(~(2x+3)^2~\) as a polynomial.
Solution.
The formula for the square of a sum says to square the first term, add twice the product of the two terms, then add the square of the second term. We replace \(a\) by \(2x\) and \(b\) by \(3\) in the formula.
\begin{align*}
(a+b)^2 \amp = a^2 ~~~~~~ + ~~~~~~ 2 a b~~~~~~ + ~~~~~~b^2\\
(2x+3)^2 \amp = (2x)^2 + ~~2 (2x)(3) + ~~~(3)^2\\
\amp ~~\blert{\text{square of}} ~~~~~ \blert{\text{twice their}} ~~~~~ \blert{\text{square of}}\\
\amp ~~\blert{\text{first term}} ~~~~~~~ \blert{\text{product}} ~~~~~ \blert{\text{second term}}\\
\amp = 4x^2 ~~~~ + ~~~~~~ 12x ~~~~~ + ~~~~~ 9
\end{align*}
Of course, you can verify that you will get the same answer for Example 7.32 if you compute the square by multiplying \((2x+3)(2x+3)\text{.}\)
Caution 7.33.
We cannot square a binomial by squaring each term separately! For example, it is NOT true that
\begin{equation*}
(2x+3)^2 = 4x^2+9~~~~~~\alert{\text{Incorrect!}}
\end{equation*}
We must use the distributive law to multiply the binomial times itself.
Reading Questions Reading Questions
3.
How do we compute \((a+b)^2\text{?}\)
Answer.
\(a^2+2ab+b^2\)
Subsection Difference of Two Squares
Now consider the product
\begin{equation*}
(a+b)(a-b) = a^2 - ab + ab - b^2
\end{equation*}
In this product, the two middle terms cancel each other, and we are left with a difference of two squares.
Difference of Two Squares.
\begin{equation*}
\blert{(a+b)(a-b)=a^2-b^2}
\end{equation*}
Example 7.34.
Multiply \(~(2y+9w)(2y-9w)~\)
Solution.
The product has the form \((a+b)(a-b)\text{,}\) with \(a\) replaced by \(2y\) and \(b\) replaced by \(9w\text{.}\) We use the difference of squares formula to write the product as a polynomial.
\begin{align*}
(a+b)(a-b) \amp = ~~a^2 ~~~~~~ - ~~~~~~b^2\\
(2y+9w)(2y-9w) \amp = (2y)^2 ~~ - ~~~(9w)^2\\
\amp ~~\blert{\text{square of}} ~~~~~ \blert{\text{square of}}\\
\amp ~~\blert{\text{first term}} ~~~~~ \blert{\text{second term}}\\
\amp = 4y^2 ~~~~ - ~~~~~ 81w^2
\end{align*}
Reading Questions Reading Questions
4.
Explain the difference between \((a-b)^2\) and \(a^2-b^2\text{.}\)
Answer.
To simplify \((a-b)^2\text{,}\) we must square the binomial.
Subsection Factoring Special Products
The three special products we have just studied are useful as patterns for factoring certain polynomials. For factoring, we view the formulas from right to left.
Special Factorizations.
- \(\displaystyle \blert{a^2+2ab+b^2=(a+b)^2}\)
- \(\displaystyle \blert{a^2-2ab+b^2=(a-b)^2}\)
- \(\displaystyle \blert{a^2-b^2=(a+b)(a-b)}\)
If we recognize one of the special forms, we can use the formula to factor it. Notice that all three special products involve two squared terms, \(a^2\) and \(b^2\text{,}\) so we first look for two squared terms in our trinomial.
Example 7.35.
Factor \(~x^2+24x+144\)
Solution.
This trinomial has two squared terms, \(x^2\) and \(144\text{.}\) These terms are \(a^2\) and \(b^2\text{,}\) so \(a=x\) and \(b=12\text{.}\) We check whether the middle term is equal to \(2ab\text{.}\)
\begin{equation*}
2ab=2(x)(12) = 24x
\end{equation*}
This is the correct middle term, so our trinomial has the form (1), with \(a=x\) and \(b=12\text{.}\) Thus,
\begin{align*}
a^2+2ab+b^2 \amp = (a+b)^2 \amp \amp \blert{\text{Replace}~a~\text{by}~x~\text{and}~b~ \text{by}~12.}\\
x^2+24x+144 \amp = (x+12)^2
\end{align*}
Reading Questions Reading Questions
5.
How can we factor \(a^2- 2ab + b^2\text{?}\)
Answer.
\((a-b)^2\)
6.
How can we factor \(a^2- b^2\text{?}\)
Answer.
\((a+b)(a-b)\)
Caution 7.36.
The sum of two squares, \(a^2+b^2\text{,}\) cannot be factored! For example,
\begin{equation*}
x^2+16,~~~~~~9x^2+4y^2,~~~~~~\text{and}~~~~~~25y^4+w^4
\end{equation*}
cannot be factored. You can check, for instance, that \(x^2+16 \not= (x+4)(x+4)\text{.}\)
Sum of Two Squares.
