The Extended Distributive Property

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

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

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.

$ab+2a+b+2$

Detailed explanation

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Test yourself on extended distributive property!

Expand the following expression:

$$ (x+4)(x+3)= $$

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Exercises to practice the distributive property

$(x-4)\times (x-2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8$

$(x+3)\times (x+6) = x^2 + 6x + 3x + 18 = x^2 + 9x + 18$

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:

$(5+8)\times (7+2)$

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:

$(5+8)\times (7+2) =$

$5\times 7+5\times 2+8\times 7+8\times 2 =$

$35+10+56+16 =$

$117$

Basic distributive property

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

Below we can see the formula:

$a\times(b+c)=ab+ac$

Here, we have multiplied $a$ 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:
$(a+b)\times(c+d)=ac+ad+bc+bd$

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

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

Step 2: Multiply $2$ 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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Test your knowledge

Question 1

$$ (a+b)(c+d)= $$ ?

Question 2

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

Question 3

$$ (a+4)(c+3)= $$

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, $4-2=4+(-2)=2$

Let's look at the exercise:

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

Step 2: Multiply $5$ 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, $-5$ times $-3$ equals $+15$ .

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

Example 3

Task:

Find the value of $X$ :

$(X+2)^2=(X+5)\times(X-2)$

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

$(X+2)^2=(X+2)\times(X+2)$

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

Now the equation looks like this:

$(X+2)\times(X+2)=(X+5)\times(X-2)$

After applying the distributive property:

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

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

We will get:

$X = - 14$

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Do you know what the answer is?

Question 1

$$ (x+13)(y+4)= $$

Question 2

Solve the following problem:

$$ (x-8)(x+y)= $$

Question 3

Solve the following problem:

$$ (12-x)(x-3)= $$

Exercises using the distributive property

Exercise 1

Assignment:

A painter has a canvas with the following dimensions:

$(23x+12)$ length

$(20x+7)$ 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.

$(23x+12)\times(20x+7)=$

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

$23x\times20x+23x\times7+12\times20x+12\times7=$

We solve accordingly

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

$460x^2+401x+84$

Answer:

$460x^2+401x+84$

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.

$3y\times(y+3z)=$

Multiply 3y by each of the terms in parentheses.

$3y\times y+3y\times3z=$

Solve accordingly

$3y^2+9yz$

Answer:

$3y^2+9yz$

Check your understanding

Question 1

Solve the following problem:

$$ (a+15)(5+a)= $$

Question 2

$$ (7+b)(a+9)= $$

Question 3

It is possible to use the distributive property to simplify the expression

$$ a(b+c) $$

Exercise 3

Task:

$(3+20)\times(12+4)=$

Solution:

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

$3\times12+3\times4+20\times12+20\times4=$

Solve accordingly

$36+12+240+80=$

We add everything together

$48+320=368$

Answer:

$368$

Exercise 4

Task:

$(12+2)\times(3+5)=$

Solution:

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

$12\times3+12\times5+2\times3+2\times5=$

Solve accordingly

$36+60+6+10=$

We add everything together

$96+16=112$

Answer:

$112$

Do you think you will be able to solve it?

Question 1

It is possible to use the distributive property to simplify the expression

$(a+b)(c\cdot g)$

Question 2

It is possible to use the distributive property to simplify the expression?

If so, what is its simplest form?

$(x+c)(4+c) =\text{?}$

Question 3

It is possible to use the distributive property to simplify the expression below?

What is its simplified form?

$$ (ab)(c d) $$

$$

Exercise 5

Task:

$(7x+4)\times(3x+4)=$

Solution:

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

$7x\times3x+7x\times4+4\times3x+4\times4=$

Solve accordingly

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

$21x^2+40x+16$

Answer:

$21x^2+40x+16$

Exercise 6

Task:

$(2x-3)\times(5x-7)$

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

$2x\times5x+2x\times(-7)+(-3)\times5x+(-3)\times(-7)=$

Solve accordingly

$10x^2-14x-15x+21=$

$10x^2-29x+21$

Answer:

$10x^2-29x+21$

Test your knowledge

Question 1

Resolve -

$$ (x-3)(x-6)= $$

Question 2

Solve the exercise:

$$ (2y-3)(y-4)= $$

Question 3

Expand the following expression:

$$ (x+4)(x+3)= $$

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:

$a\times\left(b+c\right)=a\times b+a\times c$

Distributive property of multiplication over subtraction:

$a\times\left(b-c\right)=a\times b-a\times c$

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:

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

Do you know what the answer is?

Question 1

$$ (a+b)(c+d)= $$ ?

