Extended Distributive Property: Applying the formula

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.

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:

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

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 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).

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 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.

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).

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.

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 whilst taking into account 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:

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

We will begin by performing the multiplication between the pairs of parentheses, using the mentioned distribution law. Then we will proceed to combine like terms if possible. We'll do this whilst taking into account 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:

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

We will begin by performing the multiplication between the pairs of parentheses, using the mentioned distributive law. We will combine like terms if possible whilst taking into account the correct multiplication of signs:

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

We will use the expanded 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 distribution law mentioned, and then we will proceed to combine like terms if possible. We'll do this whilst observing the correct multiplication of signs:

$(3b+7a)(-5a+2b)= \\ -15ab+6b^2-35a^2+14ab=\\ \boxed{6b^2-ab-35a^2}$Therefore, the correct answer is answer B.

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:

$(2x+3)(-5-x)\\ (\textcolor{red}{2x}+\textcolor{blue}{3})((-5)+(-x))\\$Let's begin then by opening the parentheses:

$(\textcolor{red}{2x}+\textcolor{blue}{3})((-5)+(-x))\\ \textcolor{red}{2x}\cdot (-5)+\textcolor{red}{2x}\cdot(-x)+\textcolor{blue}{3}\cdot (-5) +\textcolor{blue}{3} \cdot(-x)\\ -10x-2x^2-15-3x$

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, this is because raising any number to the power of zero yields the result 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 having zero power):
$\textcolor{purple}{-10x}\textcolor{green}{-2x^2}-15\textcolor{purple}{-3x}\\ \textcolor{green}{-2x^2} \textcolor{purple}{-10x}\textcolor{purple}{-3x}-15\\ \textcolor{green}{-2x^2}\textcolor{purple}{-13x}-15$

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 D.

Let's simplify the given expression by factoring the parentheses using the expanded distributive 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 that the sign before the term is an inseparable part of it.

We will also apply the laws of sign multiplication and thus we can present any term in parentheses to make things simpler.

$(2x-y)(4-3x)\\ (\textcolor{red}{2x}+\textcolor{blue}{(-y)})(4+(-3x))\\$Let's start then by opening the parentheses:

$(\textcolor{red}{2x}+\textcolor{blue}{(-y)})(4+(-3x))\\ \textcolor{red}{2x}\cdot 4+\textcolor{red}{2x}\cdot(-3x)+\textcolor{blue}{(-y)}\cdot 4+\textcolor{blue}{(-y)} \cdot(-3x)\\ 8x-6x^2-4y+3xy$In the operations above we used the sign multiplication laws, and the exponent law for multiplying terms with identical bases:

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

In the next step we will combine similar terms. We will define similar terms as terms in which the variables, in this case, x and y, have identical powers (in the absence of one of the unknowns from the expression, we will relate to its power as zero power, since raising any number to the power of zero will yield the result 1).

We will arrange the expression from the highest power to the lowest from left to right (we will relate to the free term as the power of zero),

Note that in the expression we received in the last step there are four different terms, since there is not even one pair of terms in which the unknowns (the variables) have the same power, so the expression we already received, is the final and most simplified expression.

We will settle for arranging it again from the highest power to the lowest from left to right:
$\textcolor{purple}{ 8x}\textcolor{green}{-6x^2}-4y\textcolor{orange}{+3xy}\\ \textcolor{green}{-6x^2}\textcolor{orange}{+3xy}\textcolor{purple}{ +8x}-4y\\$We highlighted the different terms using colors, and as already emphasized before, we made sure that the sign before the term is correct.

We thus received that the correct answer is answer D.

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 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:

$(3b+7a)(-5a+2b) \\ (\textcolor{red}{3b}+\textcolor{blue}{7a})((-5a)+2b)\\$Let's begin then with opening the parentheses:

$(\textcolor{red}{3b}+\textcolor{blue}{7a})((-5a)+2b)\\ \textcolor{red}{3b}\cdot (-5a)+\textcolor{red}{3b}\cdot2b+\textcolor{blue}{7a}\cdot (-5a) +\textcolor{blue}{7a}\cdot2b\\ -15ab+6b^2-35a^2+14ab$

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(s) (or each variable separately), in this case a and b, 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 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 zero power):
$\textcolor{purple}{-15ab}\textcolor{green}{+6b^2}-35a^2\textcolor{purple}{+14ab}\\ \textcolor{green}{6b^2}-35a^2\textcolor{purple}{-15ab}\textcolor{purple}{+14ab}\\ \textcolor{green}{6b^2}-35a^2\textcolor{purple}{-ab}$

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.

To answer this exercise, we need to understand how the extended distributive property works:

For example:

(a+1)∗(b+2)

To solve this type of exercises, the following steps must be taken:

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 together.

ab∗2a∗b∗2

We start from the first number of the exercise: 2x

2x*5x+2x*-7

10x²-14x

We will continue with the second factor: -3

-3*5x+-3*-7

-15x+21

We add all the data together:

10x²-14x-15x+21

10x²-29x+21

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:

$(\textcolor{red}{x}+\textcolor{blue}{y})(3x+2y)\\ \textcolor{red}{x}\cdot 3x+\textcolor{red}{x}\cdot2y+\textcolor{blue}{y}\cdot 3x+\textcolor{blue}{y} \cdot2y\\ 3x^2+2xy+3xy+2y^2$

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 and y, have identical exponents (in the absence of one of the variables from the expression, we'll treat its exponent as zero power, this is because raising any number to the zero power yields the result 1), we'll use the commutative law of addition, additionally we'll arrange the expression (if needed) from highest to lowest power from left to right (we'll treat the free number as zero power):
$\textcolor{purple}{ 3x^2}\textcolor{green}{+2xy}\textcolor{green}{+3xy}\textcolor{orange}{+2y^2}\\ \textcolor{purple}{ 3x^2}\textcolor{green}{+5xy}\textcolor{orange}{+2y^2}\\$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.

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 laws of sign multiplication and thus we can present any expression in parentheses, which we 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:

$(\textcolor{red}{2x}+\textcolor{blue}{3y})(2z+12m)\\ \textcolor{red}{2x}\cdot 2z+\textcolor{red}{2x}\cdot12m+\textcolor{blue}{3y}\cdot 2z+\textcolor{blue}{3y} \cdot12m\\ 4xz+24xm+6yz+36ym$

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 z,m x and y, 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),

Note that in the expression we got in the last step there are four different terms, this is because there isn't even one pair of terms where the (different) variables have the same exponent, therefore the expression we already got is the final and most simplified expression:
$\textcolor{purple}{ 4xz}\textcolor{green}{+24xm}\textcolor{black}{+6yz}\textcolor{orange}{+36ym }\\$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 have therefore concluded that the correct answer is answer D.