The sum of two squares, \(~a^2+b^2~\text{,}\) cannot be factored.
As always when factoring, we should check first for common factors.
Example 7.37.
Factor completely \(~98-28x^4+2x^8\)
Solution.
Each term has a factor of 2, so we begin by factoring out 2.
\begin{equation*}
98-28x^4+2x^8 = 2(49-14x^4+x^8)
\end{equation*}
The polynomial in parentheses has the form \((a-b)^2\text{,}\) with \(a=7\) and \(b=x^4\text{.}\) The middle term is
\begin{equation*}
-2ab=-2(7)(x^4)
\end{equation*}
We use equation (2) to write
\begin{align*}
a^2+2ab+b^2 \amp = (a-b)^2 \amp \amp \blert{\text{Replace}~a~\text{by}~7~\text{and}~b~ \text{by}~x^4.}\\
49-14x^4+x^8 \amp = (7-x^4)^2
\end{align*}
Thus,
\begin{equation*}
98-28x^4+2x^8 = 2(7-x^4)^2
\end{equation*}
Reading Questions Reading Questions
7.
What expression involving squares cannot be factored??
Answer.
\(a^2+b^2\)
Subsection Skills Warm-Up
Exercises Exercises
Exercise Group.
Express each product as a polynomial.
1.
\((z-3)^2\)2.
\((x+4)^2\)3.
\((3a+5)^2\)4.
\((2b-7)^2\)5.
\((2n-5)(2n+5)\)6.
\((4m+9)(4m-9)\)Subsubsection Answers to Skills Warm-Up
- \(\displaystyle z^2-6z+9\)
- \(\displaystyle x^2+8x+16\)
- \(\displaystyle 9a^2+30a+25\)
- \(\displaystyle 4b^2-28b+49\)
- \(\displaystyle 4n^2-25\)
- \(\displaystyle 16m^2-81\)
Subsection Lesson
Subsubsection Activity 1: Special Products
Exercises Exercises
1.
Simplify each square.
- \(\displaystyle (-6h^6)^2\)
- \(\displaystyle (12st^8)^2\)
2.
Find a monomial whose square is given.
- \(\displaystyle 64b^6\)
- \(\displaystyle 169a^4b^{16}\)
3.
Use a formula to expand each product.
- \(\displaystyle (4-3t)^2\)
- \(\displaystyle (6s+2t)^2\)
- \(\displaystyle (5x^4+4)(5x^4-4)\)
Subsubsection Activity 2: Special Factorizations
Exercises Exercises
1.
Use one of the three formulas to factor each polynomial.
- \(\displaystyle 25y^2-w^2\)
- \(\displaystyle m^6-18m^3+81\)
- \(\displaystyle 4h^2+36hk+81k^2\)
2.
Factor completely.
- \(\displaystyle x^6-16x^2\)
- \(\displaystyle 3b^9+18b^6+27b^3\)
- \(\displaystyle 2x^5y-20x^3y^3+50xy^3\)
Subsubsection Wrap-Up
Objectives.
In this Lesson we practiced the following skills:
- Identifying squares of monomials
- Using the fromulas for squares of binomials and difference of squares
- Factoring squares of binomials and difference of squares
- Knowing that the sum of squares cannot be factored
Questions.
- Explain the difference between squaring \(3b^3\) and squaring \(3+b^3\text{.}\)
- How can you check whether a trinomial might be the square of a binomial?
- Explain why we can factor the difference of two squares, but we cannot factor the sum of two squares.
Subsection Homework Preview
Exercises Exercises
Exercise Group.
Expand each product.
1.
\((4a-5b)^2\)
2.
\((4a+5b)^2\)
3.
\((4a+5b)(4a-5b)\)
Exercise Group.
Factor.
4.
\(36x^2+96x+64\)
5.
\(36x^2-96x+64\)
6.
\(36x^2-64\)
Subsubsection Answers to Homework Preview
- \(\displaystyle 16a^2-40ab+25b^2\)
- \(\displaystyle 16a^2+40ab+25b^2\)
- \(\displaystyle 16a^2-25b^2\)
- \(\displaystyle 4(3x+4)^2\)
- \(\displaystyle 4(3x-4)^2\)
- \(\displaystyle 4(3x+4)(3x-4)\)
Exercises Homework 7.4
1.
Square each monomial.
- \(\displaystyle (8t^4)^2\)
- \(\displaystyle (-12a^2)^2\)
- \(\displaystyle (10h^2k)^2\)
Answer.
- \(\displaystyle 64t^8\)
- \(\displaystyle 144a^4\)
- \(\displaystyle 100h^4k^2\)
2.
Find a monomial whose square is given.
- \(\displaystyle 16b^{16}\)
- \(\displaystyle 121z^{22}\)
- \(\displaystyle 36p^6q^{24}\)
Exercise Group.
