Download 3.6 FACTORING AND THE QUADRATIC EQUATION Textbook

118 CHAPTER 3
Algebra
3.6 FACTORING AND THE QUADRATIC EQUATION
Textbook Reference Section 5.4
CLAST OBJECTIVES
" Factor a quadratic expression
" Find the roots of a quadratic equation
A quadratic expression is a mathematical expression that can be written in the form
ax 2 + bx + c , where a ≠ 0.
A quadratic equation is a second-degree equation. In other words, the highest exponent on
the variable is two.
The standard form of a quadratic equation is ax 2 + bx + c = 0 .
Not all quadratics are factorable.
Factoring Quadratics Using the ac-Method.
1. Write the quadratic in standard form.
2. Determine the product ac.
3. Find two numbers, d and e, that multiply to form the product ac and sum to b.
4. If the two numbers do not exist, the equation is not factorable. If the two numbers do
exist, rewrite the quadratic as ax 2 + dx + ex + c .
5. Proceed as in factoring by grouping.
a. Make two pairs: ax 2 + dx and ex + c .
b. Factor each pair.
c. Factor out the greatest common factor (GCF).
Example
a) Factor: x 2 – 2x – 15
Solution
1. Identify a, b, and c. a = 1, b = -2, c = -15
2. Multiply a and c. ac = (1)(-15) = -15
3. Find two numbers whose product is –15 and sum to –2.
Factors
-1, 15
1, -15
-3, 5
3, -5
Product
-1 × 15 = -15
1 × -15 = -15
-3 × 5 = -15
3 × -5 = -15
2
Sum
-1 + 15 = 14
1 + -15 = -14
-3 + 5 = 2
3 + -5 = -2
Conclusion
No
No
No
Yes
2
4. Rewrite. x – 2x – 15 = x + 3x – 5x – 15
5. Factor by grouping:
2
2
x + 3x – 5x – 15 = (x + 3x) + (– 5x – 15)
= x(x + 3) – 5(x + 3)
= (x + 3)(x – 5)
2
Answer: x – 2x – 15 = (x + 3)(x – 5)
Note: Factoring results can be checked by multiplying the factors.
2
2
Check: (x + 3)(x – 5) = x – 5x + 3x – 15 = x – 2x – 15
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SECTION 3.6
Example
b) Factor: 3x 2 – 8x + 4
Factoring and the Quadratic Equation 119
Solution
1. Identify a, b, and c. a = 3, b = -8, c = 4
2. Multiply a and c. ac = (3)(4) = 12
3. Find two numbers whose product is 12 and sum to – 8.
Factors
-1, -12
1, 12
-2, -6
2, 6
-3, -4
3, 4
Product
Sum
Conclusion
-1 × -12 = 12
-1 + -12 = -13
No
1 × 12 = 12
1 + 12 = 13
No
-2 × -6 = 12
-2 + -6 = -8
Yes
We don’t have to continue as we
have found the correct combination.
2
2
4. Rewrite. 3x – 8x + 4 = 3x – 2x – 6x + 4
2
2
5. Factor by grouping: 3x – 2x – 6x + 4 = (3x – 2x) + (– 6x + 4)
= x (3x – 2) – 2 (3x – 2)
= (3x – 2)(x – 2)
2
Answer: 3x – 8x + 4 = (3x – 2)(x – 2)
Check Your Progress 3.6
Factor each of the following.
2. 3x 2 + 8x + 5
1. x 2 + x – 12
3. 2x 2 + 5x + 3
4. 2x 2 – 7x – 4
6. 4x 2 – 11x + 6
5. 5x 2 – 8x + 3
Special Factoring: The Difference of Two Squares
x
2
−y
2
= (x − y )
(x + y )
Examples
c) Factor: x 2 – 25
or
x
2
−y
2
= (x + y )
(x − y )
Solutions
First, recognize that the quadratic is a difference of two squares.
x 2 – 25 = x 2 – 5 2 = (x – 5) (x + 5)
Check: (x – 5) (x + 5) = x 2 – 5x + 5x – 25 = x 2 – 25
d) Factor: 4x 2 – 49
First, recognize that the quadratic is a difference of two squares.
4x 2 – 49 = (2x) 2 – 7 2 = (2x – 7) (2x + 7)
Check: (2x – 7) (2x + 7) = 4x 2 – 14x + 14x – 49 = 4x 2 – 49
e) Factor: 81x 4 – 16y 4
First, recognize that the quadratic is a difference of two squares.
81x 4 – 16y 4 = (9x 2 ) 2 – (4y 2 ) 2 = (9x 2 – 4y 2 ) (9x 2 + 4y 2 )
Notice that the first factor is also a difference of two squares. Thus,
the quadratic can be factored further.
81x 4 – 16y 4 = (9x 2 ) 2 – (4y 2 ) 2 = (9x 2 – 4y 2) (9x 2 + 4y 2 )
= (3x – 2y) (3x + 2y) (9x 2 + 4y 2 )
Answer: 81x 4 – 16y 4 = (3x – 2y) (3x + 2y) (9x 2 + 4y 2)
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120 CHAPTER 3
Algebra
Check Your Progress 3.6
Factor each of the following.
7. 25x 2 – 9
8. 49t 2 – 100
Solving Quadratic Equations
Case 1: Solving Quadratic Equations By Factoring
• The quadratic must be factorable.
