Download 216 Exercise 4 Factoring by Grouping 8–5 The General Quadratic

216
Chapter 8
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Factors and Factoring
Exercise 4
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Factoring by Grouping
Factor completely.
1.
3.
5.
7.
8.
9.
10.
11.
12.
a3 3a2 4a 12
x3 x2 x 1
x2 bx 3x 3b
3x 2y 6 xy
x2y2 3x2 4y2 12
x2 y2 2xy 4
x2 6xy 9y2 a2
m2 n2 4 4n
p2 r2 6pq 9q2
2. x3 x2 x 1
4. 2x3 x2 4x 2
6. ab a b 1
Expressions with Three or More Letters
Factor completely.
13. ax bx 3a 3b
14. x2 xy xz yz
15. a2 ac ab bc
16. ay by ab y2
17. 2a bx2 2b ax2
18. 2xy wy wz 2xz
19. 6a2 2ab 3ac bc
20. x2 bx cx bc
21. b2 bc ab ac
22. bx cx bc x2
23. x2 a2 2ab b2
24. p2 y2 x2 2xy
8–5
The General Quadratic Trinomial
When we multiply the two binomials (ax b) and (cx d), we get a trinomial with a leading
coefficient of ac, a middle coefficient of (ad bc), and a constant term of bd.
General
Quadratic
Trinomial
(ax b)(cx d ) acx2 (ad bc)x bd
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The general quadratic trinomial may be factored by trial and error or by the grouping method.
Trial and Error
To factor the general quadratic trinomial by trial and error, we look for four numbers, a, b, c,
and d, such that
ac the leading coefficient
ad bc the middle coefficient
bd the constant term
Also, the signs in the factors are found in the same way as for trinomials with a leading
coefficient of 1.
Section 8–5
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217
The General Quadratic Trinomial
Example 33: Factor 2x2 5x 3.
Solution: The leading coefficient ac is 2, and the constant term bd is 3. Try
a 1,
c 2,
b 3,
and
d1
Then ad bc 1(1) 3(2) 7. No good. It is supposed to be 5. We next try
a 1, c 2,
b 1,
and
d3
Then ad bc 1(3) 1(2) 5. This works. So
2x2 5x 3 (x 1)(2x 3)
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Grouping Method
The grouping method eliminates the need for trial and error.
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Some people have a knack for
factoring and can quickly factor
a trinomial by trial and error.
Other rely on the longer but surer
grouping method.
Example 34: Factor 3x2 16x 12.
Solution:
1. Multiply the leading coefficient and the constant term.
3(12) 36
2. Find two numbers whose product equals 36 and whose sum equals the middle coefficient, 16. Two such numbers are 2 and 18.
3. Rewrite the trinomial, splitting the middle term according to the selected factors (16x 2x 18x).
3x2 2x 18x 12
Group the first two terms together and the last two terms together.
(3x2 2x) (18x 12)
4. Remove common factors from each grouping.
x(3x 2) 6(3x 2)
5. Remove the common factor (3x 2) from the entire expression:
(3x 2)(x 6)
which are the required factors.
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It is easy to make a mistake when factoring out a negative
quantity. Thus when we factored out a 6 in going from
Step 3 to Step 4 in Example 34, we got
Common
Error
(18x 12) 6(3x 2)
but not
6(3x 2)
incorrect!
Notice that if we had grouped the terms differently in Step 3 of Example 34, we would have
arrived at the same factors as before.
3x2 18x 2x 12
3x(x 6) 2(x 6)
(3x 2)(x 6)
Sometimes an expression may be simplified before factoring, as we did in Sec. 8–3.
218
Chapter 8
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Factors and Factoring
Example 35: Factor 12(x y)6n (x y)3n z 6z2.
Solution: If we substitute a (x y)3n, our expression becomes
12a2 az 6z2
Temporarily dropping the z’s gives
12a2 a 6
which factors into
(4a 3)(3a 2)
Replacing the z’s, we get
12a2 az 6z2 (4a 3z)(3a 2z)
Finally, substituting back (x y)3n for a, we have
12(x y)6n (x y)3nz 6z2 [4(x y)3n 3z][4(x y)3n 2z]
Exercise 5
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◆◆◆
The General Quadratic Trinomial
Factor completely.
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
4x2 13x 3
5x2 11x 2
12b2 b 6
2a2 a 6
5x2 38x 21
3x2 6x 3
3x2 x 2
4x2 10x 6
4a2 4a 3
9a2 15a 14
2.
4.
6.
8.
10.
12.
14.
16.
18.
20.
5a2 8a 3
7x2 23x 6
6x2 7x 2
2x2 3x 2
4x2 7x 15
2x2 11x 12
7x2 123x 54
3x2 11x 20
9x2 27x 18
16c2 48c 35
Expressions Reducible to Quadratic Trinomials
Factor completely.
21. 49x6 14x3y 15y2
22. 18y2 42x2 24xy
23. 4x6 13x3 3
24. 5a2n 8an 3
25. 5x4n 11x2n 2
26. 12(a b)2 (a b) 6
27. 3(a x)2n 3(a x)n 6
Applications
28. An object is thrown upward with an initial velocity of 32 ft./s from a building 128 ft. above
the ground. The height s of the object above the ground at any time t is given by
s 128 32t 16t 2
Factor the right side of this equation.
Section 8–6
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219
The Perfect Square Trinomial
29. An object is thrown into the air with an initial velocity of 23 m/s. To find the
approximate time it takes for the object to reach a height of 12 m, we must solve the
quadratic equation
Width
h
5t 2 23t 12 0
Factor the left side of this equation.
30. To find the depth of cut h needed to produce a flat of a certain width on a 1-cm-radius
bar (Fig. 8–6), we must solve the equation
4h2 8h 3 0
1 cm
Factor the left side of this equation.
FIGURE 8–6
8–6
The Perfect Square Trinomial
The Square of a Binomial
In Chapter 2, we saw that the expression obtained when a binomial is squared is called a perfect
square trinomial.
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Example 36: Square the binomial (2x 3).
Solution:
(2x 3)2 4x2 6x 6x 9
4x2 12x 9
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Note that in the perfect square trinomial obtained in Example 36, the first and last terms are
the squares of the first and last terms of the binomial.
square
(2x + 3)2
square
4x2 12x 9
The middle term is twice the product of the terms of the binomial.
(2x 3)2
product
6x
twice the product
4x2 12x 9
Also, the constant term is always positive. In general, the following equations apply:
Perfect
Square
Trinomials
(a b)2 a2 2ab b2
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(a b)2 a2 2ab b2
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Factoring a Perfect Square Trinomial
We can factor a perfect square trinomial in the same way we factored the general quadratic
trinomial in Sec. 8–5. However, the work is faster if we recognize that a trinomial is a perfect
square. If it is, its factors will be the square of a binomial. The terms of that binomial are the
square roots of the first and last terms of the trinomial. The sign in the binomial will be the same
as the sign of the middle term of the trinomial.
Any quadratic trinomial can be
manipulated into the form of a
perfect square trinomial by a
procedure called completing the
square. We will use that method
in Sec. 14–2 to derive the
quadratic formula.