Extended Distributive Property - Examples, Exercises and Solutions

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.

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.

Simplify this expression by paying attention to the order of arithmetic operations which states that exponentiation precedes multiplication, division precedes addition and subtraction and that parentheses precede all of the above.

Thus, let's begin by simplifying the expressions within the parentheses, and following this, the multiplication between them.

$(12+2)\cdot(3+5)= \\ 14\cdot8=\\ 112$Therefore, the correct answer is option C.

Let's simplify the given expression, open the parentheses using the extended distribution law:

$(\textcolor{red}{t}+\textcolor{blue}{k})(c+d)=\textcolor{red}{t}c+\textcolor{red}{t}d+\textcolor{blue}{k}c+\textcolor{blue}{k}d$

Note that in the formula template for the above distribution law, we take as a default that the operation between 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, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first as an expression where addition operation exists between all terms:

$(12-x)(x-3) \\ (\textcolor{red}{12}+\textcolor{blue}{(-x)})(x+(-3))\\$Let's begin then with opening the parentheses:

$(\textcolor{red}{12}+\textcolor{blue}{(-x)})(x+(-3))\\ \textcolor{red}{12}\cdot x+\textcolor{red}{12}\cdot(-3)+\textcolor{blue}{(-x)}\cdot x +\textcolor{blue}{(-x)}\cdot(-3)\\ 12x-36-x^2 +3x$

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, we'll define like terms 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, since raising any number to the zero power yields 1), we'll use the commutative property of addition, additionally we'll arrange the expression from highest to lowest power from left to right (we'll treat the free number as having zero power):
$\textcolor{purple}{12x}\textcolor{green}{-36}-x^2\textcolor{purple}{+3x}\\ -x^2\textcolor{purple}{+12x+3x}\textcolor{green}{-36}\\ -x^2\textcolor{purple}{+15x}\textcolor{green}{-36}\\$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,

We therefore got that the correct answer is answer A (we used the commutative property of addition to verify this).

Simplify this expression paying attention to the order of arithmetic operations. Exponentiation precedes multiplication whilst division precedes addition and subtraction. Parentheses precede all of the above.

Therefore, let's first start by simplifying the expressions within the parentheses. Then we can proceed to perform the multiplication between them:

$(3+20)\cdot(12+4)=\\ 23\cdot16=\\ 368$Therefore, the correct answer is option A.

We begin by opening the parentheses using the extended distributive property to create a long addition exercise:

We then multiply the first term of the left parenthesis by the first term of the right parenthesis.

We multiply the first term of the left parenthesis by the second term of the right parenthesis.

Now we multiply the second term of the left parenthesis by the first term of the left parenthesis.

Finally, we multiply the second term of the left parenthesis by the second term of the right parenthesis.

In the following way:

$(35\times10)+(35\times5)+(4\times10)+(4\times5)=$

We solve each of the exercises within parentheses:

$350+175+40+20=$

We solve the exercise from left to right:

$350+175=525$

$525+40=565$

$565+20=585$

Let's simplify the given expression, open the parentheses using the extended distribution law:

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

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, which we'll open using the above formula, first as an expression where addition operation exists between all terms, in this expression as it's clear, all terms have a plus sign prefix, therefore we'll proceed directly to opening the parentheses,

$$Let's begin then with opening the parentheses:

$(\textcolor{red}{a}+\textcolor{blue}{15})(5+a)\\ \textcolor{red}{a}\cdot 5+\textcolor{red}{a}\cdot a+\textcolor{blue}{15}\cdot 5 +\textcolor{blue}{15}\cdot a\\ 5a+a^2+75+15a$

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

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

In the next step we'll combine like terms, we'll define like terms as terms where the variable (or variables each separately), in this case a, have identical exponents (in the absence of one of the variables from the expression, we'll consider its exponent as zero power, this is because any number raised to the power of zero equals 1), we'll use the commutative law of addition, additionally we'll arrange the expression from highest to lowest power from left to right (we'll treat the free number as power of zero):
$\textcolor{purple}{5a}\textcolor{green}{+a^2}+75\textcolor{purple}{+15a}\\ \textcolor{green}{a^2}\textcolor{purple}{+5a+15a}+75\\ \textcolor{green}{a^2}\textcolor{purple}{+20a}+75\\$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,

We therefore got that the correct answer is answer B.

