1. Equations by Factoring
Solving Quadratic
Equations by Factoring
Topic 7.1.1

2. 7.1.1 Topic Solving Quadratic Equations by Factoring
California Standards:
11.0 Students apply basic factoring techniques to second- and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials.
14.0 Students solve a quadratic equation by factoring or completing the square.
What it means for you:
You’ll solve quadratic equations by factoring.
Key words:
quadratic
factor
zero property

3. 7.1.1 Topic Solving Quadratic Equations by Factoring
In this Topic you’ll use all the factoring methods that you learned in Chapter 6 to solve quadratic equations.

4. 7.1.1 Topic Solving Quadratic Equations by Factoring
Quadratic Equations Have Degree 2
Quadratic equations contain a squared variable, but no higher powers — they have degree 2.
These are all quadratic equations, as the highest power of the variable is 2:
(i) x2 – 3x + 2 = (ii) 4x2 + 12x – 320 = (iii) y2 + 4y – 7 = 2y2 – 2y

5. 7.1.1 Topic Solving Quadratic Equations by Factoring
The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers, and a is not 0.
(i) x2 – 3x + 2 = (ii) 4x2 + 12x – 320 = (iii) y2 + 4y – 7 = 2y2 – 2y
For example, in (i) above, a = 1, b = –3, and c = 2, while in (ii), a = 4, b = 12, and c = –320.
Example (iii) above is a quadratic in y, while the others are quadratics in x.

6. 7.1.1 Topic Solving Quadratic Equations by Factoring Guided Practice
The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in these equations.
1. –x2 + 5x – 6 = 0
2. 6x2 + 31x + 35 = x2 – 12x + 9 = 0
4. 16x2 – 8x + 1 = 0 5. –x2 – 4x – 4 = 0
6. 64x2 + 48x + 9 = 0
a = –1, b = 5, c = –6
a = 6, b = 31, c = 35
a = 4, b = –12, c = 9
a = 16, b = –8, c = 1
a = –1, b = –4, c = –4
a = 64, b = 48, c = 9
Solution follows…

7. 7.1.1 Topic Solving Quadratic Equations by Factoring Guided Practice
The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in these equations.
7. 6y2 + 28y + 20 = 5 – 6y2
8. 4x2 + 6x + 1 = 3x2 + 8x 9. 4(x2 – 5x) = –25
10. 3x(3x + 4) + 8 = x(7x + 4) + 4x3 + 3 = 2(2x3 + 1)
a = 12, b = 28, c = 15
a = 1, b = –2, c = 1
a = 4, b = –20, c = 25
a = 9, b = 12, c = 4
a = 49, b = 28, c = 1
Solution follows…

8. 7.1.1 Topic Solving Quadratic Equations by Factoring
Solving is Finding Values That Make the Equality True
An equation is a statement saying that two mathematical expressions are equal.
For example, = 9, 4x + 2 = 14, and x2 – 3x + 2 = 0 are all equations.
If an equation contains a variable (an unknown quantity), then solving the equation means finding possible values of the variable that make the equation a true statement.

9. 7.1.1 Topic Solving Quadratic Equations by Factoring
Example 1
Find a solution of the equation x2 – 3x + 2 = 0.
Solution
If you evaluate the above equation using, say, x = 3, then you get: 32 – (3 × 3) + 2 = 0
But if instead you substitute x = 1, then you get: 12 – (3 × 1) + 2 = 0
This is a true statement.
So x = 1 is a solution of the equation x2 – 3x + 2 = 0.
This is not a true statement (since the left-hand side equals 2).
So x = 3 is not a solution of the equation.
Solution follows…

10. 7.1.1 Topic Solving Quadratic Equations by Factoring Guided Practice
Determine which of the two values given is a solution of the equation. 5 3 1 2 4 8
12. –x2 + 5x – 6 = 0 for x = –3 and x = x2 + 31x + 35 = 0 for x = – and x = x2 – 12x + 9 = 0 for x = and x = x2 – 8x + 1 = 0 for x = – and x = –x2 – 4x – 4 = 0 for x = 2 and x = – x2 + 48x + 9 = 0 for x = – and x =
x = 3
x = – 5 3
x = 3 2
x = 1 4
x = –2
x = – 3 8
Solution follows…

11. 7.1.1 Topic Solving Quadratic Equations by Factoring
Zero Property — if xy = 0, then x = 0 or y = 0 (or Both)
One way to solve a quadratic equation is to factor it and then make use of the following property of zero:
Zero Property
If the product mc = 0, then either: (i) m = 0, (ii) c = 0, (iii) both m = 0 and c = 0.

