Factoring Trinomials
Joseph Lee
Metropolitan Community College
Joseph Lee
Factoring Trinomials
Example 1.
Multiply.
(3x − 7)(2x + 5)
Joseph Lee
Factoring Trinomials
Example 1.
Multiply.
(3x − 7)(2x + 5)
Solution.
(3x − 7)(2x + 5) =
Joseph Lee
Factoring Trinomials
Example 1.
Multiply.
(3x − 7)(2x + 5)
Solution.
(3x − 7)(2x + 5) = 6x 2 + 15x − 14x − 35
=
Joseph Lee
Factoring Trinomials
Example 1.
Multiply.
(3x − 7)(2x + 5)
Solution.
(3x − 7)(2x + 5) = 6x 2 + 15x − 14x − 35
= 6x 2 + x − 35
Joseph Lee
Factoring Trinomials
Example 2.
Factor.
6x 2 + x − 35
Joseph Lee
Factoring Trinomials
Example 2.
Factor.
6x 2 + x − 35
Solution.
6x 2 + x − 35 = (3x − 7)(2x + 5)
Joseph Lee
Factoring Trinomials
Example 3.
Factor.
2x 2 + 3x − 5
Joseph Lee
Factoring Trinomials
Example 3.
Factor.
2x 2 + 3x − 5
Solution.
2x 2 + 3x − 5 =
Joseph Lee
Factoring Trinomials
Example 3.
Factor.
2x 2 + 3x − 5
Solution.
2x 2 + 3x − 5 = (2x
Joseph Lee
)(x
Factoring Trinomials
)
Example 3.
Factor.
2x 2 + 3x − 5
Solution.
2x 2 + 3x − 5 = (2x
)(x
)
We need factors of 5, and the only factors of 5 are 1 and 5.
Observe, our two choices:
(2x
(2x
1)(x
5)(x
5) = 2x 2
1) = 2x 2
Joseph Lee
10x x
2x 5x
Factoring Trinomials
5
5
Example 3.
Factor.
2x 2 + 3x − 5
Solution.
2x 2 + 3x − 5 = (2x
)(x
)
We need factors of 5, and the only factors of 5 are 1 and 5.
Observe, our two choices:
(2x
(2x
1)(x
5)(x
5) = 2x 2
1) = 2x 2
10x x
2x 5x
5
5
The only way to come up with 2x 2 + 3x − 5 would be
(2x
5)(x
1) = 2x 2 − 2x + 5x
Joseph Lee
Factoring Trinomials
5.
Example 3.
Factor.
2x 2 + 3x − 5
Solution.
2x 2 + 3x − 5 = (2x
)(x
)
We need factors of 5, and the only factors of 5 are 1 and 5.
Observe, our two choices:
(2x
(2x
1)(x
5)(x
5) = 2x 2
1) = 2x 2
10x x
2x 5x
5
5
The only way to come up with 2x 2 + 3x − 5 would be
(2x
5)(x
1) = 2x 2 − 2x + 5x
Finally,
(2x + 5)(x − 1) = 2x 2 + 3x − 5.
Joseph Lee
Factoring Trinomials
5.
Example 4.
Factor.
3x 2 + 11x − 20
Joseph Lee
Factoring Trinomials
Example 4.
Factor.
3x 2 + 11x − 20
Solution.
3x 2 + 11x − 20 =
Joseph Lee
Factoring Trinomials
Example 4.
Factor.
3x 2 + 11x − 20
Solution.
3x 2 + 11x − 20 = (3x
Joseph Lee
)(x
Factoring Trinomials
)
Example 4.
Factor.
3x 2 + 11x − 20
Solution.
3x 2 + 11x − 20 = (3x
)(x
We need factors of 20:
20 = 1 · 20
2 · 10
4·5
Joseph Lee
Factoring Trinomials
)
Example 4.
Factor.
3x 2 + 11x − 20
Solution.
