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Elementary Algebra

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?

Example 7.30.

  1. (w5)2=w5w5=w10
  2. (4x3)2=4x34x3=44x3x3=16x6

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 36t8.
Solution.
When we square a power, we double the exponent, so t8 is the square of t4. Because 36 is the square of 6, the monomial we want is 6t4. To check our result, we square 6t4 to see that (6t4)2=36t8.

Subsection Squares of Binomials

You can use the distributive law to verify each of the following special products.
(a+b)2=(a+b)(a+b)(ab)2=(ab)(ab)=a2+ab+ab+b2=a2abab+b2=a2+2ab+b2=a22ab+b2

Squares of Binomials.

  1. (a+b)2=a2+2ab+b2
  2. (ab)2=a22ab+b2

Reading Questions Reading Questions

2.
Explain why it is NOT true that (a+b)2=a2+b2.
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.
(a+b)2=a2      +      2ab      +      b2(2x+3)2=(2x)2+  2(2x)(3)+   (3)2  square of     twice their     square of  first term       product     second term=4x2    +      12x     +     9
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).

Caution 7.33.

We cannot square a binomial by squaring each term separately! For example, it is NOT true that
(2x+3)2=4x2+9      Incorrect!
We must use the distributive law to multiply the binomial times itself.

Reading Questions Reading Questions

3.
How do we compute (a+b)2?
Answer.
a2+2ab+b2

Subsection Difference of Two Squares

Now consider the product
(a+b)(ab)=a2ab+abb2
In this product, the two middle terms cancel each other, and we are left with a difference of two squares.

Difference of Two Squares.

(a+b)(ab)=a2b2

Example 7.34.

Multiply  (2y+9w)(2y9w) 
Solution.
The product has the form (a+b)(ab), with a replaced by 2y and b replaced by 9w. We use the difference of squares formula to write the product as a polynomial.
(a+b)(ab)=  a2            b2(2y+9w)(2y9w)=(2y)2     (9w)2  square of     square of  first term     second term=4y2         81w2

Reading Questions Reading Questions

4.
Explain the difference between (ab)2 and a2b2.
Answer.
To simplify (ab)2, 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.

  1. a2+2ab+b2=(a+b)2
  2. a22ab+b2=(ab)2
  3. a2b2=(a+b)(ab)
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, a2 and b2, so we first look for two squared terms in our trinomial.

Example 7.35.

Factor  x2+24x+144
Solution.
This trinomial has two squared terms, x2 and 144. These terms are a2 and b2, so a=x and b=12. We check whether the middle term is equal to 2ab.
2ab=2(x)(12)=24x
This is the correct middle term, so our trinomial has the form (1), with a=x and b=12. Thus,
a2+2ab+b2=(a+b)2Replace a by x and b by 12.x2+24x+144=(x+12)2

Reading Questions Reading Questions

5.
How can we factor a22ab+b2?
Answer.
(ab)2
6.
How can we factor a2b2?
Answer.
(a+b)(ab)

Caution 7.36.

The sum of two squares, a2+b2, cannot be factored! For example,
x2+16,      9x2+4y2,      and      25y4+w4
cannot be factored. You can check, for instance, that x2+16(x+4)(x+4).

Sum of Two Squares.

The sum of two squares,  a2+b2 , cannot be factored.
As always when factoring, we should check first for common factors.

Example 7.37.

Factor completely  9828x4+2x8
Solution.
Each term has a factor of 2, so we begin by factoring out 2.
9828x4+2x8=2(4914x4+x8)
The polynomial in parentheses has the form (ab)2, with a=7 and b=x4. The middle term is
2ab=2(7)(x4)
We use equation (2) to write
a2+2ab+b2=(ab)2Replace a by 7 and b by x4.4914x4+x8=(7x4)2
Thus,
9828x4+2x8=2(7x4)2

Reading Questions Reading Questions

7.
What expression involving squares cannot be factored?
Answer.
a2+b2

Subsection Skills Warm-Up

Exercises Exercises

Exercise Group.
Express each product as a polynomial.
5.
(2n5)(2n+5)
6.
(4m+9)(4m9)

Subsubsection Answers to Skills Warm-Up

  1. z26z+9
  2. x2+8x+16
  3. 9a2+30a+25
  4. 4b228b+49
  5. 4n225
  6. 16m281

Subsection Lesson

Subsubsection Activity 1: Special Products

Exercises Exercises
2.
Find a monomial whose square is given.
  1. 64b6
  2. 169a4b16
3.
Use a formula to expand each product.
  1. (43t)2
  2. (6s+2t)2
  3. (5x4+4)(5x44)

Subsubsection Activity 2: Special Factorizations

Exercises Exercises
1.
Use one of the three formulas to factor each polynomial.
  1. 25y2w2
  2. m618m3+81
  3. 4h2+36hk+81k2
2.
Factor completely.
  1. x616x2
  2. 3b9+18b6+27b3
  3. 2x5y20x3y3+50xy3

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.
  1. Explain the difference between squaring 3b3 and squaring 3+b3.
  2. How can you check whether a trinomial might be the square of a binomial?
  3. 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.
Exercise Group.
Factor.

Subsubsection Answers to Homework Preview

  1. 16a240ab+25b2
  2. 16a2+40ab+25b2
  3. 16a225b2
  4. 4(3x+4)2
  5. 4(3x4)2
  6. 4(3x+4)(3x4)

Exercises Homework 7.4

1.

Square each monomial.
  1. (8t4)2
  2. (12a2)2
  3. (10h2k)2

2.

Find a monomial whose square is given.
  1. 16b16
  2. 121z22
  3. 36p6q24

Exercise Group.

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.

Exercise Group.

For Problems 5–7, compute the product.
6.
(m9)(m9)
7.
(2b+5c)(2b+5c)

Exercise Group.

For Problems 8–13, compute the product.

Exercise Group.

For Problems 14–19, use the formula for difference of two squares to multiply.
14.
(x4)(x+4)
15.
(x+5z)(x5z)
16.
(2x3)(2x+3)
17.
(3p4)(3p+4)
18.
(2x21)(2x2+1)
19.
(h2+7t)(h27t)

Exercise Group.

For Problems 20–22, multiply. Write your answer as a polynomial.
20.
2a(3a5)2
22.
5mp2(2m2p)(2m2+p)

Exercise Group.

For Problems 23–25, factor the squares of binomials.
24.
m230m+225
25.
a64a3b+4b2

Exercise Group.

For Problems 26–31, factor.

Exercise Group.

For Problems 32–40, factor completely.
36.
2a312a2+18a
40.
162a4b82a8

41.

Is (x3)2 equivalent to x232 ? 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=a2+2ab+b2.
rectangles

43.

Use evaluation to decide whether the two expressions (a+b)2 and a2+b2 are equivalent.
a b a+b (a+b)2 a2 b2 a2+b2
2 3 0000 0000 0000 0000 0000
2 3 0000 0000 0000 0000 0000
2 3 0000 0000 0000 0000 0000

44.

Explain why you can factor x24, but you cannot factor x2+4.

45.

  1. Expand (ab)3 by multiplying.
  2. Use your formula from part (a) to expand (2x3)3
  3. Substitute a=5 and b=2 to show that (ab)3 is not equivalent to a3b3.

46.

  1. Multiply (a+b)(a2ab+b2).
  2. Factor a3+b3
  3. Factor x3+8.