Using the Distributive Property in Reverse

Prepared by Mr. DahlbergAlgebra I

Semester II

Greatest Common Factor

• Remember finding the GCF of numbers like 20 and 24?

• That means, what is the largest number that will divide evenly into both numbers.

2?

4?

6?

10?

• Does 2 work?20/2 = 10

24/2 = 12

• Does 4 work?20/4 = 5

24/4 = 6

What about 6?

What about 10?

6 and 10

• Does 6 work?20/6 = not a whole number

24/6 = 4

• Does 10 work?20/10 = 2

24/10 = not a whole number

What’s the Deal?

In today’s lesson we will

Find the GCF

Factor by Grouping

Use the Distributive property in reverse.

4x – 16

• What do both 4x and -16 have in common?

• Both have multiples of 4 as the constant or coefficient.

• If we divide both terms by 4– We get x and -4– So 4(x-4) is our answer.

Check the work

4(x-4)

=(4*x) + (4*-4)

= 4x + - 16, or 4x-16

Watch again.

5x – 35

A multiple of five is found in both.

Divide 5x by 5 to get xDivide -35 by 5 to get -7Five is factored out of both terms.5x – 35 = 5(x-7)

Try These.

7x – 49 =

7( x – 7)

3x + 36 =

3(x + 12)

Now for squared variables

4x2 + 28x

Notice that both terms have a multiple of four AND an x value.

4x2 + 28x

If we only divide four from both, we will get 4 (x2 – 7x)

Notice that both terms in the parentheses still carry at least one x value.

4(x2 + 7x)

• If we continue by dividing an x from both values in the parentheses, we will get4* x(x+7), or 4x (x+7)

Watch this.

2y2 + 20y

= 2 (y2 + 10y)

= 2* y( y+10)

= 2y(y+10)

Now let’s try factoring in one step.

5x2 +10x

Let’s factor a five and an x from both terms.

5x2 = 5x*x

10x = 5x*2

5x is divided from both terms.

5x(x + 2)

Try these.

• 7x2 – 56xWhat numbers can be

divided from both terms?

• 2n2 + 24nWhat numbers can be

divided from both terms?

• Factor out 7 and x

7x (x – 8)

• Factor out 2 and n

2n ( n + 12)

x(x+1) + 4(x+1)

• Notice that both pairs of parentheses contain the same numbers.

• We can factor our (x+1) from each term.

• We are left with x and 4

• (x+4)(x+1)

x(x+3) + 7(x+3)

• Try to factor our (x+3) from each term.

• We are left with x and 7

• (x+7)(x+3)

End of Part One.