6.5
Factoring Special Forms
In this section we revisit two special product forms that we learned in Chapter 5, the first of which was squaring a binomial.
Squaring a binomial. Here are two earlier rules for squaring a binomial. 1. (a + b)2 = a2 + 2ab + b2 2. (a − b)2 = a2 − 2ab + b2
Perfect Square Trinomials
To square a binomial such as (a + b)2, proceed as follows: 1. Square the first term: a2
2. Multiply the first and second term, then double: 2ab 3. Square the last term: b2
You Try It!
EXAMPLE 1. Expand: (2x + 3y)2
Expand: (5a + 2b)2
Solution: Using the pattern (a+ b)2
= a2
+ 2ab + b2
, we can expand (2x+ 3y)2
as follows:
(2x + 3y)2= (2x)2+ 2(2x)(3y) + (3y)2 = 4x2
+ 6xy + 9y2
Note how we square the first and second terms, then produce the middle term of our answer by multiplying the first and second terms and doubling.
Answer: 25a2
+ 20ab + 4b2
You Try It!
EXAMPLE 2. Expand: (3u2
− 5v2)2
Expand: (2s3 − 7t)2
Solution: Using the pattern (a−b)2
= a2
−2ab+b2, we can expand (3u2
−5v2)2 as follows: (3u2 − 5v2)2 = (3u2 )2 − 2(3u2)(5v2 ) + (5v2 )2 = 9u4 − 30u2v2 + 25v4
Note that the sign of the middle term is negative this time. The first and last
terms are still positive because we are squaring. Answer: 4s6
− 28s3t + 49t2 Once you’ve squared a few binomials, it’s time to do all of the work in your
head. (i) Square the first term; (ii) multiply the first and second term and double the result; and (iii) square the second term.
You Try It!
EXAMPLE 3. Expand each of the following: Expand: (5x4
− 3)2
a) (2y − 3)2
b) (4a − 3b)2
c) (x3
+ 5)2
Solution: Using the pattern (a ± b)2
= a2
± 2ab + b2, we expand each binomially mentally, writing down the answer without any intermediate steps. a) (2y − 3)2= 4y2 − 12y + 9 b) (4a − 3b)2 = 16a2 − 24ab + 9b2 c) (x3 + 5)2 = x6 + 10x3 + 25 Answer: 25x8 − 30x4+ 9 Now, because factoring is “unmultiplying,” it should be a simple matter to
reverse the process ofExample 3.
You Try It! EXAMPLE 4. Factor each of the following trinomials: Factor: 25x8
− 30x4+ 9 a) 4y2 − 12y + 9 b) 16a2 − 24ab + 9b2 c) x6 + 10x3 + 25 Solution: Because of the work already done inExample 3, it is a simple task to factor each of these trinomials.
a) 4y2 − 12y + 9 = (2y − 3)2 b) 16a2 − 24ab + 9b2 = (4a − 3b)2 c) x6 + 10x3 + 25 = (x3 + 5)2 Answer: (5x4 − 3)2
Each of the trinomials inExample 4is an example of a perfect square trinomial.
Perfect square trinomial. If a trinomial a2+ 2ab + b2 is the square of a
binomial, as in (a + b)2, then the trinomial is called a perfect square trinomial.
So, how does one recognize a perfect square trinomial? If the first and last terms of a trinomial are perfect squares, then you should suspect that you may be dealing with a perfect square trinomial. However, you also have to have the correct middle term in order to have a perfect square trinomial.
You Try It!
EXAMPLE 5. Factor each of the following trinomials: Factor: 16x2 + 72x + 81 a) 9x2 − 42x + 49 b) 49a2 + 70ab + 25b2 c) 4x2 − 37x + 9 Solution: Note that the first and last terms of each trinomial are perfect squares.
a) In the trinomial 9x2
− 42x + 49, note that (3x)2= 9x2
and 72
= 49. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 9x2
− 42x + 49 factors as follows: 9x2− 42x + 49= (3x − 7)? 2
However, we must check to see if the middle term is correct. Multiply 3x and 7, then double: 2(3x)(7) = 42x. Thus, the middle term is correct and therefore
9x2
− 42x + 49 = (3x − 7)2. b) In the trinomial 49a2
+70ab+25b2
, note that (7a)2
= 49a2
and (5b)2
= 25b2
. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 49a2
+ 70ab + 25b2
factors as follows: 49a2
+ 70ab + 25b2 ?
