Page 158 - MarceAlgebra Demystified

CHAPTER 6 Factoring 145

12 6 6 3 3 6
10: x 1 ¼ðx 1Þðx þ 1Þ¼ðx 1Þðx þ 1Þðx þ 1Þ
2
2
When the first term is not x , see if you can factor out the coefficient of x .If
2
you can, then you are left with a quadratic whose first term is x . For example
2
each term in 2x þ 16x 18 is divisible by 2:
2
2
2x þ 16x 18 ¼ 2ðx þ 8x 9Þ¼ 2ðx þ 9Þðx 1Þ.
Practice

2
1: 4x þ 28x þ 48 ¼

2
2: 3x 9x 54 ¼
2
3: 9x 9x 18 ¼
2
4: 15x 60 ¼
2
5: 6x þ 24x þ 24 ¼

Solutions

2 2
1: 4x þ 28x þ 48 ¼ 4ðx þ 7x þ 12Þ¼ 4ðx þ 4Þðx þ 3Þ
2 2
2: 3x 9x 54 ¼ 3ðx 3x 18Þ¼ 3ðx 6Þðx þ 3Þ
2 2
3: 9x 9x 18 ¼ 9ðx x 2Þ¼ 9ðx 2Þðx þ 1Þ
2 2
4: 15x 60 ¼ 15ðx 4Þ¼ 15ðx 2Þðx þ 2Þ
2 2 2
5: 6x þ 24x þ 24 ¼ 6ðx þ 4x þ 4Þ¼ 6ðx þ 2Þðx þ 2Þ¼ 6ðx þ 2Þ
2
The coefficient of the x term will not always factor away. In order to factor
2
quadratics such as 4x þ 8x þ 3 you will need to try all combinations of
factors of 4 and of 3: ð4x þ Þðx þ Þ and ð2x þ Þð2x þ Þ. The blanks
will be filled in with the factors of 3. You will need to check all of the
possibilities: ð4x þ 1Þðx þ 3Þ, ð4x þ 3Þðx þ 1Þ, and ð2x þ 1Þð2x þ 3Þ:

Example

2
4x 4x 15