4. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
5. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
6. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
7. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
8. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Often it is easier to evaluate polynomials in the factored form.
9. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 β 2x β 3 if x = 7
a. without factoring. b. by factoring it first.
Often it is easier to evaluate polynomials in the factored form.
10. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 β 2x β 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 β 2(7) β 3
Often it is easier to evaluate polynomials in the factored form.
11. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 β 2x β 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 β 2(7) β 3
= 49 β 14 β 3
Often it is easier to evaluate polynomials in the factored form.
12. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 β 2x β 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 β 2(7) β 3
= 49 β 14 β 3
= 32
Often it is easier to evaluate polynomials in the factored form.
13. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 β 2x β 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 β 2(7) β 3
= 49 β 14 β 3
= 32
x2 β 2x β 3 = (x β 3)(x+1)
Often it is easier to evaluate polynomials in the factored form.
14. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 β 2x β 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 β 2(7) β 3
= 49 β 14 β 3
= 32
x2 β 2x β 3 = (x β 3)(x+1)
We get
(7 β 3)(7 + 1)
Often it is easier to evaluate polynomials in the factored form.
15. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 β 2x β 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 β 2(7) β 3
= 49 β 14 β 3
= 32
x2 β 2x β 3 = (x β 3)(x+1)
We get
(7 β 3)(7 + 1)
= 4(8)
= 32
Often it is easier to evaluate polynomials in the factored form.
16. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:
1. to evaluate polynomials,
2. to determine the signs of the outputs,
3. most importantly, to solve polynomial-equations.
Evaluating Polynomials
Example A. Evaluate x2 β 2x β 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 β 2(7) β 3
= 49 β 14 β 3
= 32
x2 β 2x β 3 = (x β 3)(x+1)
We get
(7 β 3)(7 + 1)
= 4(8)
= 32
Often it is easier to evaluate polynomials in the factored form.
17. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
Applications of Factoring
18. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
Applications of Factoring
19. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
Applications of Factoring
20. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2]
Applications of Factoring
21. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4]
Applications of Factoring
22. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
Applications of Factoring
23. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2]
Applications of Factoring
24. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3]
Applications of Factoring
25. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3] = -9
Applications of Factoring
26. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) β 1] [(3) β 2]
Applications of Factoring
27. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) β 1] [(3) β 2] = 3 [5] [1] = 15
Applications of Factoring
28. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) β 1] [(3) β 2] = 3 [5] [1] = 15
Applications of Factoring
29. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) β 1] [(3) β 2] = 3 [5] [1] = 15
Applications of Factoring
Your turn: Double check these answers via the expanded form.
30. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) β 1] [(3) β 2] = 3 [5] [1] = 15
Applications of Factoring
Determine the signs of the outputs
Your turn: Double check these answers via the expanded form.
31. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) β 1] [(3) β 2] = 3 [5] [1] = 15
Applications of Factoring
Determine the signs of the outputs
Often we only want to know the sign of the output, i.e.
whether the output is positive or negative.
Your turn: Double check these answers via the expanded form.
32. Example B. Evaluate 2x3 β 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 β 5x2 + 2x = x(2x2 β 5x + 2)
= x(2x β 1)(x β 2)
For x = -2:
(-2)[2(-2) β 1] [(-2) β 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) β 1] [(-1) β 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) β 1] [(3) β 2] = 3 [5] [1] = 15
Applications of Factoring
Determine the signs of the outputs
Often we only want to know the sign of the output, i.e.
whether the output is positive or negative. It is easy to do this
using the factored form.
Your turn: Double check these answers via the expanded form.
33. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Applications of Factoring
34. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Applications of Factoring
35. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1)
Applications of Factoring
36. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + .
Applications of Factoring
37. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
Applications of Factoring
38. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1)
Applications of Factoring
39. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1) is (β)(+) = β .
Applications of Factoring
40. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1) is (β)(+) = β . So the outcome is negative.
Applications of Factoring
41. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1) is (β)(+) = β . So the outcome is negative.
Applications of Factoring
Solving Equations
42. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1) is (β)(+) = β . So the outcome is negative.
Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations.
43. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1) is (β)(+) = β . So the outcome is negative.
Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations. These are equations of the form
polynomial = polynomial
44. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1) is (β)(+) = β . So the outcome is negative.
Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations. These are equations of the form
polynomial = polynomial
To solve these equations, we use the following obvious fact.
45. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1) is (β)(+) = β . So the outcome is negative.
Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations. These are equations of the form
polynomial = polynomial
To solve these equations, we use the following obvious fact.
Fact: If A*B = 0,
then either A = 0 or B = 0
46. Example C. Determine the outcome is + or β for x2 β 2x β 3
if x = -3/2, -1/2.
Factor x2 β 2x β 3 = (x β 3)(x + 1)
Hence for x = -3/2:
(-3/2 β 3)(-3/2 + 1) is (β)(β) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 β 3)(-1/2 + 1) is (β)(+) = β . So the outcome is negative.
Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations. These are equations of the form
polynomial = polynomial
To solve these equations, we use the following obvious fact.
