Solving Equations by Factoring
Joseph Lee
Metropolitan Community College
Joseph Lee
Solving Equations by Factoring
Example 1.
Solve.
x 2 − 5x + 6 = 0
Joseph Lee
Solving Equations by Factoring
Example 1.
Solve.
x 2 − 5x + 6 = 0
Solution.
x 2 − 5x + 6 = 0
Joseph Lee
Solving Equations by Factoring
Example 1.
Solve.
x 2 − 5x + 6 = 0
Solution.
x 2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
Joseph Lee
Solving Equations by Factoring
Example 1.
Solve.
x 2 − 5x + 6 = 0
Solution.
x 2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
Thus, by the zero factor property, either
x −2= 0
or
Joseph Lee
x −3= 0
Solving Equations by Factoring
Example 1.
Solve.
x 2 − 5x + 6 = 0
Solution.
x 2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
Thus, by the zero factor property, either
x −2= 0
or
x= 2
Joseph Lee
x −3= 0
x= 3
Solving Equations by Factoring
Example 1.
Solve.
x 2 − 5x + 6 = 0
Solution.
x 2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
Thus, by the zero factor property, either
x −2= 0
or
x= 2
x −3= 0
x= 3
The solution set is {2, 3}.
Joseph Lee
Solving Equations by Factoring
Example 2.
Solve.
3x 2 − 10x − 8 = 0
Joseph Lee
Solving Equations by Factoring
Example 2.
Solve.
3x 2 − 10x − 8 = 0
Solution.
3x 2 − 10x − 8 = 0
Joseph Lee
Solving Equations by Factoring
Example 2.
Solve.
3x 2 − 10x − 8 = 0
Solution.
3x 2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
Joseph Lee
Solving Equations by Factoring
Example 2.
Solve.
3x 2 − 10x − 8 = 0
Solution.
3x 2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
Thus, by the zero factor property, either
3x + 2 = 0
Joseph Lee
or
x −4= 0
Solving Equations by Factoring
Example 2.
Solve.
3x 2 − 10x − 8 = 0
Solution.
3x 2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
Thus, by the zero factor property, either
3x + 2 = 0
3x = −2
Joseph Lee
or
x −4= 0
x= 4
Solving Equations by Factoring
Example 2.
Solve.
3x 2 − 10x − 8 = 0
Solution.
3x 2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
Thus, by the zero factor property, either
3x + 2 = 0
or
3x = −2
x= −
x −4= 0
x= 4
2
3
Joseph Lee
Solving Equations by Factoring
Example 2.
Solve.
3x 2 − 10x − 8 = 0
Solution.
3x 2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
Thus, by the zero factor property, either
3x + 2 = 0
3x = −2
or
x −4= 0
x= 4
2
3
2
The solution set is − , 4 .
3
x= −
Joseph Lee
Solving Equations by Factoring
Example 3.
Solve.
3x 3 − 3x 2 − 6x = 0
Joseph Lee
Solving Equations by Factoring
Example 3.
Solve.
3x 3 − 3x 2 − 6x = 0
Solution.
3x 3 − 3x 2 − 6x = 0
Joseph Lee
Solving Equations by Factoring
Example 3.
Solve.
3x 3 − 3x 2 − 6x = 0
Solution.
3x 3 − 3x 2 − 6x = 0
3x(x 2 − x − 2) = 0
Joseph Lee
Solving Equations by Factoring
Example 3.
Solve.
3x 3 − 3x 2 − 6x = 0
Solution.
3x 3 − 3x 2 − 6x = 0
3x(x 2 − x − 2) = 0
3x(x + 1)(x − 2) = 0
Joseph Lee
Solving Equations by Factoring
Example 3.
Solve.
3x 3 − 3x 2 − 6x = 0
Solution.
3x 3 − 3x 2 − 6x = 0
3x(x 2 − x − 2) = 0
3x(x + 1)(x − 2) = 0
Thus, by the zero factor property, either
3x = 0
or
x +1= 0
Joseph Lee
or
x −2= 0
Solving Equations by Factoring
Example 3.
Solve.
3x 3 − 3x 2 − 6x = 0
Solution.
3x 3 − 3x 2 − 6x = 0
3x(x 2 − x − 2) = 0
3x(x + 1)(x − 2) = 0
Thus, by the zero factor property, either
3x = 0
x= 0
or
x +1= 0
or
x −2= 0
x = −1
Joseph Lee
Solving Equations by Factoring
x= 2
Example 3.
Solve.
3x 3 − 3x 2 − 6x = 0
Solution.
3x 3 − 3x 2 − 6x = 0
3x(x 2 − x − 2) = 0
3x(x + 1)(x − 2) = 0
Thus, by the zero factor property, either
3x = 0
or
x +1= 0
or
x −2= 0
x = −1
x= 0
The solution set is {−1, 0, 2}.
Joseph Lee
Solving Equations by Factoring
x= 2
Example 4.
Solve.