For Problems 3–4, write the area of the square in two different ways:
- as the sum of four smaller areas,
- as one large square, using the formula Area = (length)\(^2\text{.}\)
3.
Answer.
- \(\displaystyle x^2+8x+8x+64\)
- \(\displaystyle (x+8)^2\)
4.
Exercise Group.
For Problems 5–7, compute the product.
5.
\((n+1)(n+1)\)Answer.
\(n^2+2n+1\)
6.
\((m-9)(m-9)\)7.
\((2b+5c)(2b+5c)\)Answer.
\(4b^2+20bc+25c^2\)
Exercise Group.
For Problems 8–13, compute the product.
8.
\((x+1)^2\)9.
\((2x-3)^2\)Answer.
\(4x^2-12+9\)
10.
\((5x+2y)^2\)11.
\((3a+b)^2\)Answer.
\(9a^2+6ab+b^2\)
12.
\((7b^3-6)^2\)13.
\((2h+5k^4)^2\)Answer.
\(4h^2+20hk^4+25k^8\)
Exercise Group.
For Problems 14–19, use the formula for difference of two squares to multiply.
14.
\((x-4)(x+4)\)15.
\((x+5z)(x-5z)\)Answer.
\(x^2-25z^2\)
16.
\((2x-3)(2x+3)\)17.
\((3p-4)(3p+4)\)Answer.
\(9p^2-16\)
18.
\((2x^2-1)(2x^2+1)\)19.
\((h^2+7t)(h^2-7t)\)Answer.
\(h^4-29t^2\)
Exercise Group.
For Problems 20–22, multiply. Write your answer as a polynomial.
20.
\(-2a(3a-5)^2\)21.
\(4x^2(2x+6y)^2\)Answer.
\(16x^4+96x^3y+144x^2y^2\)
22.
\(5mp^2(2m^2-p)(2m^2+p)\)Exercise Group.
For Problems 23–25, factor the squares of binomials.
23.
\(y^2+6y+9\)Answer.
\((y+3)^2\)
24.
\(m^2-30m+225\)25.
\(a^6-4a^3b+4b^2\)Answer.
\((a^3-2b)^2\)
Exercise Group.
For Problems 26–31, factor.
26.
\(z^2-64\)27.
\(1-g^2\)Answer.
\((1-g)(1+g)\)
28.
\(-225+a^2\)29.
\(x^2-9\)Answer.
\((x-3)(x+3)\)
30.
\(36-a^2b^2\)31.
\(64y^2-49x^2\)Answer.
\((8y-7x)(8y+7x)\)
Exercise Group.
For Problems 32–40, factor completely.
32.
\(a^4+10a^2+25\)33.
\(36y^8-49\)Answer.
\((6y^4+7)(6y^4-7)\)
34.
\(16x^6-9y^4\)35.
\(3a^2-75\)Answer.
\(3(a+5)(a-5)\)
36.
\(2a^3-12a^2+18a\)37.
\(9x^7-81x^3\)Answer.
\(9x^3(x^2+3)(x^2-3)\)
38.
\(12h^2+3k^6\)39.
\(81x^8-y^4\)Answer.
\((9x^4+y^2)(3x^2+y)(3x^2-y)\)
40.
\(162a^4b^8-2a^8\)41.
Is \(x-3)^2\) equivalent to \(x^2-3^2\text{?}\) Explain why or why not, and give a numerical example to justify your answer.
Answer.
No. Let \(x=1\text{,}\) then \((x-1)^2=(-2)^2=4\text{,}\) but \(x^2-3^2=1-9=-8\)
42.
Use areas to explain why the figure illustrates the product \((a+b)^2 = a^2+2ab+b^2.\)
43.
Use evaluation to decide whether the two expressions \((a+b)^2\) and \(a^2+b^2\) are equivalent.
| \(a\) | \(b\) | \(a+b\) | \((a+b)^2\) | \(a^2\) | \(b^2\) | \(a^2+b^2\) |
| \(2\) | \(3\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) |
| \(-2\) | \(-3\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) |
| \(2\) | \(-3\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) | \(\hphantom{0000}\) |
Answer.
No.
44.
Explain why you can factor \(x^2-4\text{,}\) but you cannot factor \(x^2+4\text{.}\)
45.
- Expand \((a-b)^3\) by multiplying.
- Use your formula from part (a) to expand \((2x-3)^3\)
- Substitute \(a=5\) and \(b=2\) to show that \((a-b)^3\) is not equivalent to \(a^3-b^3\text{.}\)
Answer.
- \(\displaystyle a^3-3a^2b+3ab^2-b^3\)
- \(\displaystyle 8x^3-36x^2+54x-27\)
- \((a-b)^3=27\text{,}\) \(a^3-b^3=117\)
46.
- Multiply \((a+b)(a^2-ab+b^2)\text{.}\)
- Factor \(a^3+b^3\)
- Factor \(x^3+8\text{.}\)
