The document discusses applications of factoring polynomials. It provides examples of how factoring can be used to evaluate polynomials by substituting values into the factored form. Factoring is also useful for determining the sign of outputs and for solving polynomial equations, which is described as the most important application of factoring. Examples are given to demonstrate evaluating polynomials both with and without factoring, and checking the answers obtained from factoring using the expanded form.

1. Applications of Factoring

2. Applications of Factoring
There are many applications of the factored forms of
polynomials.

3. Applications of Factoring
There are many applications of the factored forms of
polynomials. Following are some of them:

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 – 2x – 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 – 2x – 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 – 2(7) – 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 – 2x – 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 – 2(7) – 3
= 49 – 14 – 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 – 2x – 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 – 2(7) – 3
= 49 – 14 – 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 – 2x – 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 – 2(7) – 3
= 49 – 14 – 3
= 32
x2 – 2x – 3 = (x – 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 – 2x – 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 – 2(7) – 3
= 49 – 14 – 3
= 32
x2 – 2x – 3 = (x – 3)(x+1)
We get
(7 – 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 – 2x – 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 – 2(7) – 3
= 49 – 14 – 3
= 32
x2 – 2x – 3 = (x – 3)(x+1)
We get
(7 – 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 – 2x – 3 if x = 7
a. without factoring. b. by factoring it first.
We get
72 – 2(7) – 3
= 49 – 14 – 3
= 32
x2 – 2x – 3 = (x – 3)(x+1)
We get
(7 – 3)(7 + 1)
= 4(8)
= 32
Often it is easier to evaluate polynomials in the factored form.

17. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
Applications of Factoring

18. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
Applications of Factoring

19. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
Applications of Factoring

20. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2]
Applications of Factoring

21. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4]
Applications of Factoring

22. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
Applications of Factoring

23. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2]
Applications of Factoring

24. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3]
Applications of Factoring

25. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
Applications of Factoring

26. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 2]
Applications of Factoring

27. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring

28. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring

29. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring
Your turn: Double check these answers via the expanded form.

30. Example B. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 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 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 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 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 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 – for x2 – 2x – 3
if x = -3/2, -1/2.
Applications of Factoring

34. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Applications of Factoring

35. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1)
Applications of Factoring

36. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + .
Applications of Factoring

37. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
Applications of Factoring

38. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1)
Applications of Factoring

39. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – .
Applications of Factoring

40. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring

41. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . So the outcome is negative.
Applications of Factoring
Solving Equations

42. Example C. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . 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 – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . 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 – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . 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 – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . 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 – for x2 – 2x – 3
if x = -3/2, -1/2.
Factor x2 – 2x – 3 = (x – 3)(x + 1)
Hence for x = -3/2:
(-3/2 – 3)(-3/2 + 1) is (–)(–) = + . So the outcome is positive.
And for x = -1/2:
(-1/2 – 3)(-1/2 + 1) is (–)(+) = – . 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.

47. Example D.
a. If 3(x – 2) = 0,
Applications of Factoring

48. Example D.
a. If 3(x – 2) = 0, then (x – 2) = 0,
Applications of Factoring

49. Example D.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring

50. Example D.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x – 2) = 0,

51. Example D.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,

52. Example D.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1

53. Example D.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
b. If (x + 1)(x – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2

54. Example D.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Applications of Factoring
To solve polynomial equation,
b. If (x + 1)(x – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2

55. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2

56. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2

57. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2

58. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0

59. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor
(x – 3)(x + 1) = 0

60. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor
(x – 3)(x + 1) = 0
There are two linear
x–factors. We may extract
one answer from each.

61. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
There are two linear
x–factors. We may extract
one answer from each.

62. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3
There are two linear
x–factors. We may extract
one answer from each.

63. Example D.
a. If 3(x – 2) = 0, then (x – 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 – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2
Example E. Solve for x
a. x2 – 2x – 3 = 0 Factor
(x – 3)(x + 1) = 0
Hence x – 3 = 0 or x + 1 = 0
x = 3 or x = -1
There are two linear
x–factors. We may extract
one answer from each.

