The Extended Distributive Property

πŸ†Practice extended distributive property

The distributive property allows us to remove parentheses and simplify an expression, even if there is more than one set of parentheses.

In order to get rid of of the parentheses, we will multiply each term of the first parentheses by each term of the second parentheses, paying special attention to the addition/ subtraction signs.

For example:

(5+8)Γ—(7+2)(5+8)\times (7+2)

Using the distributive property, we can simplify the expression.

All we need to do is to multiply each of the terms in the first parentheses by each of the terms in the second parentheses:

(5+8)Γ—(7+2)= (5+8)\times (7+2) =

5Γ—7+5Γ—2+8Γ—7+8Γ—2=5\times 7+5\times 2+8\times 7+8\times 2 =

35+10+56+16=35+10+56+16 =

117117

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Test yourself on extended distributive property!

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\( (x+2)(x-4)= \)

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Example

(xβˆ’4)Γ—(xβˆ’2)=x2βˆ’2xβˆ’4x+8=x2βˆ’6x+8(x-4)\times (x-2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8

(x+3)Γ—(x+6)=x2+6x+3x+18=x2+9x+18(x+3)\times (x+6) = x^2 + 6x + 3x + 18 = x^2 + 9x + 18


The extended distributive property allows us to solve exercises with two sets of parentheses that are multiplied by eachother.

For example: (a+1)Γ—(b+2) (a+1)\times(b+2)

To find the solution, we will go through the following steps:

  • Step 1: Multiply the first term in the first parentheses by each of the terms in the second parentheses.
  • Step 2: Multiply the second term in the first parentheses by each of the terms in the second parentheses.
  • Step 3: Associate like terms.

ab+2a+b+2 ab+2a+b+2


Basic distributive property

Let's take a moment to remember our basic distributive property.

Below we can see the formula:

aΓ—(b+c)=ab+ac a\times(b+c)=ab+ac

Here, we have multiplied a a by each of the terms inside the parentheses, keeping the same order.

Extended distributive property

Now we will apply the same concept in the extended distributive property. This allows us to solve exercises with two sets of parentheses.

For example:
(a+b)Γ—(c+d)=ac+ad+bc+bd (a+b)\times(c+d)=ac+ad+bc+bd

How does the extended distributive property work?

  • Step 1: Multiply the first term in the first parentheses by each of the terms in the second parentheses.
  • Step 2: Multiply the second term in the first parentheses by each of the terms in the second parentheses.
  • Step 3: Associate like terms.

Example 1

1- The distributive property

Step 1: Multiply A A by each of the terms included in the second parentheses.

2 - The distributive property

Step 2: Multiply 2 2 by each of the terms included in the second parentheses.

3- The distributive property

Step 3: Order the terms and combine like terms, if any:

4- The distributive property

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Example 2: What do we do with a minus sign?

So, what do we do when we see a minus sign in one or both of the parentheses? Do we do anything different?

The method is the same! The only difference is that we need to make sure to put out minus/ negative signs in the right places when we distribute.

It can helpful to remember that a "minus sign" is the same as saying "plus a negative number."

For example, 4βˆ’2=4+(βˆ’2)=2 4-2=4+(-2)=2

Let's look at the exercise:

5 - The distributive property

Step 1: Multiply A A by each of the terms included inside the second parentheses.

6 - The distributive property

Step 2: Multiply 5 5 by each of the terms included inside the second parentheses.

7 - The distributive property

Pay attention to the signs of each of the terms! For example, we will see that, βˆ’5 -5 times βˆ’3 -3 equals +15 +15 .

In this case, there are no terms that we want to combine.


Example 3

Task:

Find the value of X X :

(X+2)2=(X+5)Γ—(Xβˆ’2) (X+2)^2=(X+5)\times(X-2)

Let's look at the left side of the equation and simplify:

(X+2)2=(X+2)Γ—(X+2) (X+2)^2=(X+2)\times(X+2)

Now we can use the extended distributive property on each side of the equation.

Now the equation looks like this:

(X+2)Γ—(X+2)=(X+5)Γ—(Xβˆ’2) (X+2)\times(X+2)=(X+5)\times(X-2)

After applying the distributive property:

X2+2X+2X+4=X2–2X+5X–10 X^2 + 2X + 2X + 4 = X^2 – 2X + 5X – 10

Let's reduce, combine like terms and arrange the equation.

We will get:

X=βˆ’14 X = - 14


Do you know what the answer is?

Exercises using the distributive property

Exercise 1

Assignment:

A painter has a canvas with the following dimensions:

(23x+12) (23x+12) length

(20x+7) (20x+7) width

What is the area the painter needs to paint?

Solution:

We multiply the length of the canvas by the width to find the area.

(23x+12)Γ—(20x+7)= (23x+12)\times(20x+7)=

Multiply each term in the first parentheses by each term in the second parentheses.