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!

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).

Let's simplify the given expression, open 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, which we'll open using the above formula, first as an expression where addition operation exists between all terms, in this expression as it's clear, all terms have a plus sign prefix, therefore we'll proceed directly to opening the parentheses,

$$Let's begin then with opening the parentheses:

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

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, we'll define like terms 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 since raising any number to the power of zero yields 1), we'll use the commutative property of addition, additionally 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,

We therefore got that the correct answer is answer C.

Let's simplify the given expression, open 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, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first as an expression where addition operation exists between all terms:

$(x-6)(x+8)\\ \downarrow\\ \big(\textcolor{red}{x}+\textcolor{blue}{(-6)}\big)(x+8)\\$Let's begin then with opening the parentheses:

$\big(\textcolor{red}{x}+\textcolor{blue}{(-6)}\big)(x+8)\\ \textcolor{red}{x}\cdot x+\textcolor{red}{x}\cdot8+\textcolor{blue}{(-6)}\cdot x+\textcolor{blue}{(-6)}\cdot8\\ x^2+8x-6x-48$

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, we'll define like terms 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, this is because raising any number to the zero power yields the result 1), we'll use the commutative property of addition, additionally we'll arrange the expression from highest to lowest power from left to right (we'll treat the free number as having zero power):
$\textcolor{purple}{x^2}\textcolor{green}{+8x-6x}\textcolor{orange}{-48}\\ \textcolor{purple}{x^2}\textcolor{green}{+2x}\textcolor{orange}{-48}\\$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,

We therefore got that the correct answer is answer A.

Let's simplify the given expression, open the parentheses using the expanded 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, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first as an expression where addition operation exists between all terms:

$(x-8)(x+y)\\ (\textcolor{red}{x}+\textcolor{blue}{(-8)})(x+y)\\$Let's begin then with opening the parentheses:

$(\textcolor{red}{x}+\textcolor{blue}{(-8)})(x+y)\\ \textcolor{red}{x}\cdot x+\textcolor{red}{x}\cdot y+\textcolor{blue}{(-8)}\cdot x +\textcolor{blue}{(-8)}\cdot y\\ x^2+xy-8x -8y$

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

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

Note that in the expression we got in the last stage there are four different terms, this is because there isn't even one pair of terms where the variables (different ones) have the same exponent, additionally the expression is already organized therefore the expression we got is the final and most simplified form:
$\textcolor{purple}{ x^2}\textcolor{green}{+xy}-8x \textcolor{orange}{-8y}\\$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,

We therefore concluded that the correct answer is answer A.

In order to simplify the given expression we must use the expanded distributive law as seen below:

$(a+b)(c+d)=ac+ad+bc+bd$

First, we'll perform the multiplication between the pairs of parentheses, using the mentioned distributive law, and then we'll combine like terms if possible. We'll do this while paying attention to the correct multiplication of signs:

$(-7-4y)(5x+6)=\\ -35x-42-20xy-24y$Therefore, the correct answer is answer A.

We will use the extended distribution law as seen below in order to simplify the given expression:

$(a+b)(c+d)=ac+ad+bc+bd$

First, we'll perform the multiplication between the pairs of parentheses, using the mentioned distribution law, and then we'll combine like terms if possible. We'll do this while paying attention to the correct multiplication of signs:

$(2x+3)(-5-x)= \\ -10x-2x^2-15-3x=\\ \boxed{-2x^2-13x-15}$Therefore, the correct answer is answer D.

We will use the expanded distributive law seen below in order to simplify the given expression:

$(a+b)(c+d)=ac+ad+bc+bd$

First, we'll perform the multiplication between the pairs of parentheses, using the mentioned distributive law, and then we'll combine like terms if possible. We'll do this while paying attention to the correct multiplication of signs:

$(2x-y)(4-3x)= \\ \boxed{8x-6x^2-4y+3xy}$Therefore, the correct answer is answer C.