3x 2 + 11x − 20 = (3x
)(x
)
60x
x
3x 20x
30x 2x
6x 10x
15x 4x
12x 5x
20
20
20
20
20
20
We need factors of 20:
20 = 1 · 20
2 · 10
4·5
We have the following choices:
(3x
(3x
(3x
(3x
(3x
(3x
1)(x 20) = 3x 2
20)(x 1) = 3x 2
2)(x 10) = 3x 2
10)(x 2) = 3x 2
4)(x 5) = 3x 2
5)(x 4) = 3x 2
Joseph Lee
Factoring Trinomials
Example 5.
Factor.
3x 2 + 11x − 20
Solution.
3x 2 + 11x − 20 = (3x
)(x
)
We need factors of 20:
20 = 1 · 20
2 · 10
4·5
We have the following choices:
(3x 1)(x 20) =
(3x 20)(x 1) =
(3x 2)(x 10) =
(3x 10)(x 2) =
(3x − 4)(x + 5) =
(3x 5)(x 4) =
Joseph Lee
3x 2 60x
3x 2 3x
3x 2 30x
3x 2 6x
3x 2 + 15x
3x 2 12x
x
20x
2x
10x
− 4x
5x
Factoring Trinomials
20
20
20
20
− 20
20
Example 6.
Factor.
5x 2 − 13x + 6
Joseph Lee
Factoring Trinomials
Example 6.
Factor.
5x 2 − 13x + 6
Solution.
5x 2 − 13x + 6 =
Joseph Lee
Factoring Trinomials
Example 6.
Factor.
5x 2 − 13x + 6
Solution.
5x 2 − 13x + 6 = (5x
Joseph Lee
)(x
Factoring Trinomials
)
Example 6.
Factor.
5x 2 − 13x + 6
Solution.
5x 2 − 13x + 6 = (5x
)(x
We need factors of 6:
6= 1·6
2·3
Joseph Lee
Factoring Trinomials
)
Example 6.
Factor.
5x 2 − 13x + 6
Solution.
5x 2 − 13x + 6 = (5x
)(x
)
We need factors of 6:
6= 1·6
2·3
We have the following choices:
(5x
(5x
(5x
(5x
1)(x
6)(x
2)(x
3)(x
6) =
1) =
3) =
2) =
Joseph Lee
5x 2
5x 2
5x 2
5x 2
30x x 6
5x 6x 6
15x 2x 6
10x 3x 6
Factoring Trinomials
Example 6.
Factor.
5x 2 − 13x + 6
Solution.
5x 2 − 13x + 6 = (5x
)(x
)
We need factors of 6:
6= 1·6
2·3
We have the following choices:
(5x
(5x
(5x
(5x
1)(x
6)(x
2)(x
3)(x
6) =
1) =
3) =
2) =
5x 2
5x 2
5x 2
5x 2
30x x 6
5x 6x 6
15x 2x 6
10x 3x 6
Our choice is:
(5x − 3)(x − 2) = 5x 2 − 10x − 3x + 6
Joseph Lee
Factoring Trinomials
Example 7.
Factor.
4x 2 + 12x − 7
Joseph Lee
Factoring Trinomials
Example 7.
Factor.
4x 2 + 12x − 7
Solution. Note we have two possibilities to begin with:
4x 2 + 12x − 7 =
Joseph Lee
Factoring Trinomials
Example 7.
Factor.
4x 2 + 12x − 7
Solution. Note we have two possibilities to begin with:
4x 2 + 12x − 7 = (4x
= (2x
Joseph Lee
)(x
)(2x
Factoring Trinomials
)
)
Example 7.
Factor.
4x 2 + 12x − 7
Solution. Note we have two possibilities to begin with:
4x 2 + 12x − 7 = (4x
= (2x
)(x
)(2x
The only factors of 7, however, are 1 and 7.
Joseph Lee
Factoring Trinomials
)
)
Example 7.
Factor.
4x 2 + 12x − 7
Solution. Note we have two possibilities to begin with:
4x 2 + 12x − 7 = (4x
= (2x
)(x
)(2x
)
)
The only factors of 7, however, are 1 and 7.
Thus, we have the following choices:
(4x 1)(x
(4x 7)(x
(2x 1)(2x
7) = 4x 2
1) = 4x 2
7) = 4x 2
Joseph Lee
28x x 7
4x 7x 7
14x 2x 7
Factoring Trinomials
Example 7.
