Example ( x − 4 ) × ( x − 2 ) = x 2 − 2 x − 4 x + 8 = x 2 − 6 x + 8 (x-4)\times (x-2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8 ( x − 4 ) × ( x − 2 ) = x 2 − 2 x − 4 x + 8 = x 2 − 6 x + 8
( x + 3 ) × ( x + 6 ) = x 2 + 6 x + 3 x + 18 = x 2 + 9 x + 18 (x+3)\times (x+6) = x^2 + 6x + 3x + 18 = x^2 + 9x + 18 ( x + 3 ) × ( x + 6 ) = x 2 + 6 x + 3 x + 18 = x 2 + 9 x + 18
The extended distributive property allows us to solve exercises with two sets of parentheses that are multiplied by eachother.
For example: ( a + 1 ) × ( b + 2 ) (a+1)\times(b+2) ( a + 1 ) × ( 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. a b + 2 a + b + 2 ab+2a+b+2 ab + 2 a + b + 2
Basic distributive property
Let's take a moment to remember our basic distributive property.
Below we can see the formula:
a × ( b + c ) = a b + a c a\times(b+c)=ab+ac a × ( b + c ) = ab + a c
Here, we have multiplied a 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 ) × ( c + d ) = a c + a d + b c + b d (a+b)\times(c+d)=ac+ad+bc+bd ( a + b ) × ( c + d ) = a c + a d + b c + b d
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
Step 1: Multiply A A A by each of the terms included in the second parentheses.
Step 2: Multiply 2 2 2 by each of the terms included in the second parentheses.
Step 3: Order the terms and combine like terms, if any:
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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, 4 − 2 = 4 + ( − 2 ) = 2 4-2=4+(-2)=2 4 − 2 = 4 + ( − 2 ) = 2
Let's look at the exercise:
Step 1: Multiply A A A by each of the terms included inside the second parentheses.
Step 2: Multiply 5 5 5 by each of the terms included inside the second parentheses.
Pay attention to the signs of each of the terms! For example, we will see that, − 5 -5 − 5 times − 3 -3 − 3 equals + 15 +15 + 15 .
In this case, there are no terms that we want to combine.
Example 3 Task:
Find the value of X X X :
( X + 2 ) 2 = ( X + 5 ) × ( X − 2 ) (X+2)^2=(X+5)\times(X-2) ( X + 2 ) 2 = ( X + 5 ) × ( X − 2 )
Let's look at the left side of the equation and simplify:
( X + 2 ) 2 = ( X + 2 ) × ( X + 2 ) (X+2)^2=(X+2)\times(X+2) ( X + 2 ) 2 = ( X + 2 ) × ( X + 2 )
Now we can use the extended distributive property on each side of the equation.
Now the equation looks like this:
( X + 2 ) × ( X + 2 ) = ( X + 5 ) × ( X − 2 ) (X+2)\times(X+2)=(X+5)\times(X-2) ( X + 2 ) × ( X + 2 ) = ( X + 5 ) × ( X − 2 )
After applying the distributive property:
X 2 + 2 X + 2 X + 4 = X 2 – 2 X + 5 X – 10 X^2 + 2X + 2X + 4 = X^2 – 2X + 5X – 10 X 2 + 2 X + 2 X + 4 = X 2 –2 X + 5 X –10
Let's reduce, combine like terms and arrange the equation.
We will get:
X = − 14 X = - 14 X = − 14
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Do you know what the answer is?
Exercises using the distributive property Exercise 1 Assignment:
A painter has a canvas with the following dimensions:
( 23 x + 12 ) (23x+12) ( 23 x + 12 ) length
( 20 x + 7 ) (20x+7) ( 20 x + 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.
( 23 x + 12 ) × ( 20 x + 7 ) = (23x+12)\times(20x+7)= ( 23 x + 12 ) × ( 20 x + 7 ) =
Multiply each term in the first parentheses by each term in the second parentheses.
23 x × 20 x + 23 x × 7 + 12 × 20 x + 12 × 7 = 23x\times20x+23x\times7+12\times20x+12\times7= 23 x × 20 x + 23 x × 7 + 12 × 20 x + 12 × 7 =
We solve accordingly
460 x 2 + 161 x + 240 x + 84 = 460x^2+161x+240x+84= 460 x 2 + 161 x + 240 x + 84 =
460 x 2 + 401 x + 84 460x^2+401x+84 460 x 2 + 401 x + 84
Answer:
460 x 2 + 401 x + 84 460x^2+401x+84 460 x 2 + 401 x + 84
Exercise 2 Task:
Find the area of the following rectangle:
Leave variables in your answer.
Solution:
To find the area we multiply the width by the length.
3 y × ( y + 3 z ) = 3y\times(y+3z)= 3 y × ( y + 3 z ) =
Multiply 3y by each of the terms in parentheses.
3 y × y + 3 y × 3 z = 3y\times y+3y\times3z= 3 y × y + 3 y × 3 z =
Solve accordingly
3 y 2 + 9 y z 3y^2+9yz 3 y 2 + 9 yz
Answer:
3 y 2 + 9 y z 3y^2+9yz 3 y 2 + 9 yz
Exercise 3 Task:
( 3 + 20 ) × ( 12 + 4 ) = (3+20)\times(12+4)= ( 3 + 20 ) × ( 12 + 4 ) =
Solution:
We multiply each of the terms in the first parentheses by the terms in the second parentheses.