64. b. 2x(x + 1) = 4x + 3(1 – x)
Applications of Factoring

65. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
Applications of Factoring

66. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3
Applications of Factoring

67. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
Applications of Factoring

68. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0
Applications of Factoring

69. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0
Applications of Factoring

70. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
Applications of Factoring
or x – 1 = 0

71. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
Applications of Factoring
or x – 1 = 0

72. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0

73. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1

74. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x

75. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x

76. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0

77. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0

78. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0

79. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
There are three linear
x–factors. We may extract
one answer from each.

80. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
x = 0
There are three linear
x–factors. We may extract
one answer from each.

81. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
x = 0 or 2x + 3 = 0
There are three linear
x–factors. We may extract
one answer from each.

82. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
x = 0 or 2x + 3 = 0 or 2x – 3 = 0
There are three linear
x–factors. We may extract
one answer from each.

83. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
x = 0 or 2x + 3 = 0 or 2x – 3 = 0
2x = -3
There are three linear
x–factors. We may extract
one answer from each.

84. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
x = 0 or 2x + 3 = 0 or 2x – 3 = 0
2x = -3
x = -3/2
There are three linear
x–factors. We may extract
one answer from each.

85. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
x = 0 or 2x + 3 = 0 or 2x – 3 = 0
2x = -3 2x = 3
x = -3/2
There are three linear
x–factors. We may extract
one answer from each.

86. b. 2x(x + 1) = 4x + 3(1 – x) Expand
2x2 + 2x = 4x + 3 – 3x
2x2 + 2x = x + 3 Set one side 0
2x2 + 2x – x – 3 = 0
2x2 + x – 3 = 0 Factor
(2x + 3)(x – 1) = 0 Get the answers
2x + 3 = 0
2x = -3
x = -3/2
Applications of Factoring
or x – 1 = 0
x = 1
c. 8x(x2 – 1) = 10x Expand
8x3 – 8x = 10x Set one side 0
8x3 – 8x – 10x = 0
8x3 – 18x = 0 Factor
2x(2x + 3)(2x – 3) = 0
x = 0 or 2x + 3 = 0 or 2x – 3 = 0
2x = -3 2x = 3
x = -3/2 x = 3/2
There are three linear
x–factors. 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 – 3x – 4, x = –2, 3, 5 2. x2 – 2x – 15, x = –1, 4, 7
3. x2 – 2x – 1, x = ½ ,–2, –½ 4. x3 – 2x2, x = –2, 2, 4
5. x3 – 4x2 – 5x, x = –4, 2, 6 6. 2x3 – 3x2 + x, x = –3, 3, 5
B. Determine if the output is positive or negative using the
factored form.
7. x2 – 3x – 4, x = –2½, –2/3, 2½, 5¼
8. –x2 + 2x + 8, x = –2½, –2/3, 2½, 5¼
9. x3 – 2x2 – 8x, x = –4½, –3/4, ¼, 6¼,
11. 4x2 – x3, x = –1.22, 0.87, 3.22, 4.01
12. 18x – 2x3, x = –4.90, –2.19, 1.53, 3.01
10. 2x3 – 3x2 – 2x, x = –2½, –3/4, ¼, 3¼,

88. C. Solve the following equations. Check the answers.
Applications of Factoring
18. x2 – 3x = 10 20. x(x – 2) = 24 21. 2x2 = 3(x + 1) – 1
28. x3 – 2x2 = 0
22. x2 = 4
25. 2x(x – 3) + 4 = 2x – 4
29. x3 – 2x2 – 8x = 0
31. 4x2 = x3
30. 2x2(x – 3) = –4x
26. x(x – 3) + x + 6 = 2x2 + 3x
13. x2 – 3x – 4 = 0 14. x2 – 2x – 15 = 0 15. x2 + 7x + 12 = 0
16. –x2 – 2x + 8 = 0 17. 9 – x2 = 0 18. 2x2 – x – 1 = 0
27. x(x + 4) + 9 = 2(2 – x)
23. 8x2 = 2 24. 27x2 – 12 = 0
32. 4x = x3 33. 4x2 = x4
34. 7x2 = –4x3 – 3x 35. 5 = (x + 2)(2x + 1)
36. (x – 1)2 = (x + 1)2 – 4 37. (x + 1)2 = x2 + (x – 1)2
38. (x + 2)2 – (x + 1)2= x2 39. (x + 3)2 – (x + 2)2 = (x + 1)2