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Section 7.5 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?

Example 7.30.
  1. \((w^5)^2 = w^5 \cdot w^5 = w^{10}\)
  2. \((4x^3)^2 = 4x^3 \cdot 4x^3 = 4 \cdot 4 \cdot x^3 \cdot x^3 = 16x^6\)
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?

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.
  1. \(\blert{(a+b)^2=a^2+2ab+b^2}\)
  2. \(\blert{(a-b)^2=a^2-2ab+b^2}\)

Reading Questions Reading Questions

2.

Explain why it is NOT true that \((a+b)^2=a^2+b^2\text{.}\)

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 \((a+b)^2\text{?}\)

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 \(a^2-b^2\text{.}\)

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.
  1. \(\blert{a^2+2ab+b^2=(a+b)^2}\)
  2. \(\blert{a^2-2ab+b^2=(a-b)^2}\)
  3. \(\blert{a^2-b^2=(a+b)(a-b)}\)

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 \(a^2- 2ab + b^2\text{?}\)

6.

How can we factor \(a^2- b^2\text{?}\)

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??

Subsection Skills Warm-Up

Exercises Exercises

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.
\((4m+9)(4m-9)\)

Solutions Answers to Skills Warm-Up

Exercises Exercises

Exercises Homework 7.4

1.

Square each monomial.

  1. \((8t^4)^2\)
  2. \((-12a^2)^2\)
  3. \((10h^2k)^2\)
2.

Find a monomial whose square is given.

  1. \(16b^{16}\)
  2. \(121z^{22}\)
  3. \(36p^6q^{24}\)

For Problems 3–4, write the area of the square in two different ways:

  1. as the sum of four smaller areas,
  2. as one large square, using the formula Area = (length)\(^2\text{.}\)

3.
rectangle
4.
rectangle

For Problems 5–7, compute the product.

5.
\((n+1)(n+1)\)
6.
\((m-9)(m-9)\)
7.
\((2b+5c)(2b+5c)\)

For Problems 8–13, compute the product.

8.
\((x+1)^2\)
9.
\((2x-3)^2\)
10.
\((5x+2y)^2\)
11.
\((3a+b)^2\)
12.
\((7b^3-6)^2\)
13.
\((2h+5k^4)^2\)

For Problems 14–19, use the formula for difference of two squares to multiply.

14.
\((x-4)(x+4)\)
15.
\((x+5z)(x-5z)\)
16.
\((2x-3)(2x+3)\)
17.
\((3p-4)(3p+4)\)
18.
\((2x^2-1)(2x^2+1)\)
19.
\((h^2+7t)(h^2-7t)\)

For Problems 20–22, multiply. Write your answer as a polynomial.

20.
\(-2a(3a-5)^2\)
21.
\(4x^2(2x+6y)^2\)
22.
\(5mp^2(2m^2-p)(2m^2+p)\)

For Problems 23–25, factor the squares of binomials.

23.
\(y^2+6y+9\)
24.
\(m^2-30m+225\)
25.
\(a^6-4a^3b+4b^2\)

For Problems 26–31, factor.

26.
\(z^2-64\)
27.
\(1-g^2\)
28.
\(-225+a^2\)
29.
\(x^2-9\)
30.
\(36-a^2b^2\)
31.
\(64y^2-49x^2\)

For Problems 32–40, factor completely.

32.
\(a^4+10a^2+25\)
33.
\(36y^8-49\)
34.
\(16x^6-9y^4\)
35.
\(3a^2-75\)
36.
\(2a^3-12a^2+18a\)
37.
\(9x^7-81x^3\)
38.
\(12h^2+3k^6\)
39.
\(81x^8-y^4\)
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.

42.

Use areas to explain why the figure illustrates the product \((a+b)^2 = a^2+2ab+b^2.\)

rectangles
43.

Use evaluation to decide whether the two expressions \((a+b)^2\) and \(a^2+b^2\) are equivalent.

\(a\) \(b\) \(a+b\) \((a+b)^2\) \(a^2\) \(b^2\) \(a^2+b^2\)
\(2\) \(3\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
\(-2\) \(-3\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
\(2\) \(-3\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
44.

Explain why you can factor \(x^2-4\text{,}\) but you cannot factor \(x^2+4\text{.}\)

45.
  1. Expand \((a-b)^3\) by multiplying.
  2. Use your formula from part (a) to expand \((2x-3)^3\)
  3. Substitute \(a=5\) and \(b=2\) to show that \((a-b)^3\) is not equivalent to \(a^3-b^3\text{.}\)
46.
  1. Multiply \((a+b)(a^2-ab+b^2)\text{.}\)
  2. Factor \(a^3+b^3\)
  3. Factor \(x^3+8\text{.}\)