Factor.
4x 2 + 12x − 7
Solution. Note we have two possibilities to begin with:
4x 2 + 12x − 7 = (4x
= (2x
)(x
)(2x
)
)
The only factors of 7, however, are 1 and 7.
Thus, we have the following choices:
(4x 1)(x
(4x 7)(x
(2x 1)(2x
7) = 4x 2
1) = 4x 2
7) = 4x 2
28x x 7
4x 7x 7
14x 2x 7
Our choice is:
(2x − 1)(2x + 7) = 4x 2 + 14x − 2x − 7
Joseph Lee
Factoring Trinomials
Example 8.
Factor.
6x 2 − 27x + 12
Joseph Lee
Factoring Trinomials
Example 8.
Factor.
6x 2 − 27x + 12
Solution. We should first observe that we may factor out a
common factor.
6x 2 − 27x + 12 =
Joseph Lee
Factoring Trinomials
Example 8.
Factor.
6x 2 − 27x + 12
Solution. We should first observe that we may factor out a
common factor.
6x 2 − 27x + 12 = 3(2x 2 − 9x + 4)
=
Joseph Lee
Factoring Trinomials
Example 8.
Factor.
6x 2 − 27x + 12
Solution. We should first observe that we may factor out a
common factor.
6x 2 − 27x + 12 = 3(2x 2 − 9x + 4)
= 3(2x
)(x
)
Joseph Lee
Factoring Trinomials
Example 8.
Factor.
6x 2 − 27x + 12
Solution. We should first observe that we may factor out a
common factor.
6x 2 − 27x + 12 = 3(2x 2 − 9x + 4)
= 3(2x
)(x
)
We need factors of 4:
4= 1·4
2·2
Joseph Lee
Factoring Trinomials
Example 8.
Factor.
6x 2 − 27x + 12
Solution. We should first observe that we may factor out a
common factor.
6x 2 − 27x + 12 = 3(2x 2 − 9x + 4)
= 3(2x
)(x
)
We need factors of 4:
4= 1·4
2·2
We have the following choices:
3(2x
3(2x
3(2x
1)(x
4)(x
2)(x
4) = 3(2x 2
1) = 3(2x 2
2) = 3(2x 2
Joseph Lee
8x
2x
4x
x 4)
4x 4)
2x 4)
Factoring Trinomials
Example 8.
Factor.
6x 2 − 27x + 12
Solution. We should first observe that we may factor out a
common factor.
6x 2 − 27x + 12 = 3(2x 2 − 9x + 4)
= 3(2x
)(x
)
We need factors of 4:
4= 1·4
2·2
We have the following choices:
3(2x
3(2x
3(2x
1)(x
4)(x
2)(x
4) = 3(2x 2
1) = 3(2x 2
2) = 3(2x 2
8x
2x
4x
x 4)
4x 4)
2x 4)
Our choice is:
3(2x − 1)(x − 4) = 3(2x 2 − 8x − x + 4)
Joseph Lee
Factoring Trinomials
Example 9.
Factor.
(x + 7)2 − 6(x + 7) − 16
Joseph Lee
Factoring Trinomials
Example 9.
Factor.
(x + 7)2 − 6(x + 7) − 16
Solution. Let u = x + 7.
(x + 7)2 − 6(x + 7) − 16 =
Joseph Lee
Factoring Trinomials
Example 9.
Factor.
(x + 7)2 − 6(x + 7) − 16
Solution. Let u = x + 7.
(x + 7)2 − 6(x + 7) − 16 = u 2 − 6u − 16
=
Joseph Lee
Factoring Trinomials
Example 9.
Factor.
(x + 7)2 − 6(x + 7) − 16
Solution. Let u = x + 7.
(x + 7)2 − 6(x + 7) − 16 = u 2 − 6u − 16
= (u − 8)(u + 2)
=
Joseph Lee
Factoring Trinomials
Example 9.
Factor.
(x + 7)2 − 6(x + 7) − 16
Solution. Let u = x + 7.
(x + 7)2 − 6(x + 7) − 16 = u 2 − 6u − 16
= (u − 8)(u + 2)
= [(x + 7) − 8][(x + 7) + 2]
=
Joseph Lee
Factoring Trinomials
Example 9.