2x 2 = 5x + 3
Joseph Lee
Solving Equations by Factoring
Example 4.
Solve.
2x 2 = 5x + 3
Solution.
2x 2 = 5x + 3
Joseph Lee
Solving Equations by Factoring
Example 4.
Solve.
2x 2 = 5x + 3
Solution.
2x 2 = 5x + 3
2x 2 − 5x − 3 = 0
Joseph Lee
Solving Equations by Factoring
Example 4.
Solve.
2x 2 = 5x + 3
Solution.
2x 2 = 5x + 3
2x 2 − 5x − 3 = 0
(2x + 1)(x − 3) = 0
Joseph Lee
Solving Equations by Factoring
Example 4.
Solve.
2x 2 = 5x + 3
Solution.
2x 2 = 5x + 3
2x 2 − 5x − 3 = 0
(2x + 1)(x − 3) = 0
Thus, by the zero factor property, either
2x + 1 = 0
Joseph Lee
or
x −3= 0
Solving Equations by Factoring
Example 4.
Solve.
2x 2 = 5x + 3
Solution.
2x 2 = 5x + 3
2x 2 − 5x − 3 = 0
(2x + 1)(x − 3) = 0
Thus, by the zero factor property, either
2x + 1 = 0
x= −
or
1
2
Joseph Lee
x −3= 0
x= 3
Solving Equations by Factoring
Example 4.
Solve.
2x 2 = 5x + 3
Solution.
2x 2 = 5x + 3
2x 2 − 5x − 3 = 0
(2x + 1)(x − 3) = 0
Thus, by the zero factor property, either
2x + 1 = 0
1
2
1
The solution set is − , 3 .
2
x= −
Joseph Lee
or
x −3= 0
x= 3
Solving Equations by Factoring
Example 5.
Solve.
(x − 3)(x − 5) = −1
Joseph Lee
Solving Equations by Factoring
Example 5.
Solve.
(x − 3)(x − 5) = −1
Solution.
(x − 3)(x − 5) = −1
Joseph Lee
Solving Equations by Factoring
Example 5.
Solve.
(x − 3)(x − 5) = −1
Solution.
(x − 3)(x − 5) = −1
x 2 − 8x + 15 = −1
Joseph Lee
Solving Equations by Factoring
Example 5.
Solve.
(x − 3)(x − 5) = −1
Solution.
(x − 3)(x − 5) = −1
x 2 − 8x + 15 = −1
x 2 − 8x + 16 = 0
Joseph Lee
Solving Equations by Factoring
Example 5.
Solve.
(x − 3)(x − 5) = −1
Solution.
(x − 3)(x − 5) = −1
x 2 − 8x + 15 = −1
x 2 − 8x + 16 = 0
(x − 4)2 = 0
Joseph Lee
Solving Equations by Factoring
Example 5.
Solve.
(x − 3)(x − 5) = −1
Solution.
(x − 3)(x − 5) = −1
x 2 − 8x + 15 = −1
x 2 − 8x + 16 = 0
(x − 4)2 = 0
Thus, by the zero factor property,
x −4= 0
Joseph Lee
Solving Equations by Factoring
Example 5.
Solve.
(x − 3)(x − 5) = −1
Solution.
(x − 3)(x − 5) = −1
x 2 − 8x + 15 = −1
x 2 − 8x + 16 = 0
(x − 4)2 = 0
Thus, by the zero factor property,
x −4= 0
x= 4
Joseph Lee
Solving Equations by Factoring
Example 5.
Solve.
(x − 3)(x − 5) = −1
Solution.
(x − 3)(x − 5) = −1
x 2 − 8x + 15 = −1
x 2 − 8x + 16 = 0
(x − 4)2 = 0
Thus, by the zero factor property,
x −4= 0
x= 4
The solution set is {4}.
Joseph Lee
Solving Equations by Factoring
Example 6.
Solve.
3x(x − 2) = 4(x + 1) + 4
Joseph Lee
Solving Equations by Factoring
Example 6.
Solve.
3x(x − 2) = 4(x + 1) + 4
Solution.
3x(x − 2) = 4(x + 1) + 4
Joseph Lee
Solving Equations by Factoring
Example 6.
Solve.
3x(x − 2) = 4(x + 1) + 4
Solution.
3x(x − 2) = 4(x + 1) + 4
3x 2 − 6x = 4x + 4 + 4
Joseph Lee
Solving Equations by Factoring
Example 6.
Solve.
3x(x − 2) = 4(x + 1) + 4
Solution.
3x(x − 2) = 4(x + 1) + 4
3x 2 − 6x = 4x + 4 + 4
3x 2 − 6x = 4x + 8
Joseph Lee
Solving Equations by Factoring
Example 6.
Solve.
3x(x − 2) = 4(x + 1) + 4
Solution.
3x(x − 2) = 4(x + 1) + 4
3x 2 − 6x = 4x + 4 + 4
3x 2 − 6x = 4x + 8
3x 2 − 10x − 8 = 0
Joseph Lee
Solving Equations by Factoring
Example 6.
