The document discusses applications of factoring expressions. The main purposes of factoring an expression E into a product E=AB is to utilize properties of multiplication. The most important application of factoring is to solve polynomial equations by setting each factor equal to 0 based on the zero-product property. Examples are provided to demonstrate solving polynomial equations by factoring, setting each factor equal to 0, and extracting the solutions.

1. Applications of Factoring

2. The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
Applications of Factoring

3. Applications of Factoring
The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
i. If the product AB = 0, then A or B must be 0.

4. Applications of Factoring
The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
i. If the product AB = 0, then A or B must be 0.
ii. The sign of the product can be determined by the signs of
A and B.

5. Applications of Factoring
The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
i. If the product AB = 0, then A or B must be 0.
ii. The sign of the product can be determined by the signs of
A and B.
Base on these, we may extract a lot more information about a
formula from it’s factored form than it’s expanded form.

6. Applications of Factoring
Solving Equations
The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
i. If the product AB = 0, then A or B must be 0.
ii. The sign of the product can be determined by the signs of
A and B.
Solving Equations
Base on these, we may extract a lot more information about a
formula from it’s factored form than it’s expanded form.

7. Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations.
The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
i. If the product AB = 0, then A or B must be 0.
ii. The sign of the product can be determined by the signs of
A and B.
Solving Equations
Base on these, we may extract a lot more information about a
formula from it’s factored form than it’s expanded form.

8. Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations. These are equations of the form
polynomial = polynomial
The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
i. If the product AB = 0, then A or B must be 0.
ii. The sign of the product can be determined by the signs of
A and B.
Solving Equations
Base on these, we may extract a lot more information about a
formula from it’s factored form than it’s expanded form.
The most important application for factoring is to solve
polynomial equations. These are equations of the form

9. 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.
The Zero-Product Rule:
If A*B = 0, then either A = 0 or B = 0
The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
i. If the product AB = 0, then A or B must be 0.
ii. The sign of the product can be determined by the signs of
A and B.
Solving Equations
Base on these, we may extract a lot more information about a
formula from it’s factored form than it’s expanded form.
The most important application for factoring is to solve
polynomial equations. These are equations of the form

10. 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.
The Zero-Product Rule:
If A*B = 0, then either A = 0 or B = 0
For example, if 3x = 0, then x must 0 (because 3 is not 0).
The main purposes of factoring an expression E into a product
E = AB is to utilize the two special properties of multiplication.
i. If the product AB = 0, then A or B must be 0.
ii. The sign of the product can be determined by the signs of
A and B.
Solving Equations
Base on these, we may extract a lot more information about a
formula from it’s factored form than it’s expanded form.
The most important application for factoring is to solve
polynomial equations. These are equations of the form

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

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

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

14. Example A.
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,

15. Example A.
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,

16. Example A.
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

17. Example A.
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

18. Example A.
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

19. Example A.
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

20. Example A.
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

21. Example A.
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

22. Example A.
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 B. Solve for x
a. x2 – 2x – 3 = 0

23. Example A.
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 B. Solve for x
a. x2 – 2x – 3 = 0 Factor
(x – 3)(x + 1) = 0

24. Example A.
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 B. 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.

25. Example A.
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 B. 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.

26. Example A.
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 B. 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.

27. Example A.
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 B. 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.

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

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

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

31. 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

32. 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

33. 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

34. 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

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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.

44. 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.

45. 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.

46. 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.

47. 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.

48. 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.

49. 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.

50. 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.

51. Applications of Factoring
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input
Evaluating Polynomials

52. Applications of Factoring
Evaluating Polynomials
Often it is easier to evaluate polynomials in the factored form.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

53. Applications of Factoring
Evaluating Polynomials
Example C. 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.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

54. Applications of Factoring
Evaluating Polynomials
Example C. 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.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

55. Applications of Factoring
Evaluating Polynomials
Example C. 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.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

56. Applications of Factoring
Evaluating Polynomials
Example C. 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.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

57. Applications of Factoring
Evaluating Polynomials
Example C. 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.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

58. Applications of Factoring
Evaluating Polynomials
Example C. 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.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

59. Applications of Factoring
Evaluating Polynomials
Example C. 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.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

60. Applications of Factoring
Evaluating Polynomials
Example C. 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.
Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
• determine the sign of the output of a given input

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

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

63. Example C. 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

64. Example C. 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

65. Example C. 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

66. Example C. 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

67. Example C. 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

68. Example C. 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

69. Example C. 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

70. Example C. 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

71. Example C. 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

72. Example C. 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

73. Example C. 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.

74. Example C. 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.

75. Example C. 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.

76. Example C. 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.

77. Example D. Determine the outcome is + or – for x2 – 2x – 3
if x = -3/2, -1/2.
Applications of Factoring

78. Example D. 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

79. Example D. 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

80. Example D. 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

81. Example D. 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

82. Example D. 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

83. Example D. 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

84. 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,

85. 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