Extended Distributive Property: Applying the formula

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

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!

$ac+3a+4c+12$

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

Let's simplify the given expression, open 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 option A.

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

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

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

$10x^2-29x+21$

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

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.

$-6x^2+3xy +8x-4y$

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

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:

$350+175=525$

$525+40=565$

$565+20=585$

585

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

$15x-36-x^2$

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

$a^2+20a+75$

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

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

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

$ab+7a+9b+63$

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

$x^2+7x+12$

$(x-6)(x+8)=$

$x^2+2x-48$

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

$x^2-2x-8$

$(x-6)(x+2)=$

$x^2-4x-12$

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

$x^2+xy-8x-8y$

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

$x^2-y^2$

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

$xy+4x+13y+52$

Solve the following exercise

$(2x-3)(5x-7)=$

^{$10x^2-29x+21$}

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

$-2x^2-13x-15$

Solve the following exercise

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

_{$-2x^2-13x-15$}

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

$3x^2+5xy+2y^2$