Factoring

Selasa, 05 April 2022

The distributive property, a(b + c) = ab + ac, can be used to factor a quantity from two or more terms. In the formula ab + ac = a(b + c), a is factored from (or divided into) ab and ac. The first step in factoring is to decide what quantity you want to factor from each term. Second write each term as a product of the factor and something else (this step will become unnecessary once you are experienced). Third apply the distribution property in reverse.

Practice

1. 4x – 10y =

2. 5x² + 15 =

3. 15x³y²z⁷ – 30xy²z⁴ + 6x⁴y²z⁶ =

Solutions

1. 4x – 10y = 2 · 2x – 2 · 5y = 2(2x – 5y)

2. 5x² + 15 = 5 · x² + 5 · 3 = 5(x² – 3)

3. 15x³y²z⁷ – 30xy²z⁴ + 6x⁴y²z⁶ = 3xy²z⁴· 5x²z³ – 3xy²z⁴ · 10 + 3xy²z⁴ · 2x³z²

= 3xy²z⁴(5x²z³ – 10 + 2x³z²)

Factoring a negative quantity has the same effect on signs within parentheses as distributing a negative quantity does—every sign changes. Negative quantities are factored in the next examples and practice problems.

Practice

1. 28xy² – 14x =

2. – 18y² + 6xy =

3. 20xyz² – 5yz =

Solutions

1. 28xy² – 14x = – 7x(– 4y² + 2)

2. – 18y² + 6xy = – 6y(3y – x)

3. 20xyz² – 5yz = – 5yz(– 4xz + 1)

The associative and distributive properties can be confusing. The associative property states (ab)c = a(bc). This property says that when multiplying three (or more) quantities you can multiply the first two then the third or multiply the second two then the first. For example, it might be tempting to write 5(x + 1)(y – 1) = (5x + 5)(5y – 15). But (5x + 5)(5y – 15) = [5(x + 1)][5(y – 3)] = 25(x + 1)(y – 3). The ‘‘5’’ can be grouped either with ‘‘x + 1’’ or with ‘‘y – 3’’ but not both: [5(x + 1)(y – 3) = (5x + 5)(y – 3) or (x + 1)[5(y – 3)] = (x + 1)(5y – 1).

Factors themselves can have more than one term. For instance 3(x + 4) – x(x + 4) has x + 4 as a factor in each term, so x + 4 can be factored from 3(x + 4) and x(x + 4):

3(x + 4) – x(x + 4) = 3(3 – x)(x + 4)

Practice

1. 2(x – y) + 3y(x – y) =

2. 6x(4 – 3x) – 2y(4 – 3x) – 5(4 – 3x) =

3. 3(x – 2y)⁴ + 2x(x – 2y)⁴ =

Solutions

1. 2(x – y) + 3y(x – y) = (2 + 3y)(x – y)

2. 6x(4 – 3x) – 2y(4 – 3x) – 5(4 – 3x) = (6x – 2y – 5)(4 – 3x)

3. 3(x – 2y)⁴ + 2x(x – 2y)⁴ = (3 + 2x)(x – 2y)⁴

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