= (7a + 5b)2
However, we must check to see if the middle term is correct. Multiply 7a and 5b, then double: 2(7a)(5b) = 70ab. Thus, the middle term is correct and therefore 49a2 + 70ab + 25b2 = (7a + 5b)2 . List of Squares n n2 0 0 1 1 2 4 3 9 4 16 5 25 6 36 7 49 8 64 9 81 10 100 11 121 12 144 13 169 14 196 15 225 16 256 17 289 18 324 19 361 20 400 21 441 22 484 23 529 24 576 25 625 c) In the trinomial 4x2
− 37x + 9, note that (2x)2= 4x2and (3)2= 9. Hence,
the first and last terms are perfect squares. Taking the square roots, we suspect that 4x2
− 37x + 9 factors as follows: 4x2− 37x + 9= (2x − 3)? 2
However, we must check to see if the middle term is correct. Multiply 2x and 3, then double: 2(2x)(3) = 12x. However, this is not the middle term of 4x2
− 37x + 9, so this factorization is incorrect! We must find another way to factor this trinomial.
Comparing 4x2
− 37x + 9 with ax2+ bx + c, we need a pair of integers whose product is ac = 36 and whose sum is b = −37. The integer pair −1 and −36 comes to mind. Replace the middle term as a sum of like terms using this ordered pair.
4x2
− 37x + 9 = 4x2− x − 36x + 9 −37x = −x − 36x.
= x(4x − 1) − 9(4x − 1) Factor by grouping.
= (x − 9)(4x − 1) Factor out 4x − 1.
This example clearly demonstrates how important it is to check the middle term.
Answer: (4x + 9)2
Remember the first rule of factoring!
The first rule of factoring. The first step to perform in any factoring problem is factor out the GCF.
You Try It! EXAMPLE 6. Factor each of the following trinomials: Factor: −4x3
− 24x2− 36x a) 2x3 y + 12x2 y2 + 18xy3 b) −4x5 + 32x4 − 64x3 Solution: Remember, first factor out the GCF.
a) In the trinomial 2x3y + 12x2y2+ 18xy3, we note that the GCF of 2x3y,
12x2y2, and 18xy3 is 2xy. We first factor out 2xy.
2x3 y + 12x2 y2 + 18xy3 = 2xy(x2 + 6xy + 9y2 )
We now note that the first and last terms of the resulting trinomial factor are perfect squares, so we take their square roots and factors as follows.
= 2xy(x + 3y)2
Of course, the last factorization is correct only if the middle term is correct. Because 2(x)(3y) = 6xy matches the middle term of x2
+ 6xy + 9y2
, we do have a perfect square trinomial and our result is correct.
b) In the trinomial −4x5
+ 32x4
− 64x3, we note that the GCF of 4x5
, 32x4
, and 64x3
is 4x3
. We first factor out 4x3
.
−4x5+ 32x4− 64x3= 4x3(−x2+ 8x − 16) However, the first and third terms of −x2
+ 8x − 16 are negative, and thus are not perfect squares. Let’s begin again, this time factoring out −4x3.
−4x5+ 32x4
− 64x3= −4x3
(x2
− 8x + 16) This time the first and third terms of x2
− 8x + 16 are perfect squares. We take their square roots and write:
= −4x3
(x − 4)2
Again, this last factorization is correct only if the middle term is correct. Because 2(x)(4) = 8x, we do have a perfect square trinomial and our result is correct.
Answer: −4x(x + 3)2
The Difference of Squares
The second special product form we learned in Chapter 5 was the difference of squares.
The difference of squares. Here is the difference of squares rule. (a + b)(a − b) = a2− b2
If you are multiplying two binomials which have the exact same terms in the “First” positions and the exact same terms in the “Last” positions, but one set is separated by a plus sign while the other set is separated by a minus sign, then multiply as follows:
1. Square the first term: a2
2. Square the second term: b2
You Try It!
EXAMPLE 7. Expand each of the following: Expand: (4x − 3y)(4x + 3y)
a) (3x + 5)(3x − 5) b) (a3
− 2b3)(a3+ 2b3)
Solution: We apply the difference of squares pattern to expand each of the given problems.
a) In (3x + 5)(3x − 5), we have the exact same terms in the “First” and “Last” positions, with the first set separated by a plus sign and the second set separated by a minus sign.
a) Square the first term: (3x)2= 9x2
b) Square the second term: 52= 25
c) Place a minus sign between the two squares. Hence:
(3x + 5)(3x − 5) = 9x2
− 25 b) In (a3
− 2b3)(a3+ 2b3), we have the exact same terms in the “First” and
“Last” positions, with the first set separated by a minus sign and the second set separated by a plus sign.
a) Square the first term: (a3)2= a6
b) Square the second term: (2b3)2= 4b6
c) Place a minus sign between the two squares. Hence: (a3 − 2b3)(a3 + 2b3 ) = a6 − 4b6 Answer: 16x2 − 9y2 Because factoring is “unmultiplying,” is should be a simple matter to reverse
the process ofExample 7.
