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!

Resolve -

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

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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

Solve the exercise:

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

Question 2

Solve the exercise:

$$ (3x-1)(x+2)= $$

Question 3

Solve the exercise:

$$ (5x-2)(3+x)= $$

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

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

Question 2

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

Question 3

$(35+4)\times(10+5)=$

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

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

Question 2

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

Question 3

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

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

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

Question 2

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

Question 3

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

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

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

Question 2

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

Question 3

Resolve -

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

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

Solve the exercise:

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

Question 2

Solve the exercise:

$$ (3x-1)(x+2)= $$

Question 3

Solve the exercise:

$$ (5x-2)(3+x)= $$

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

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

Question 2

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

Question 3

$(35+4)\times(10+5)=$

Examples with solutions for Extended Distributive Property

Exercise #1

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

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 #2

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

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 #3

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

$a(b+c)$

Video Solution

Step-by-Step Solution

We can simplify the given expression using the extended distributive law if we remember that adding 0 to any number yields the same number, meaning that:

$x+0=x$ Let's apply this to the given expression to treat the term multiplying the parentheses as a binomial:

$a(b+c) \\ (a+0)(b+c)$ Now let's open the parentheses in the expression we got using the extended distributive law:

$(\textcolor{red}{t}+\textcolor{blue}{k})(x+y)=\textcolor{red}{t}x+\textcolor{red}{t}y+\textcolor{blue}{k}x+\textcolor{blue}{k}y$ and we get:

$(\textcolor{red}{a}+\textcolor{blue}{0})(b+c)\\ \textcolor{red}{a}b+\textcolor{red}{a}c+\textcolor{blue}{0}b+\textcolor{blue}{0}c$ We'll continue and remember that multiplying 0 by any number will always yield 0 and we get:

$\textcolor{red}{a}b+\textcolor{red}{a}c+\textcolor{blue}{0}b+\textcolor{blue}{0}c \\ \textcolor{red}{a}b+\textcolor{red}{a}c$ Therefore the correct answer is - yes, we can simplify the given expression using the extended distributive law (it's not necessary - but possible), meaning the correct answer is answer B.

Answer

No, the answer $ab+ac$

Exercise #4

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

If so, what is its simplest form?

$(a+c)(4+c)$

Video Solution

Step-by-Step Solution

We simplify the given expression by opening the parentheses using the extended distributive property:

$(\textcolor{red}{x}+\textcolor{blue}{y})(t+d)=\textcolor{red}{x}t+\textcolor{red}{x}d+\textcolor{blue}{y}t+\textcolor{blue}{y}d$ Keep in mind that in the distributive property formula mentioned above, we assume that the operation between the terms inside the parentheses is an addition operation, therefore, of course, we will not forget that the sign of the term's coefficient is ery important.

We will also apply the rules of multiplication of signs, so we can present any expression within parentheses that's opened with the distributive property as an expression with addition between all the terms.

In this expression we only have addition signs in parentheses, therefore we go directly to opening the parentheses,

We start by opening the parentheses:

$(\textcolor{red}{x}+\textcolor{blue}{c})(4+c)\\ \textcolor{red}{x}\cdot 4+\textcolor{red}{x}\cdot c+\textcolor{blue}{c}\cdot 4+\textcolor{blue}{c} \cdot c\\ 4x+xc+4c+c^2$ To simplify this expression, we use the power law for multiplication between terms with identical bases:

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

In the next step like terms come into play.

We define like terms as terms in which the variables (in this case, x and c) have identical powers (in the absence of one of the variables from the expression, we will refer to its power as zero power, this is because raising any number to the power of zero results in 1).

We will also use the substitution property, and we will order the expression from the highest to the lowest power from left to right (we will refer to the regular integer as the power of zero),

Keep in mind that in this new expression there are four different terms, this is because there is not even one pair of terms in which the variables (different) have the same power. Also it is already ordered by power, therefore the expression we have is the final and most simplified expression: $\textcolor{purple}{4x}\textcolor{green}{+xc}\textcolor{black}{+4c}\textcolor{orange}{+c^2 }\\ \textcolor{orange}{c^2 }\textcolor{green}{+xc}\textcolor{purple}{+4x}\textcolor{black}{+4c}\\$ We highlight the different terms using colors and, as emphasized before, we make sure that the main sign of the term is correct.

We use the substitution property for multiplication to note that the correct answer is option A.

Answer

Yes, the meaning is $4x+cx+4c+c^2$

Exercise #5

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

What is its simplified form?

$(ab)(c d)$

Video Solution

Step-by-Step Solution

Let's remember the extended distributive property:

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$ Note that the operation between the terms inside the parentheses is a multiplication operation:

$(a b)(c d)$ Unlike in the extended distributive property previously mentioned, which is addition (or subtraction, which is actually the addition of the term with a minus sign),

Also, we notice that since there is a multiplication among all the terms, both inside the parentheses and between the parentheses, this is a simple multiplication and the parentheses are actually not necessary and can be remoed. We get:

$(a b)(c d)= \\ abcd$ Therefore, opening the parentheses in the given expression using the extended distributive property is incorrect and produces an incorrect result.

Therefore, the correct answer is option d.

Answer

No, $abcd$ .

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