Question 2

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

Question 3

$$ (a+4)(c+3)= $$

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 $\left(x+3\right)\left(x-8\right)=$

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

$\left(x+3\right)\left(x-8\right)=x^2-8x+3x-24$

Reducing like terms we get

$\left(x+3\right)\left(x-8\right)=x^2-5x-24$

Answer

$x^2-5x-24$

Example 2

Task:

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

Using the extended distributive property we get:

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

Reducing like terms:

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

Answer

$6x^2-13x+5$

Check your understanding

Question 1

$$ (x+13)(y+4)= $$

Question 2

Solve the following problem:

$$ (x-8)(x+y)= $$

Question 3

Solve the following problem:

$$ (12-x)(x-3)= $$

Examples with solutions for Extended Distributive Property

Exercise #1

Expand the following expression:

$(x+4)(x+3)=$

Video Solution

Step-by-Step Solution

Let's simplify the given expression by opening the parentheses using the extended distribution law:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

Note that in the formula template for the above distribution law, we take by default that the operation between the terms inside the parentheses is addition. Therefore we won't forget of course that the sign preceding the term is an inseparable part of it. We will also apply the rules of sign multiplication and thus we can present any expression in parentheses. We'll open the parentheses using the above formula, first as an expression where an addition operation exists between all terms. In this expression it's clear that all terms have a plus sign prefix. Therefore we'll proceed directly to opening the parentheses,

Let's begin:

$(\textcolor{red}{x}+\textcolor{blue}{4})(x+3)\\ \textcolor{red}{x}\cdot x+\textcolor{red}{x}\cdot3+\textcolor{blue}{4}\cdot x +\textcolor{blue}{4}\cdot3\\ x^2+3x+4x+12$

In calculating the above multiplications, we used the multiplication table and the laws of exponents for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

In the next step we'll combine like terms, which we define as terms where the variable (or variables each separately), in this case x, have identical exponents .(In the absence of one of the variables from the expression, we'll consider its exponent as zero power given that raising any number to the power of zero yields 1) We'll apply the commutative property of addition, furthermore we'll arrange (if needed) the expression from highest to lowest power from left to right (we'll treat the free number as having zero power):
$\textcolor{purple}{x^2}\textcolor{green}{+3x}\textcolor{green}{+4x}+12\\ \textcolor{purple}{x^2}\textcolor{green}{+7x}+12$ In the combining of like terms performed above, we highlighted the different terms using colors, and as emphasized before, we made sure that the sign preceding the term is an inseparable part of it,

Thus the correct answer is C.

Answer

$x^2+7x+12$

Exercise #2

$(a+b)(c+d)=$ ?

Video Solution

Step-by-Step Solution

Let's simplify the expression by opening the parentheses using the distributive property:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$

Therefore, the correct answer is (a).

Answer

$\text{ac + ad}+bc+bd$

Exercise #3

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

Video Solution

Step-by-Step Solution

To solve this problem, we'll apply the FOIL method for multiplying binomials:

First: Multiply the first terms in each binomial: $(2x)(x) = 2x^2$ .
Outer: Multiply the outer terms in the product: $(2x)(3) = 6x$ .
Inner: Multiply the inner terms: $(y)(x) = xy$ .
Last: Multiply the last terms: $(y)(3) = 3y$ .

Next, we combine these results to form the expanded expression:

$2x^2 + 6x + xy + 3y$ .

Since terms $6x$ and $xy$ are not like terms, they cannot be combined, resulting in the final expression:

$2x^2 + xy + 6x + 3y$ .

Upon reviewing the multiple-choice options, the correct answer is the expanded expression, choice 4: $2x^2 + xy + 6x + 3y$ .

Answer

$2x^2+xy+6x+3y$

Exercise #4

$(a+4)(c+3)=$

Video Solution

Step-by-Step Solution

When we encounter a multiplication exercise of this type, we know that we must use the distributive property.

Step 1: Multiply the first factor of the first parentheses by each of the factors of the second parentheses.

Step 2: Multiply the second factor of the first parentheses by each of the factors of the second parentheses.

Step 3: Group like terms.

a * (c+3) =

a*c + a*3

4 * (c+3) =

4*c + 4*3

ac+3a+4c+12

There are no like terms to simplify here, so this is the solution!

Answer

$ac+3a+4c+12$

Exercise #5

$(x+13)(y+4)=$

Video Solution

Step-by-Step Solution

To solve this problem, we'll perform a step-by-step expansion of the expression $(x+13)(y+4)$ using the distributive property:

Step 1: Multiply the first terms $(x \cdot y) = xy$ .
Step 2: Multiply the outer terms $(x \cdot 4) = 4x$ .
Step 3: Multiply the inner terms $(13 \cdot y) = 13y$ .
Step 4: Multiply the last terms $(13 \cdot 4) = 52$ .

After completing these steps, combine the results:

$xy + 4x + 13y + 52$

This is the final expanded form of the expression. By comparing with the given choices, the correct answer is:

$xy + 4x + 13y + 52$

Therefore, the correct choice is option 3.

Answer

$xy+4x+13y+52$

Start practice

The Extended Distributive Property

Test yourself on extended distributive property!

Exercises to practice the distributive property

Example 1

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

Example 3

Exercises using the distributive property

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Review questions

What is the distributive property of multiplication?

What is the distributive property of division?

What is the extended distributive property?

Where is the extended distributive property used?

Example 1

Example 2

Examples with solutions for Extended Distributive Property

Exercise #1

Video Solution

Step-by-Step Solution

Answer

Exercise #2

Video Solution

Step-by-Step Solution

Answer

Exercise #3

Video Solution

Step-by-Step Solution

Answer

Exercise #4

Video Solution

Step-by-Step Solution

Answer

Exercise #5

Video Solution

Step-by-Step Solution

Answer

More Questions

Extended Distributive Property