You Try It!
EXAMPLE 8. Factor each of the following: Factor: 81x2
− 49 a) 9x2
− 25 b) a6
− 4b6
Solution: Because of the work already done in Example 7, it is a simple matter to factor (or “unmultiply”) each of these problems.
a) 9x2 − 25 = (3x + 5)(3x − 5) b) a6 − 4b6= (a3 − 2b3)(a3 + 2b3 )
In each case, note how we took the square roots of each term, then separated one set with a plus sign and the other with a minus sign. Because of the commutative property of multiplication, it does not matter which one you make plus and which one you make minus.
Answer: (9x + 7)(9x − 7)
Always remember the first rule of factoring.
The first rule of factoring. The first step to perform in any factoring problem is factor out the GCF.
You Try It!
EXAMPLE 9. Factor: x3
− 9x Factor: 4x4
− 16x2
Solution: In x3
− 9x, the GCF of x3 and 9x is x. Factor out x. x3− 9x = x(x2− 9)
Note that x2
− 9 is now the difference of two perfect squares. Take the square roots of x2
and 9, which are x and 3, then separate one set with a plus sign and the other set with a minus sign.
= x(x + 3)(x − 3) Answer: 4x2
(x + 2)(x − 2)
Factoring Completely
Sometimes after one pass at factoring, factors remain that can be factored further. You must continue to factor in this case.
You Try It!
EXAMPLE 10. Factor: x4
− 16 Factor: x4
− 81
Solution: In x4
− 16, we have the difference of two squares: (x2)2= x4 and
42= 16. First, we take the square roots, then separate one set with a plus sign
and the other set with a minus sign. x4
− 16 = (x2+ 4)(x2
Note that x2
+4 is the sum of two squares and does not factor further. However, x2
− 4 is the difference of two squares. Take the square roots, x and 2, then separate one set with a plus sign and the other set with a minus sign.
= (x2
+ 4)(x + 2)(x − 2)
Done. We cannot factor further. Answer:
(x2
+ 9)(x + 3)(x − 3)
Nonlinear Equations Revisited
Remember, if an equation is nonlinear, the first step is to make one side equal to zero by moving all terms to one side of the equation. Once you’ve completed this important first step, factor and apply the zero product property to find the solutions.
You Try It!
EXAMPLE 11. Solve for x: 25x2= 169 Solve for x: 16x2= 121
Solution: Make one side equal to zero, factor, then apply the zero product property.
25x2
= 169 Original equation.
25x2
− 169 = 0 Subtract 169 from both sides.
Note that we have two perfect squares separated by a minus sign. This is the difference of squares pattern. Take the square roots, making one term plus and one term minus.
(5x + 13)(5x − 13) = 0 Use difference of squares to factor.
Use the zero product property to complete the solution, setting each factor equal to zero and solving the resulting equations.
5x + 13 = 0 or 5x − 13 = 0 x = −13
5 x =
13 5 Hence, the solutions of 25x2
= 169 are x = −13/5 and x = 13/5. We encourage
readers to check each of these solutions. Answer: −11/4, 11/4
You Try It!
EXAMPLE 12. Solve for x: 49x2
+ 81 = 126x Solve for x:
25x2
= 80x − 64
Solution: Make one side equal to zero, factor, then apply the zero product property.
One can also argue that the only number whose square is zero is the number zero. Hence, one can go directly from
(7x − 9)2 = 0 to
7x − 9 = 0.
Hence, the only solution of 49x2
+ 81 = 126x is x = 9/7.
49x2
+ 81 = 126x Original equation.
49x2
− 126x + 81 = 0 Subtract 126x from both sides.
Note that the first and last terms of the trinomial are perfect squares. Hence, it make sense to try and factor as a perfect square trinomial, taking the square roots of the first and last terms.
(7x − 9)2
= 0 Factor as a perfect square trinomial.
Of course, be sure to check the middle term. Because −2(7x)(9) = −126x, the middle term is correct. Because (7x − 9)2
= (7x − 9)(7x − 9), we can use the zero product property to set each factor equal to zero and solve the resulting equations. 7x − 9 = 0 or 7x − 9 = 0 x = 9 7 x = 9 7 Hence, the only solution of 49x2
+ 81 = 126x is x = 9/7. We encourage readers to check this solution.
Answer: 8/5
You Try It!