12. 7.1.1 Topic Solving Quadratic Equations by Factoring
Example 2
Solve x2 + 2x – 15 = 0 by factoring.
Solution
x2 + 2x – 15 = (x – 3)(x + 5)
So if (x – 3)(x + 5) = 0, then by the zero property, either (x – 3) = 0 or (x + 5) = 0.
So either x = 3 or x = –5.
Solution follows…

13. 7.1.1 Topic Solving Quadratic Equations by Factoring
Example 3
Solve 2x2 + 3x – 20 = 0 by factoring.
Solution
2x2 + 3x – 20 = (2x – 5)(x + 4) = 0
So either x = or x = –4. 5 2
Solution follows…

14. 7.1.1 Topic Solving Quadratic Equations by Factoring Guided Practice
Solve each of these quadratic equations by using the zero property.
x = – , – 7 2 5 3
18. (2x + 7)(3x + 5) = 0
19. (x – 5)(x – 1) = 0
x2 – 1 = 0
a2 – 25 = 0
22. 4x2 + 8x + 3 = 0
23. 2x2 – 17x – 9 = 0
x = 5, 1
x = , – 1 7
a = , – 5 8
x = – , – 3 2 1
x = – , 9 1 2
Solution follows…

15. 7.1.1 Topic Solving Quadratic Equations by Factoring Guided Practice
Solve each of these quadratic equations by using the zero property.
x = – , 3 1 4
24. 4x2 – 11x – 3 = 0
x2 – x – 2 = 0
26. 2x2 + 11x + 12 = 0
x2 – 27x + 5 = 0
28. 3x2 – 17x – 28 = 0
29. 2x2 – x – 28 = 0
x = , – 1 2 5
x = – , –4 3 2
x = , 5 2 1
x = – , 7 4 3
x = – , 4 7 2
Solution follows…

16. 7.1.1 Topic Solving Quadratic Equations by Factoring
Using Factoring to Solve Quadratic Equations
First arrange the terms in the quadratic equation so that you have zero on one side.
Then factor the nonzero expression (if possible).
Once done, you can use the zero property to find the solutions.

17. 7.1.1 Topic Solving Quadratic Equations by Factoring
Example 4
Solve x2 – 6x – 7 = 0.
Solution
The right-hand side of the equation is already zero, so you can just factor the left-hand side:
x2 – 6x – 7 = (x + 1)(x – 7)
So (x + 1)(x – 7) = 0.
Using the zero property, either x + 1 = 0 or x – 7 = 0. So either x = –1 or x = 7.
Solution follows…

18. 7.1.1 Topic Solving Quadratic Equations by Factoring
Example 5
Solve x2 + 2x – 11 = –3.
Solution
This time, you have to arrange the equation so you have zero on one side.
x2 + 2x – 11 = –3
By adding 3 to both sides,
the right-hand side becomes 0.
Þ x2 + 2x – 8 = 0
Now you can factor
the left-hand side:
x2 + 2x – 8 = (x + 4)(x – 2) = 0
Since you have two expressions multiplied to give zero, you can use the zero property. That is, either x + 4 = 0 or x – 2 = 0.
So either x = –4 or x = 2.
Solution follows…

19. 7.1.1 Topic Solving Quadratic Equations by Factoring
Example 6
Solve 3x = 45x.
Solution
Once again, the first thing to do is get zero on one side:
3x = 45x
Þ 3x2 – 45x = 0
The left-hand side can be factored, which means you can rewrite this as: 3(x2 – 15x + 56) = 0,
or 3(x – 7)(x – 8) = 0
So using the zero property, either x – 7 = 0 or x – 8 = 0. So either x = 7 or x = 8.
Solution follows…

20. 7.1.1 Topic Solving Quadratic Equations by Factoring Guided Practice
Solve each of these equations.
30. x2 – 2x – 15 = 0
31. x2 – 7x – 18 = 0
32. k2 + 10k + 24 = 0
33. 4m2 + 4m – 15 = 0
34. 8k2 – 14k = 49
35. 15k2 + 28k = –5
x = –3, 5
x = –2, 9
k = –6, –4
m = , – 3 2 5
k = – , 7 4 2
k = – , – 5 3 1
Solution follows…

