1. Copyright © Cengage Learning. All rights reserved.4
Quadratic Functions
Copyright © Cengage Learning. All rights reserved.

2. 4.5 Solving Equations by Factoring
Copyright © Cengage Learning. All rights reserved.

3. Objectives Solve a polynomial equation by factoring.
Find a quadratic equation from a graph.

4. Solving by Factoring

5. Solving by Factoring
Factoring can be used to solve some polynomial equations. The basis of factoring is to take a polynomial and break it into simpler pieces that are multiplied together.
Standard Form Factored Form
f (x) = x2 + 8x f (x) = (x + 3)(x + 5)
If we multiply the factored form out and simplify, it will become the standard form.

6. Solving by Factoring f (x) = (x + 3)(x + 5) f (x) = x2 + 5x + 3x + 15
The main reason factoring an expression is useful is the product property of zero, which says that when you multiply any number by zero, the answer is zero.

7. Solving by Factoring
The zero factor property follows from the product property of zero and states that if two numbers are multiplied together and the answer is zero, then one or both of those factors must be zero.

8. Solving by Factoring
If we factor a polynomial into simpler parts, we will be able to solve the simpler equations.
It is very important to note that we can use the factored form to solve an equation only if it is set equal to zero. If the quadratic equation is not equal to zero, we must first move everything to one side so that it equals zero and then factor.

9. Example 1 – Solving polynomial equations by factoring
Solve the following by factoring.
a. x2 + 7x – 40 = 20
b. 2x = 21x
c. x3 + 8x2 + 15x = 0
Solution:
x2 + 7x – 60 = 0
Set the quadratic equal to
zero and factor.
Steps 2 and 3 are in the margin on
the right.

10. Example 1 – Solution x2 – 5x + 12x – 60 = 0 (x2 – 5x) + (12x – 60) = 0
cont’d
x2 – 5x + 12x – 60 = 0
(x2 – 5x) + (12x – 60) = 0
x(x – 5) + 12(x – 5) = 0
(x – 5)(x + 12) = 0
x – 5 = x + 12 = 0
x = x = –12
Step 4.
Step 5.
Set the factors equal to zero
and finish solving.
Check both answers.

11. Example 1 – Solution (5)2 + 7(5) – 40 20 20 = 20
cont’d
(5)2 + 7(5) –
20 = 20
(–12)2 + 7(–12) –
Both answers work.

12. Example 1 – Solution b. 2x2 + 27 = 21x 2x2 – 21x + 27 = 0
cont’d
b x = 21x
2x2 – 21x + 27 = 0
2x2 – 3x – 18x + 27 = 0
(2x2 – 3x) + (–18x + 27) = 0
x(2x – 3) – 9(2x – 3) = 0
(2x – 3)(x – 9) = 0
Set the quadratic equal to
zero and factor.
Steps 2 and 3 are in the
margin on the right.
Step 4.
Step 5.
Set the factors equal to zero and
finish solving.

13. Example 1 – Solution 2x – 3 = 0 x – 9 = 0 x = 9 31.5 = 31.5
cont’d
2x – 3 = x – 9 = 0
x = 9
31.5 = 31.5
2(9) (9)
189 = 189
Check both answers.
Both answers work.

14. Example 1 – Solution c. x3 + 8x2 + 15x = 0 x(x2 + 8x + 15) = 0
cont’d
c. x3 + 8x2 + 15x = 0
x(x2 + 8x + 15) = 0
x(x2 + 3x + 5x + 15) = 0
x[(x2 + 3x) + (5x + 15)] = 0
x[x(x + 3) + 5(x + 3)] = 0
x(x + 3) + (x + 5) = 0
Take out the x in common and
factor the remaining quadratic.
The factored-out x will remain at
the front of each step until the
factoring is complete.

15. Example 1 – Solution x = 0 x + 3 = 0 x + 5 = 0 x = 0 x = –3 x = –5
cont’d
x = 0 x + 3 = 0 x + 5 = 0
x = x = – x = –5
(0)3 + 8(0)2 + 15(0) 0
0 = 0
(–3)3 + 8(–3)2 + 15(–3) 0
Now rewrite into three equations
and solve each separately.
Check all three answers in the original equation.

16. Example 1 – Solution (–5)3 + 8(–5)2 + 15(–5) 0 0 = 0 cont’d
(–5)3 + 8(–5)2 + 15(–5) 0
0 = 0
All three answers work.

17. Finding an Equation from the Graph

18. Finding an Equation from the Graph
The zero factor property can also be used to help find equations by looking at their graphs.
As we solved different equations by factoring, we always set the equation equal to zero and then factor.
If we set a function equal to zero and solve, we are looking for the horizontal intercepts of the function.