EXAMPLE 13. Solve for x: 2x3
+ 3x2 = 50x + 75 Solve for x: 5x3 + 36 = x2 + 180x
Solution: Make one side equal to zero, factor, then apply the zero product property. 2x3 + 3x2 = 50x + 75 Original equation. 2x3 + 3x2
− 50x − 75 = 0 Make one side zero.
This is a four-term expression, so we try factoring by grouping. Factor x2 out
of the first two terms, and −25 out of the second two terms. x2
(2x + 3) − 25(2x + 3) = 0 Factor by grouping.
(x2
Complete the factorization by using the difference of squares to factor x2
− 25. (x + 5)(x − 5)(2x + 3) = 0 Use difference of squares to factor.
Finally, use the zero product property. Set each factor equal to zero and solve for x.
x + 5 = 0 or x − 5 = 0 or 2x + 3 = 0
x = −5 x = 5 x = −3
2 Hence, the solutions of 2x3
+ 3x2
= 50x + 75 are x = −5, x = 5, and x = −3/2.
We encourage readers to check each of these solutions. Answer: −6, 6, 1/5
Let’s solve another nonlinear equation, matching the algebraic and
graphi-cal solutions.
You Try It! EXAMPLE 14. Solve the equation x3= 4x, both algebraically and graphi- Solve the equation x3= 16x
both algebraically and graphically, then compare your answers.
cally, then compare your answers.
Solution: Note that we have a power of x larger than one, so the equation x3
= 4x is nonlinear. Make one side zero and factor. x3= 4x Original equation.
x3
− 4x = 0 Nonlinear. Make one side zero.
x(x2
− 4) = 0 Factor out GCF.
x(x + 2)(x − 2) = 0 Apply difference of squares.
Note that we now have a product of three factors that equals zero. The zero product property says that at least one of these factors must equal zero.
x = 0 or x + 2 = 0 or x − 2 = 0
x = −2 x = 2
Hence, the solutions of x3
= 4x are x = 0, x = −2, and x = 2.
Graphical solution. Load y = x3 and y = 4x into Y1 and Y2 in the
Y= menu of your calculator. Select 6:ZStandard from the ZOOM menu to produce the graph inFigure 6.26.
Although the image in Figure 6.26 shows all three points of intersection, adjusting the WINDOW parameters as shown in Figure 6.27, then pressing the GRAPH button will produce a nicer view of the points of intersection, as shown in the figure on the right inFigures 6.27.
Figure 6.26: Sketching y = x3and y = 4x.
Figure 6.27: Adjusting the viewing window.
Use the 5:intersect tool from the CALC menu to find the three points of intersection. Press the ENTER key in response to “First curve,” then press ENTER again in response to “Second curve,” then use the left-arrow key to move your cursor close to the leftmost point of intersection and press ENTER in response to “Guess.” The result is shown in the first image on the left in Figure 6.28. Repeat the process to find the remaining points of intersection. The results are shown in the last two images inFigure 6.28.
Figure 6.28: Finding the points of intersection.
Thus, the graphical solutions are x = −2, x = 0, and x = 2.
Reporting the solution on your homework: Duplicate the image in your calculator’s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves.
• Label the horizontal and vertical axes with x and y, respectively (see Figure 6.29).
• Place your WINDOW parameters at the end of each axis (seeFigure 6.29). • Label the graph with its equation (see Figure 6.29).
• Drop dashed vertical lines through each x-intercept. Shade and label the x-values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation x3
= 4x (seeFigure 6.29). −5 5 15 −15 x y −2 0 2 y = 4x y = x3
Figure 6.29: Reporting your graphical solution on your homework.
Finally, note that the graphical solutions x = −2, x = 0, and x = 2 match our algebraic solutions exactly.