23xΓ—20x+23xΓ—7+12Γ—20x+12Γ—7=23x\times20x+23x\times7+12\times20x+12\times7=

We solve accordingly

460x2+161x+240x+84= 460x^2+161x+240x+84=

460x2+401x+84 460x^2+401x+84

Answer:

460x2+401x+84 460x^2+401x+84


Exercise 2

Task:

Find the area of the following rectangle:

Leave variables in your answer.

Exercise 2 Calculating the area of the rectangle

Solution:

To find the area we multiply the width by the length.

3yΓ—(y+3z)= 3y\times(y+3z)=

Multiply 3y by each of the terms in parentheses.

3yΓ—y+3yΓ—3z= 3y\times y+3y\times3z=

Solve accordingly

3y2+9yz 3y^2+9yz

Answer:

3y2+9yz 3y^2+9yz


Check your understanding

Exercise 3

Task:

(3+20)Γ—(12+4)= (3+20)\times(12+4)=

Solution:

We multiply each of the terms in the first parentheses by the terms in the second parentheses.

3Γ—12+3Γ—4+20Γ—12+20Γ—4= 3\times12+3\times4+20\times12+20\times4=

Solve accordingly

36+12+240+80= 36+12+240+80=

We add everything together

48+320=368 48+320=368

Answer:

368 368


Exercise 4

Task:

(12+2)Γ—(3+5)= (12+2)\times(3+5)=

Solution:

We multiply each of the terms in the first parentheses by the terms of the second parentheses.

12Γ—3+12Γ—5+2Γ—3+2Γ—5= 12\times3+12\times5+2\times3+2\times5=

Solve accordingly

36+60+6+10= 36+60+6+10=

We add everything together

96+16=112 96+16=112

Answer:

112 112


Do you think you will be able to solve it?

Exercise 5

Task:

(7x+4)Γ—(3x+4)=(7x+4)\times(3x+4)=

Solution:

We multiply each of the terms of the first parentheses by the terms of the second parentheses.

7xΓ—3x+7xΓ—4+4Γ—3x+4Γ—4= 7x\times3x+7x\times4+4\times3x+4\times4=

Solve accordingly

21x2+28x+12x+16= 21x^2+28x+12x+16=

21x2+40x+16 21x^2+40x+16

Answer:

21x2+40x+16 21x^2+40x+16


Exercise 6

Task:

(2xβˆ’3)Γ—(5xβˆ’7) (2x-3)\times(5x-7)

We multiply each of the terms of the first parentheses by the terms of the second parentheses.

2xΓ—5x+2xΓ—(βˆ’7)+(βˆ’3)Γ—5x+(βˆ’3)Γ—(βˆ’7)=2x\times5x+2x\times(-7)+(-3)\times5x+(-3)\times(-7)=

Solve accordingly

10x2βˆ’14xβˆ’15x+21= 10x^2-14x-15x+21=

10x2βˆ’29x+21 10x^2-29x+21

Answer:

10x2βˆ’29x+21 10x^2-29x+21


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Review questions

What is the distributive property of multiplication?

The distributive property of multiplication is a rule in mathematics that says that multiplying the sum of two (or more) numbers is the same as multiplying the numbers separately and adding/ subtracting them together.

Distributive property of multiplication over addition:

aΓ—(b+c)=aΓ—b+aΓ—c a\times\left(b+c\right)=a\times b+a\times c

Distributive property of multiplication over subtraction:

aΓ—(bβˆ’c)=aΓ—bβˆ’aΓ—c a\times\left(b-c\right)=a\times b-a\times c


What is the distributive property of division?

Just as in the distributive property of multiplication, the distributive property of division (also over addition or subtraction) helps us to simplify an expression.

We can express it as follows:

(a+b):c=a:c+b:c \left(a+b\right):c=a:c+b:c


Do you know what the answer is?

What is the extended distributive property?

The extended distributive property uses the same concept as the basic distributive property to simplify expressions with two sets of parentheses.


Where is the extended distributive property used?

Example 1

Task:

Solve (x+3)(xβˆ’8)= \left(x+3\right)\left(x-8\right)=

We will use the extended distributive property, multiplying each of the terms as follows:

(x+3)(xβˆ’8)=x2βˆ’8x+3xβˆ’24 \left(x+3\right)\left(x-8\right)=x^2-8x+3x-24

Reducing like terms we get

(x+3)(xβˆ’8)=x2βˆ’5xβˆ’24 \left(x+3\right)\left(x-8\right)=x^2-5x-24

Answer

x2βˆ’5xβˆ’24 x^2-5x-24


Example 2

Task:

(2xβˆ’1)(3xβˆ’5)= \left(2x-1\right)\left(3x-5\right)=

Using the extended distributive property we get:

(2xβˆ’1)(3xβˆ’5)=6x2βˆ’10xβˆ’3x+5 \left(2x-1\right)\left(3x-5\right)=6x^2-10x-3x+5

Reducing like terms:

(2xβˆ’1)(3xβˆ’5)=6x2βˆ’13x+5 \left(2x-1\right)\left(3x-5\right)=6x^2-13x+5

Answer

6x2βˆ’13x+5 6x^2-13x+5


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