21. 7.1.1 Topic Solving Quadratic Equations by Factoring Guided Practice
Solve each of these equations.
36. 6y2 + 28y + 20 = 5 – 6y2
37. 4x2 + 6x + 1 = 3x2 + 8x
38. 4(x2 – 5x) = –25
39. 3x(3x + 4) + 8 = 4
40. x(x + 4) + 9 = 5
41. x(x – 5) + 3 = –3
y = – , – 5 6 3 2
x = 1
x = 5 2
x = – 2 3
x = –2
x = 2, 3
Solution follows…

22. 7.1.1 Topic Solving Quadratic Equations by Factoring Guided Practice
Solve each of these equations.
42. 6x(3x – 4) – 7 = –15
43. 3 = 2 – 12x(3x – 1)
44. 7x(7x + 2) + 4x3 + 3 = 2(2x3 + 1)
45. (2x + 9)2(x + 3)(x + 1)–1(x + 3)–1(x + 1) = 0
46. 2x(3x + 3) + 4(x + 1) = 1 + 2x + 2x2
x = 2 3
x = 1 6
x = – 1 7
x = – 9 2
x = – , – 1 2 3
Solution follows…

23. 7.1.1 Topic Solving Quadratic Equations by Factoring
Independent Practice
The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in the quadratic equations below.
1. 4x2 + 20x + 9 = 0
2. x2 – 9x + 8 = x2 + 5x = x
4. x(2x + 3) = 5(2x + 3) 5. y(2y + 7) = 9(2y + 7)
6. (x + 2)(x – 2) = 3x
a = 4, b = 20, c = 9
a = 1, b = –9, c = 8
a = 2, b = –9, c = –35
a = 2, b = –7, c = –15
a = 2, b = –11, c = –63
a = 1, b = –3, c = –4
Solution follows…

24. 7.1.1 Topic Solving Quadratic Equations by Factoring
Independent Practice
Use the zero product property to solve these equations.
7. (2y + 9)(2y – 3) = 0 8. (2a + 5)(a – 11) = 0 9. (y – 3)2(y – 5)(y – 3)–1 = (y – 4)3(2y – 9)2(y – 4)–3(y – 7)(2y – 9)–1 = 0
y = – , 9 2 3
a = – , 11 5 2
y = 3, 5
y = , 7 9 2
Solution follows…

25. 7.1.1 Topic Solving Quadratic Equations by Factoring
Independent Practice
Solve the following equations.
11. 4x2 + 20x + 9 = 0
12. x2 – 9x + 8 = 0
13. 2x2 + 5x = x
14. x(2x + 3) = 5(2x + 3)
x = – , – 1 2 9
x = 1, 8
x = – , 7 5 2
x = – , 5 3 2
Solution follows…

26. 7.1.1 Topic Solving Quadratic Equations by Factoring
Independent Practice
Solve the following equations.
15. y(2y + 7) = 9(2y + 7)
16. (x + 2)(x – 2) = 3x
17. 2x(2x – 5) = 3(2x – 5)
18. 2x(3x – 1) + 7 = 7(2 – 3x)
y = – , 9 7 2
x = 4, –1
x = , 5 2 3
x = – , 7 2 1 3
Solution follows…

27. 7.1.1 Topic Solving Quadratic Equations by Factoring
Independent Practice
19. The product of two consecutive positive numbers is 30. Find the numbers.
20. The product of two consecutive positive odd numbers is 35. Find the numbers.
21. The area of a rectangle is 35 ft2. If the width of the rectangle is x ft and the length is (3x + 16) ft, find the value of x.
5, 6
5, 7
x = 5 3
Solution follows…

28. 7.1.1 Topic Solving Quadratic Equations by Factoring
Independent Practice
22. The area of a rectangle is 75 cm2. If the length of the rectangle is (4x + 25) cm and the width is 2x cm, find the dimensions of the rectangle.
23. Eylora has x pet goldfish and Leo has (4x – 25). If the product of the numbers of Eylora’s and Leo’s goldfish is 21, how many goldfish does Leo have?
24. Scott fixed x computers and Meimei fixed (5x – 7) computers. If the product of the number each fixed is 6, who fixed more computers?
cm by 30 cm 5 2
3 goldfish
Meimei fixed more computers
Solution follows…

29. 7.1.1 Topic Solving Quadratic Equations by Factoring Round Up
Don’t forget that you need to rearrange the equation until you’ve got a zero on one side before you can factor a quadratic.
Not all quadratics can be factored like this, as you’ll see in the next Topic.