3 × 12 + 3 × 4 + 20 × 12 + 20 × 4 = 3\times12+3\times4+20\times12+20\times4= 3 × 12 + 3 × 4 + 20 × 12 + 20 × 4 =
Solve accordingly
36 + 12 + 240 + 80 = 36+12+240+80= 36 + 12 + 240 + 80 =
We add everything together
48 + 320 = 368 48+320=368 48 + 320 = 368
Answer:
368 368 368
Exercise 4 Task:
( 12 + 2 ) × ( 3 + 5 ) = (12+2)\times(3+5)= ( 12 + 2 ) × ( 3 + 5 ) =
Solution:
We multiply each of the terms in the first parentheses by the terms of the second parentheses.
12 × 3 + 12 × 5 + 2 × 3 + 2 × 5 = 12\times3+12\times5+2\times3+2\times5= 12 × 3 + 12 × 5 + 2 × 3 + 2 × 5 =
Solve accordingly
36 + 60 + 6 + 10 = 36+60+6+10= 36 + 60 + 6 + 10 =
We add everything together
96 + 16 = 112 96+16=112 96 + 16 = 112
Answer:
112 112 112
Do you think you will be able to solve it?
Question 3 a \( \text{ac+ad}+bc+bd \)
Exercise 5 Task:
( 7 x + 4 ) × ( 3 x + 4 ) = (7x+4)\times(3x+4)= ( 7 x + 4 ) × ( 3 x + 4 ) =
Solution:
We multiply each of the terms of the first parentheses by the terms of the second parentheses.
7 x × 3 x + 7 x × 4 + 4 × 3 x + 4 × 4 = 7x\times3x+7x\times4+4\times3x+4\times4= 7 x × 3 x + 7 x × 4 + 4 × 3 x + 4 × 4 =
Solve accordingly
21 x 2 + 28 x + 12 x + 16 = 21x^2+28x+12x+16= 21 x 2 + 28 x + 12 x + 16 =
21 x 2 + 40 x + 16 21x^2+40x+16 21 x 2 + 40 x + 16
Answer:
21 x 2 + 40 x + 16 21x^2+40x+16 21 x 2 + 40 x + 16
Exercise 6 Task:
( 2 x − 3 ) × ( 5 x − 7 ) (2x-3)\times(5x-7) ( 2 x − 3 ) × ( 5 x − 7 )
We multiply each of the terms of the first parentheses by the terms of the second parentheses.
2 x × 5 x + 2 x × ( − 7 ) + ( − 3 ) × 5 x + ( − 3 ) × ( − 7 ) = 2x\times5x+2x\times(-7)+(-3)\times5x+(-3)\times(-7)= 2 x × 5 x + 2 x × ( − 7 ) + ( − 3 ) × 5 x + ( − 3 ) × ( − 7 ) =
Solve accordingly
10 x 2 − 14 x − 15 x + 21 = 10x^2-14x-15x+21= 10 x 2 − 14 x − 15 x + 21 =
10 x 2 − 29 x + 21 10x^2-29x+21 10 x 2 − 29 x + 21
Answer:
10 x 2 − 29 x + 21 10x^2-29x+21 10 x 2 − 29 x + 21
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:
a × ( b + c ) = a × b + a × c a\times\left(b+c\right)=a\times b+a\times c a × ( b + c ) = a × b + a × c
Distributive property of multiplication over subtraction:
a × ( b − c ) = a × b − a × c a\times\left(b-c\right)=a\times b-a\times c a × ( b − c ) = a × b − a × 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 ( a + b ) : 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 ) ( x − 8 ) = \left(x+3\right)\left(x-8\right)= ( x + 3 ) ( x − 8 ) =
We will use the extended distributive property, multiplying each of the terms as follows:
( x + 3 ) ( x − 8 ) = x 2 − 8 x + 3 x − 24 \left(x+3\right)\left(x-8\right)=x^2-8x+3x-24 ( x + 3 ) ( x − 8 ) = x 2 − 8 x + 3 x − 24
Reducing like terms we get
( x + 3 ) ( x − 8 ) = x 2 − 5 x − 24 \left(x+3\right)\left(x-8\right)=x^2-5x-24 ( x + 3 ) ( x − 8 ) = x 2 − 5 x − 24
Answer
x 2 − 5 x − 24 x^2-5x-24 x 2 − 5 x − 24
Example 2 Task:
( 2 x − 1 ) ( 3 x − 5 ) = \left(2x-1\right)\left(3x-5\right)= ( 2 x − 1 ) ( 3 x − 5 ) =
Using the extended distributive property we get:
( 2 x − 1 ) ( 3 x − 5 ) = 6 x 2 − 10 x − 3 x + 5 \left(2x-1\right)\left(3x-5\right)=6x^2-10x-3x+5 ( 2 x − 1 ) ( 3 x − 5 ) = 6 x 2 − 10 x − 3 x + 5
Reducing like terms:
( 2 x − 1 ) ( 3 x − 5 ) = 6 x 2 − 13 x + 5 \left(2x-1\right)\left(3x-5\right)=6x^2-13x+5 ( 2 x − 1 ) ( 3 x − 5 ) = 6 x 2 − 13 x + 5
Answer
6 x 2 − 13 x + 5 6x^2-13x+5 6 x 2 − 13 x + 5