Factor.
(x + 7)2 − 6(x + 7) − 16
Solution. Let u = x + 7.
(x + 7)2 − 6(x + 7) − 16 = u 2 − 6u − 16
= (u − 8)(u + 2)
= [(x + 7) − 8][(x + 7) + 2]
= (x − 1)(x + 9)
Joseph Lee
Factoring Trinomials
Example 10.
Factor.
4x 6 − 4x 3 − 15
Joseph Lee
Factoring Trinomials
Example 10.
Factor.
4x 6 − 4x 3 − 15
Solution. Let u = x 3 .
4x 6 − 4x 3 − 15 =
Joseph Lee
Factoring Trinomials
Example 10.
Factor.
4x 6 − 4x 3 − 15
Solution. Let u = x 3 .
4x 6 − 4x 3 − 15 = 4u 2 − 4u − 15
=
Joseph Lee
Factoring Trinomials
Example 10.
Factor.
4x 6 − 4x 3 − 15
Solution. Let u = x 3 .
4x 6 − 4x 3 − 15 = 4u 2 − 4u − 15
= (2u
)(2u
=
Joseph Lee
Factoring Trinomials
)
Example 10.
Factor.
4x 6 − 4x 3 − 15
Solution. Let u = x 3 .
4x 6 − 4x 3 − 15 = 4u 2 − 4u − 15
= (2u
)(2u
= (2u + 3)(2u − 5)
=
Joseph Lee
Factoring Trinomials
)
Example 10.
Factor.
4x 6 − 4x 3 − 15
Solution. Let u = x 3 .
4x 6 − 4x 3 − 15 = 4u 2 − 4u − 15
= (2u
)(2u
)
= (2u + 3)(2u − 5)
= (2x 3 + 3)(2x 3 − 5)
Joseph Lee
Factoring Trinomials
Example 11.
Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Joseph Lee
Factoring Trinomials
Example 11.
Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Solution. Let u = 2n2 + 1.
4(2n2 + 1)2 − 7(2n2 + 1) + 3 =
Joseph Lee
Factoring Trinomials
Example 11.
Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Solution. Let u = 2n2 + 1.
4(2n2 + 1)2 − 7(2n2 + 1) + 3 = 4u 2 − 7u + 3
=
Joseph Lee
Factoring Trinomials
Example 11.
Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Solution. Let u = 2n2 + 1.
4(2n2 + 1)2 − 7(2n2 + 1) + 3 = 4u 2 − 7u + 3
= (4u
)(u
=
Joseph Lee
Factoring Trinomials
)
Example 11.
Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Solution. Let u = 2n2 + 1.
4(2n2 + 1)2 − 7(2n2 + 1) + 3 = 4u 2 − 7u + 3
= (4u
)(u
= (4u − 3)(u − 1)
=
Joseph Lee
Factoring Trinomials
)
Example 11.
Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Solution. Let u = 2n2 + 1.
4(2n2 + 1)2 − 7(2n2 + 1) + 3 = 4u 2 − 7u + 3
= (4u
)(u
)
= (4u − 3)(u − 1)
= [4(2n2 + 1) − 3][(2n2 + 1) − 1]
=
Joseph Lee
Factoring Trinomials
Example 11.
Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Solution. Let u = 2n2 + 1.
4(2n2 + 1)2 − 7(2n2 + 1) + 3 = 4u 2 − 7u + 3
= (4u
)(u
)
= (4u − 3)(u − 1)
= [4(2n2 + 1) − 3][(2n2 + 1) − 1]
= (8n2 + 4 − 3)(2n2 )
=
Joseph Lee
Factoring Trinomials
Example 11.
Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Solution. Let u = 2n2 + 1.
4(2n2 + 1)2 − 7(2n2 + 1) + 3 = 4u 2 − 7u + 3
= (4u
)(u
)
= (4u − 3)(u − 1)
= [4(2n2 + 1) − 3][(2n2 + 1) − 1]
= (8n2 + 4 − 3)(2n2 )
= (8n2 + 1)(2n2 )
Joseph Lee
Factoring Trinomials