Solve.
3x(x − 2) = 4(x + 1) + 4
Solution.
3x(x − 2) = 4(x + 1) + 4
3x 2 − 6x = 4x + 4 + 4
3x 2 − 6x = 4x + 8
3x 2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
Joseph Lee
Solving Equations by Factoring
Example 6.
Solve.
3x(x − 2) = 4(x + 1) + 4
Solution.
3x(x − 2) = 4(x + 1) + 4
3x 2 − 6x = 4x + 4 + 4
3x 2 − 6x = 4x + 8
3x 2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
2
The solution set is − , 4 .
3
Joseph Lee
Solving Equations by Factoring
Pythagorean Theorem
If a right triangle has legs of lengths a and b and hypotenuse of
length c, then
a2 + b 2 = c 2 .
c
b
a
Joseph Lee
Solving Equations by Factoring
Example 7.
x +2
x
x +1
Joseph Lee
Solving Equations by Factoring
Example 7.
x +2
x
x +1
Solution.
x 2 + (x + 1)2 = (x + 2)2
Joseph Lee
Solving Equations by Factoring
Example 7.
x +2
x
x +1
Solution.
x 2 + (x + 1)2 = (x + 2)2
x 2 + x 2 + 2x + 1 = x 2 + 4x + 4
Joseph Lee
Solving Equations by Factoring
Example 7.
x +2
x
x +1
Solution.
x 2 + (x + 1)2 = (x + 2)2
x 2 + x 2 + 2x + 1 = x 2 + 4x + 4
2x 2 + 2x + 1 = x 2 + 4x + 4
Joseph Lee
Solving Equations by Factoring
Example 7.
x +2
x
x +1
Solution.
x 2 + (x + 1)2 =
x 2 + x 2 + 2x + 1 =
2x 2 + 2x + 1 =
x 2 − 2x − 3 =
Joseph Lee
(x + 2)2
x 2 + 4x + 4
x 2 + 4x + 4
0
Solving Equations by Factoring
Example 7.
x +2
x
x +1
Solution.
x 2 + (x + 1)2 =
x 2 + x 2 + 2x + 1 =
2x 2 + 2x + 1 =
x 2 − 2x − 3 =
(x − 3)(x + 1) =
Joseph Lee
(x + 2)2
x 2 + 4x + 4
x 2 + 4x + 4
0
0
Solving Equations by Factoring
Example 7.
x +2
x
x +1
Solution.
x 2 + (x + 1)2 =
x 2 + x 2 + 2x + 1 =
2x 2 + 2x + 1 =
x 2 − 2x − 3 =
(x − 3)(x + 1) =
(x + 2)2
x 2 + 4x + 4
x 2 + 4x + 4
0
0
The solution set is {−1, 3}. Since x represents the length of a leg
of the triangle, we know x = 3. Thus, the sides of the triangle are
3, 4, and 5.
Joseph Lee
Solving Equations by Factoring
Example 8.
x
x −8
x −1
Joseph Lee
Solving Equations by Factoring
Example 8.
x
x −8
x −1
Solution.
(x − 8)2 + (x − 1)2 = x 2
Joseph Lee
Solving Equations by Factoring
Example 8.
x
x −8
x −1
Solution.
(x − 8)2 + (x − 1)2 = x 2
x 2 − 16x + 64 + x 2 − 2x + 1 = x 2
Joseph Lee
Solving Equations by Factoring
Example 8.
x
x −8
x −1
Solution.
(x − 8)2 + (x − 1)2 = x 2
x 2 − 16x + 64 + x 2 − 2x + 1 = x 2
2x 2 − 18x + 65 = x 2
Joseph Lee
Solving Equations by Factoring
Example 8.
x
x −8
x −1
Solution.
(x − 8)2 + (x − 1)2 =
x 2 − 16x + 64 + x 2 − 2x + 1 =
2x 2 − 18x + 65 =
x 2 − 18x + 65 =
Joseph Lee
x2
x2
x2
0
Solving Equations by Factoring
Example 8.
x
x −8
x −1
Solution.
(x − 8)2 + (x − 1)2 =
x 2 − 16x + 64 + x 2 − 2x + 1 =
2x 2 − 18x + 65 =
x 2 − 18x + 65 =
(x − 5)(x − 13) =
Joseph Lee
x2
x2
x2
0
0
Solving Equations by Factoring
Example 8.
x
x −8
x −1
Solution.
(x − 8)2 + (x − 1)2 =
x 2 − 16x + 64 + x 2 − 2x + 1 =
2x 2 − 18x + 65 =
x 2 − 18x + 65 =
(x − 5)(x − 13) =
x2
x2
x2
0
0
The solution set is {5, 13}. Since x − 8 represents the length of a
leg of the triangle, we know x = 13 (otherwise x − 8 = −3). Thus,
the sides of the triangle are 5, 12, and 13.
Joseph Lee
Solving Equations by Factoring