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 solving polynomial equations by using the zero-product property. Factoring polynomials makes it easier to evaluate them for given inputs and determine the sign of outputs. Examples are provided to illustrate solving equations by factoring and evaluating polynomials in factored form.

1. Applications of Factoring

2. 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.

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 its factored form than its 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.
Base on these, we may extract a lot more information about a
formula from its factored form than its expanded form.

7. Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations or 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.
Base on these, we may extract a lot more information about a
formula from its factored form than its expanded form.

8. Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations or equations of the form
polynomial = polynomial
To solve these equations, we use the following obvious fact.
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 its factored form than its expanded form.

9. Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations or 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.
Base on these, we may extract a lot more information about a
formula from its factored form than its expanded form.

10. Applications of Factoring
Solving Equations
The most important application for factoring is to solve
polynomial equations or 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.
Base on these, we may extract a lot more information about a
formula from its factored form than its expanded form.

11. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations

12. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
b. If (x + 1)(x – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,

13. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
b. If (x + 1)(x – 2) = 0,
then either (x + 1) = 0 or (x – 2) = 0,
x = –1 or x = 2

14. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
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
Example B. Solve for x
a. x2 – 2x = 3

15. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
To solve polynomial equation,
1. set one side of the equation to be 0 and 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
Example B. Solve for x
a. x2 – 2x = 3
x2 – 2x – 3 = 0

16. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
To solve polynomial equation,
1. set one side of the equation to be 0 and 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
Example B. Solve for x
a. x2 – 2x = 3
x2 – 2x – 3 = 0 factor
(x – 3)(x + 1) = 0

17. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
To solve polynomial equation,
1. set one side of the equation to be 0 and 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
x2 – 2x – 3 = 0 factor
(x – 3)(x + 1) = 0

18. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
To solve polynomial equation,
1. set one side of the equation to be 0 and 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
x2 – 2x – 3 = 0 factor
(x – 3)(x + 1) = 0
There are two linear
x–factors. We may extract
one answer from each.

19. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
To solve polynomial equation,
1. set one side of the equation to be 0 and 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
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.

20. Example A.
a. If 3(x – 2) = 0, then (x – 2) = 0, so x must be 2.
Solving Equations
To solve polynomial equation,
1. set one side of the equation to be 0 and 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
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.

21. Following are two other important applications of the factored
forms of polynomials:
• to evaluate polynomials
Solving Equations

22. 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
Solving Equations

23. 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
Solving Equations

24. 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
Solving Equations

25. 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
Solving Equations

26. 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
Solving Equations

27. 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
Solving Equations

28. 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 have that
(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
Solving Equations

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

30. 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

31. 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

32. 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

33. 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

34. 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

35. 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.

36. 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.

37. 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.

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

39. 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

40. 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

41. 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

42. 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 (–)(+) = – . So the outcome is negative.
Applications of Factoring

43. 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,

44. 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