19. Finding an Equation from the Graph
When we solve using factoring, we are actually finding the horizontal intercepts of the graph related to that equation.
We can use the connection between the horizontal intercepts and the factors of an expression to find an equation using its graph.
The horizontal intercepts of a graph also represent the zeros of the function.

20. Finding an Equation from the Graph
In this graph, the horizontal intercepts are (–2, 0) and (–8, 0), so the zeros of the equation are and x = –2 and x = –8. Taking the zeros and working backwards through the process of solving by factoring, we will get an equation for the graph.
x = – x = –2
x + 8 = x + 2 = 0
0 = (x + 8)(x + 2)
y = (x + 8)(x + 2)

21. Finding an Equation from the Graph
This equation in factored form could be missing a constant that was factored out.
To find this constant, we can use another point from the graph and solve for the missing constant.
If we let a represent the missing constant and use the point (–3, 5), we can solve for a.
y = a(x + 8)(x + 2)

22. Finding an Equation from the Graph
Substituting in the –1 for a and simplifying, we get
y = –1(x + 8)(x + 2)
y = –x2 – 10x – 16

23. Finding an Equation from the Graph

24. Example 4 – Finding a quadratic equation from the graph
Use the graph to find an equation for the quadratic.
a b. c.

25. Example 1(a) – Solution
Step 1: Find the horizontal intercepts of the graph and write them as zeros of the equation.
The horizontal intercepts for this graph are located at (–4, 0) and (3, 0), so we know that x = –4 and x = 3 are the zeros of the quadratic equation.

26. Example 1(a) – Solution
cont’d
Step 2: Undo the solving process to find the factors of the equation. Therefore, we have:
x = –4 x = 3
x + 4 = x – 3 = 0
This gives us the factors x + 4 and x – 3 and the equation
y = (x + 4) (x – 3)

27. Example 1(a) – Solution
cont’d
Step 3: Use another point from the graph to find any constant factors. Using the point (1, –10), we get
y = a(x + 4)(x – 3)
–10 = a(1 + 4)(1 – 3)
–10 = a(5)(–2)
–10 = –10a
1 = a
The constant factor is 1.

28. Example 1(a) – Solution
cont’d
Step 4: Multiply out the factors and simplify. Therefore, we have the equation
y = 1(x + 4)(x – 3)
y = x2 + x – 12
The y-intercept for this quadratic would be (0, –12), which agrees with the graph, so we have the correct equation.

29. Example 1(b) – Solution
cont’d
Step 1: Find the horizontal intercepts of the graph and write them as zeros of the equation.
The horizontal intercepts for this graph are (2, 0) and (5, 0), so we know that x = 2 and x = 5 are the zeros of the quadratic equation.

30. Example 1(b) – Solution
cont’d
Step 2: Undo the solving process to find the factors of the equation. Therefore, we have
x = 2 x = 5
x – 2 = x – 5 = 0
This gives us the factors (x – 2) and (x – 5) and the equation
y = (x – 2)(x – 5)

31. Example 1(b) – Solution
cont’d
Step 3: Use another point from the graph to find any constant factors. Using the point (7, 10), we get
y = a(x – 2)(x – 5)
10 = a(7 – 2)(7 – 5)
10 = a(5)(2)
10 = 10a
1 = a
The constant factor is 1.

32. Example 1(b) – Solution
cont’d
Step 4: Multiply out the factors and simplify. Therefore, we have the equation
y = 1(x – 2)(x – 5)
y = x2 – 7x + 10
The y-intercept for this quadratic would be (0, 10), which agrees with the graph, so we have the correct equation.

33. Example 1(c) – Solution
cont’d
Step 1: Find the horizontal intercepts of the graph and write them as zeros of the equation.
The horizontal intercepts for this graph are (2, 0) and (5, 0), so we know that x = 2 and x = 5 are the zeros of the quadratic equation.

34. Example 1(c) – Solution
cont’d
Step 2: Undo the solving process to find the factors of the equation. Therefore, we have
x = x = 5
x – 2 = 0 x – 5 = 0
This gives us the factors (x – 2) and (x – 5) and the equation
y = (x – 2)(x – 5)

35. Example 1(c) – Solution
cont’d
Step 3: Use another point from the graph to find any constant factors. Using the point (10, –80), we get
y = a(x – 2)(x – 5)
–80 = a(10 – 2)(10 – 5)
–80 = 40a
–2 = a
The constant factor is –2.

36. Example 1(c) – Solution
cont’d
Step 4: Multiply out the factors and simplify. Therefore, we have the equation
y = –2(x – 2)(x – 5)
y = –2x2 + 14x – 20
The y-intercept for this quadratic would be (0, –20), which agrees with the graph, so we have the correct equation.