Fact: If A*B = 0,
then either A = 0 or B = 0
For example, if 3x = 0, then x must be equal to 0.
48. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0,
Applications of Factoring
49. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
50. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x β 2) = 0,
51. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
52. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1
53. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
54. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
55. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
56. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
57. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
3. get the answers.
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
58. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
3. get the answers.
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
Example E. Solve for x
a. x2 β 2x β 3 = 0
59. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
3. get the answers.
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
Example E. Solve for x
a. x2 β 2x β 3 = 0 Factor
(x β 3)(x + 1) = 0
60. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
3. get the answers.
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
Example E. Solve for x
a. x2 β 2x β 3 = 0 Factor
(x β 3)(x + 1) = 0
There are two linear
xβfactors. We may extract
one answer from each.
61. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
3. get the answers.
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
Example E. Solve for x
a. x2 β 2x β 3 = 0 Factor
(x β 3)(x + 1) = 0
Hence x β 3 = 0 or x + 1 = 0
There are two linear
xβfactors. We may extract
one answer from each.
62. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
3. get the answers.
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
Example E. Solve for x
a. x2 β 2x β 3 = 0 Factor
(x β 3)(x + 1) = 0
Hence x β 3 = 0 or x + 1 = 0
x = 3
There are two linear
xβfactors. We may extract
one answer from each.
63. Example D.
a. If 3(x β 2) = 0, then (x β 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
1. set one side of the equation to be 0, move all the terms to
the other side.
2. factor the polynomial,
3. get the answers.
b. If (x + 1)(x β 2) = 0,
then either (x + 1) = 0 or (x β 2) = 0,
x = β1 or x = 2
Example E. Solve for x
a. x2 β 2x β 3 = 0 Factor
(x β 3)(x + 1) = 0
Hence x β 3 = 0 or x + 1 = 0
x = 3 or x = -1
There are two linear
xβfactors. We may extract
one answer from each.
64. b. 2x(x + 1) = 4x + 3(1 β x)
Applications of Factoring
67. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
Applications of Factoring
68. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0
Applications of Factoring
69. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0
Applications of Factoring
70. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
Applications of Factoring
or x β 1 = 0
71. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
Applications of Factoring
or x β 1 = 0
72. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
73. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
74. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x
75. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x
76. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
77. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0
78. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
79. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
There are three linear
xβfactors. We may extract
one answer from each.
80. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
x = 0
There are three linear
xβfactors. We may extract
one answer from each.
81. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
x = 0 or 2x + 3 = 0
There are three linear
xβfactors. We may extract
one answer from each.
82. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
x = 0 or 2x + 3 = 0 or 2x β 3 = 0
There are three linear
xβfactors. We may extract
one answer from each.
83. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
x = 0 or 2x + 3 = 0 or 2x β 3 = 0
2x = -3
There are three linear
xβfactors. We may extract
one answer from each.
84. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
x = 0 or 2x + 3 = 0 or 2x β 3 = 0
2x = -3
x = -3/2
There are three linear
xβfactors. We may extract
one answer from each.
85. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
x = 0 or 2x + 3 = 0 or 2x β 3 = 0
2x = -3 2x = 3
x = -3/2
There are three linear
xβfactors. We may extract
one answer from each.
86. b. 2x(x + 1) = 4x + 3(1 β x) Expand
2x2 + 2x = 4x + 3 β 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x β x β 3 = 0
2x2 + x β 3 = 0 Factor
(2x + 3)(x β 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x β 1 = 0
x = 1
c. 8x(x2 β 1) = 10x Expand
8x3 β 8x = 10x Set one side 0
8x3 β 8x β 10x = 0
8x3 β 18x = 0 Factor
2x(2x + 3)(2x β 3) = 0
x = 0 or 2x + 3 = 0 or 2x β 3 = 0
2x = -3 2x = 3
x = -3/2 x = 3/2
There are three linear
xβfactors. We may extract
one answer from each.
87. Exercise A. Use the factored form to evaluate the following
expressions with the given input values.
Applications of Factoring
1. x2 β 3x β 4, x = β2, 3, 5 2. x2 β 2x β 15, x = β1, 4, 7
3. x2 β 2x β 1, x = Β½ ,β2, βΒ½ 4. x3 β 2x2, x = β2, 2, 4
5. x3 β 4x2 β 5x, x = β4, 2, 6 6. 2x3 β 3x2 + x, x = β3, 3, 5
B. Determine if the output is positive or negative using the
factored form.
7. x2 β 3x β 4, x = β2Β½, β2/3, 2Β½, 5ΒΌ
8. βx2 + 2x + 8, x = β2Β½, β2/3, 2Β½, 5ΒΌ
9. x3 β 2x2 β 8x, x = β4Β½, β3/4, ΒΌ, 6ΒΌ,
11. 4x2 β x3, x = β1.22, 0.87, 3.22, 4.01
12. 18x β 2x3, x = β4.90, β2.19, 1.53, 3.01
10. 2x3 β 3x2 β 2x, x = β2Β½, β3/4, ΒΌ, 3ΒΌ,