1. Write the equation in standard form: ax 2 + bx + c = 0
2. Factor the quadratic.
3. Set each factor equal to zero (0).
4. Solve the resulting equations to find the solutions to the quadratic equation.
Examples
f) Solve: 3x 2 – 17x + 10 = 0
Solutions
Notice that the quadratic equation is already in standard form.
Factor the quadratic on the left side of the equation.
3x 2 – 17x + 10 = 0
(3x – 2)(x – 5) = 0
Set each factor equal to zero and solve.
3x − 2 = 0
x−5 = 0
2
x=
x=5
3
Check by substituting the x – values in the original equation.
3(5) 2 – 17(5) + 0 = 3(25) – 85 + 10 = 75 – 85 + 10
= -10 + 10 = 0
⎛2⎞
3⎜ ⎟
⎝3⎠
g) Solve: 9x 2 – 25 = 0
2
4 34
⎛2⎞
⎛ 4 ⎞ 34
− 17⎜ ⎟ + 10 = 3⎜ ⎟ −
+ 10 = −
+ 10
3 3
⎝3⎠
⎝9⎠ 3
− 30
=
+ 10 = −10 + 10 = 0
3
Notice that the quadratic equation is already in standard form.
Factor the quadratic on the left side of the equation.
9x 2 – 25 = 0
( 3x ) 2 - ( 5 ) 2 = 0
( 3x – 5 )(3x + 5) = 0
Set each factor equal to zero and solve.
3x − 5 = 0
3x + 5 = 0
5
5
x=
x=−
3
3
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SECTION 3.6
Example
h) Solve: 4x 2 + 9x = -2
Factoring and the Quadratic Equation 121
Solution
First, write the quadratic equation in standard form by adding
2 to both sides. Then, factor the quadratic on the left side of
the equation.
4y 2 + 9y = -2
2
4y + 9y + 2 = -2+2
4y 2 + 9y + 2 = 0
(4y + 1)(y + 2) = 0
Set each factor equal to zero and solve.
4y +1 = 0
y+2=0
1
y=−
y = −2
4
Check Your Progress 3.6
Find the solutions.
9. x 2 – x = 2
10. 3x 2 + 10x – 8 = 0
11. 2x 2 – 5x – 3 = 0
12. x 2 + 8x = -15
13. 36t 2 – 25 = 0
14. 3x 2 – 5x = 2
Case 2: Solving Quadratic Equations Using the Quadratic Formula
• We can use the quadratic formula to find the roots to any quadratic
equation.
1. Write the equation in standard form: ax 2 + bx + c = 0
2. Identify a, b, and c.
3. Use the quadratic formula: x =
−b ±
4. Simplify.
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b
2a
2
− 4ac
122 CHAPTER 3
Algebra
Examples
i) Find the real roots:
12x 2 – 5x – 3 = 0
Solutions
Notice that the quadratic equation is already in standard form.
a = 12, b = −5, c = −3 . Substitute these values into the quadratic
x=
formula and simplify.
x=
=
− (− 5) ±
b
2
− 4ac
2a
(− 5) 2 − 4(12)(− 3) 5 ±
=
2(12 )
25 + 144 5 ± 169
=
24
24
5 + 13 18 3
=
=
24
24 4
x=
5 ± 13
24
x=
j) Find the real roots:
2x 2 + x = 15
−b ±
5 − 13 − 8 − 1
=
=
24
24
3
First, write the quadratic equation in standard form by subtracting 15 from
both sides of the equation. 2x 2 + x – 15 = 0
a = 2, b = 1, c = −15 . Substitute these values into the quadratic formula
and simplify.
x=
=
− (1) ±
x=
−b ±
b
2
− 4ac
2a
(1) 2 − 4(2)(− 15) − 1
=
2(2 )
±
1 + 120
4
=
− 1 ± 121
4
1 ± 11
4
x=
− 1 + 11 10 5
=
=
4
4 2
x=
− 1 − 11 − 12
=
= −3
4
4
Check Your Progress 3.6
For Questions 15 – 20, find the roots to the equation.
16. x 2 – 6x – 3 = 0
15. x 2 – 10x – 3 = 0
17. 3x 2 + 2x – 4 = 0
18. 2m 2 + 6m = 3
19. 7x 2 + 6x - 4 = 0
20. 25x 2 + 10x = -1
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SECTION 3.6
See If You
Remember
SECTIONS 3.1 – 3.5
1. Simplify: 5π + 2 − 9π
3. Simplify:
5. Simplify:
Factoring and the Quadratic Equation 123
2. Simplify: 250 − 2 10
(
4. 6.1 × 10
7
4
) (3.2 × 10 9 )
−
63
5
⎛1
⎞
− 6 ⎜ + 4⎟
6
⎝4
⎠
6. Simplify: 8 y − y × 3 + 14 y
7. Solve: 3(t + 1) = 2[t + 3(t − 2 )]
8. Name the property: x + (y + z) = x + (z + y)
9. Solve: 4(y – 1) + 5y < 7y – 4
10. Find f (3), if f (x) = -x 2 + 2x – 5
11. Find the solution set.
x − 4 y = 15
4 x − 3 y = 21
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2
÷7×4
124 CHAPTER 3
Algebra
12. Graph: y = -2x + 3
13. Graph: x – y = 5
14. Graph the solution: -x + y > 4
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