Answer: −4, 0, 4
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Exercises
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In Exercises1-8, expand each of the given expressions. 1. (8r − 3t)2 2. (6a + c)2 3. (4a + 7b)2 4. (4s + t)2 5. (s3 − 9)2 6. (w3 + 7)2 7. (s2 + 6t2 )2 8. (7u2 − 2w2)2
In Exercises9-28, factor each of the given expressions. 9. 25s2 + 60st + 36t2 10. 9u2 + 24uv + 16v2 11. 36v2 − 60vw + 25w2 12. 49b2 − 42bc + 9c2 13. a4+ 18a2b2+ 81b4 14. 64u4 − 144u2w2 + 81w4 15. 49s4 − 28s2t2 + 4t4 16. 4a4 − 12a2c2 + 9c4 17. 49b6 − 112b3+ 64 18. 25x6 − 10x3+ 1 19. 49r6 + 112r3 + 64 20. a6 − 16a3+ 64 21. 5s3 t − 20s2t2+ 20st3 22. 12r3 t − 12r2t2+ 3rt3 23. 8a3c + 8a2c2+ 2ac3 24. 18x3 z − 60x2 z2 + 50xz3 25. −48b3 + 120b2 − 75b 26. −45c3 + 120c2 − 80c 27. −5u5 − 30u4− 45u3 28. −12z5 − 36z4− 27z3
In Exercises29-36, expand each of the given expressions. 29. (21c + 16)(21c − 16) 30. (19t + 7)(19t − 7) 31. (5x + 19z)(5x − 19z) 32. (11u + 5w)(11u − 5w) 33. (3y4+ 23z4)(3y4 − 23z4) 34. (5x3+ z3)(5x3 − z3) 35. (8r5+ 19s5)(8r5 − 19s5) 36. (3u3 + 16v3 )(3u3 − 16v3)
In Exercises37-60, factor each of the given expressions. 37. 361x2 − 529 38. 9b2 − 25 39. 16v2 − 169 40. 81r2 − 169 41. 169x2 − 576y2 42. 100y2 − 81z2 43. 529r2 − 289s2 44. 49a2 − 144b2 45. 49r6 − 256t6 46. 361x10 − 484z10 47. 36u10 − 25w10 48. a6 − 81c6 49. 72y5 − 242y3 50. 75y5 − 147y3 51. 1444a3 b − 324ab3 52. 12b3 c − 1875bc3 53. 576x3 z − 1156xz3 54. 192u3 v − 507uv3 55. 576t4 − 4t2 56. 4z5 − 256z3 57. 81x4 − 256 58. 81x4 − 1 59. 81x4 − 16 60. x4 − 1
In Exercises61-68, factor each of the given expressions completely. 61. z3 + z2 − 9z − 9 62. 3u3 + u2 − 48u − 16 63. x3 − 2x2y − xy2 + 2y3 64. x3 + 2x2 z − 4xz2 − 8z3 65. r3 − 3r2t − 25rt2 + 75t3 66. 2b3 − 3b2c − 50bc2 + 75c3 67. 2x3 + x2 − 32x − 16 68. r3 − 2r2− r + 2
In Exercises69-80, solve each of the given equations for x. 69. 2x3+ 7x2= 72x + 252 70. 2x3+ 7x2= 32x + 112 71. x3+ 5x2= 64x + 320 72. x3+ 4x2= 49x + 196 73. 144x2 + 121 = 264x 74. 361x2 + 529 = 874x 75. 16x2= 169 76. 289x2= 4 77. 9x2= 25 78. 144x2= 121 79. 256x2 + 361 = −608x 80. 16x2 + 289 = −136x
In Exercises81-84, perform each of the following tasks:
i) Use a strictly algebraic technique to solve the given equation.
ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown inExample 12.
81. x3= x 82. x3= 9x 83. 4x3= x 84. 9x3= x
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Answers
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1. 64r2 − 48rt + 9t2 3. 16a2 + 56ab + 49b2 5. s6 − 18s3+ 81 7. s4 + 12s2 t2 + 36t4 9. (5s + 6t)2 11.(6v − 5w)2 13.(a2+ 9b2)2 15.(7s2 − 2t2)2 17.(7b3 − 8)2 19.(7r3 + 8)2 21.5st(s − 2t)2 23.2ac(2a + c)2 25. −3b(4b − 5)2 27. −5u3(u + 3)2 29.441c2 − 256 31.25x2 − 361z2 33.9y8 − 529z8 35.64r10 − 361s10 37.(19x + 23)(19x − 23) 39.(4v + 13)(4v − 13) 41.(13x + 24y)(13x − 24y) 43.(23r + 17s)(23r − 17s) 45.(7r3 + 16t3 )(7r3 − 16t3) 47.(6u5 + 5w5 )(6u5 − 5w5) 49.2y3 (6y + 11)(6y − 11) 51.4ab(19a + 9b)(19a − 9b) 53.4xz(12x + 17z)(12x − 17z) 55.4t2 (12t + 1)(12t − 1) 57.(9x2 + 16)(3x + 4)(3x − 4) 59.(9x2 + 4)(3x + 2)(3x − 2) 61.(z + 3)(z − 3)(z + 1) 63.(x + y)(x − y)(x − 2y) 65.(r + 5t)(r − 5t)(r − 3t) 67.(x + 4)(x − 4)(2x + 1)69.x = −6, 6, −7 2 71.x = −8, 8, −5 73.x = 11 12 75.x = −13 4 , 13 4 77.x = −5 3, 5 3 79.x = −19 16 81.x = 0, −1, 1 83.x = 0, −1/2, 1/2
