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# Factoring by Grouping Help

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By McGraw-Hill Professional
Updated on Sep 26, 2011

## Combining Terms as Common Factors

Sometimes you can combine two or more terms at a time in such a way that each term has an algebraic expression as a common factor.

#### Examples

3x2 – 3 + x3x

If 3 is factored from the first two terms and x is factored from the last two terms, we would have two terms with a factor of x2 – 1.

3x2 – 3 + x3x = 3(x2 – 1)+ x (x2 – 1) = (3 + x )(x2 – 1)

You could also combine the first and third terms and the second and fourth terms.

3x2 – 3 + x3x = 3x2 + x3 – 3 – x = x2 (3 + x) – (3 + x)

= (x2 – 1)(3 + x)

3xy – 2y + 3x2 – 2x = y (3x – 2)+ x (3x – 2) = (y + x)(3x – 2)

5x2 – 25 – x2 y + 5y = 5(x2 – 5) – y (x2 – 5) = (5 – y)(x2 – 5)

4x4 + x3 – 4x – 1 = x3 (4x + 1) – (4x + 1) = (x3 – 1)(4x + 1)

### Factoring by Grouping Practice Problems

#### Practice

1. 6xy2 + 4xy + 9xy + 6x =

2. x3 + x2x – 1 =

3. 15xy + 5x + 6y + 2 =

4. 2x4 – 6xx3y + 3y =

5. 9x3 + 18x2x – 2 =

#### Solutions

1. 6xy2 + 4xy + 9xy + 6x = 2xy (3y + 2) + 3x (3y + 2)

= (2xy + 3x )(3y + 2) = x (2y + 3)(3y + 2)

2. x3 + x2x – 1 = x2 (x + 1) – 1(x + 1) = (x2 – 1)(x + 1)

3. 15xy + 5x + 6y + 2 = 5x (3y + 1)+ 2(3y + 1)

= (5x + 2)(3y + 1)

4. 2x4 – 6xx3y + 3y = 2x (x3 – 3) – y (x3 – 3) = (2xy )(x3 – 3)

5. 9x3 + 18x2x – 2 = 9x 2(x + 2) – 1(x + 2) = (9x2 – 1)(x + 2)

Practice problems for this concept can be found at: Algebra Factoring Practice Test.

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