**It is a general term for various tools and techniques that will help us solve more complex exercises in the future.**

**It is a general term for various tools and techniques that will help us solve more complex exercises in the future.**

Powers are a shorthand way of writing the multiplication of a number by itself several times.

**For example:**

$4^5=4\times4\times4\times4\times4$

$4$ is the number that is multiplied by itself. It is called the "Base of power".

$5$ represents the number of times the base is multiplied by itself and it is called the "Exponent".

Question 1

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

Question 2

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

Question 3

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

Question 4

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

Question 5

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

If so, what is its simplest form?

\( (a+c)(4+c) \)

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

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

Therefore, let's first start by simplifying the expressions within parentheses, then we perform the multiplication between them:

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

368

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

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

Therefore, let's start by simplifying the expressions within parentheses, then perform the multiplication between them:

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

112

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

We will open the parentheses using the extended distributive property and create a long addition exercise:

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

Then 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

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

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!

$ac+3a+4c+12$

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

If so, what is its simplest form?

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

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

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

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

Question 1

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

What is its simplified form?

\( (ab)(c d) \)

\( \)

Question 2

Find the common factor:

\( ab+bc \)

Question 3

Find the common factor:

\( 7a+14b \)

Question 4

Find the biggest common factor:

\( 12x+16y \)

Question 5

Decompose the following expression into factors:

\( 20ab-4ac \)

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

What is its simplified form?

$(ab)(c d)$

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.__

No, $abcd$.

Find the common factor:

$ab+bc$

$ab+bc=a\times b+b\times c$

Let's consider that the common factor is b, so we will remove it:

$b(ab+bc)=$

We divide by b:$b(\frac{ab}{b}+\frac{bc}{b})=$

$b(a+c)$

$b(a+c)$

Find the common factor:

$7a+14b$

We divide 14 in a multiplication exercise to help us simplify accordingly:$7\times a+7\times b\times2=$

We extract the common factor 7:

$7(a+2\times b)=7(a+2b)$

$7(a+2b)$

Find the biggest common factor:

$12x+16y$

We break down the coefficients 12 and 16 into multiplication exercises with a multiplier factor to then simplify:

$3\times4\times x+4\times4\times y$

We extract 4 which is the common factor:

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

$4(3x+4y)$

Decompose the following expression into factors:

$20ab-4ac$

We will break down the coefficient of 20 into a multiplication exercise that will help us simplify:$5\times4\times a\times b-4\times a\times c$

We extract 4a as a common factor:$4a(5\times b-c)=4a(5b-c)$

$4a(5b-c)$

Question 1

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

Question 2

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

Question 3

Find the common factor:

\( 25y-100xy^2 \)

Question 4

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

\( (17+c)(5+a+3) \)

Question 5

Decompose the following expression into factors:

\( 15a^2+10a+5 \)

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

Let's simplify the given expression, **develop parentheses using the expanded distributive law:**

$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$Let's pay attention that in the expression on the right side of the distributive law **we take as a factor the first term that the parentheses are multiplied by**, so we will not forget that ** the sign before the term is an inseparable part of it,** and we will also apply the laws of sign multiplication and thus we can present any expression in parentheses, which we will develop using the above expression, starting as an expression in which the multiplication operation takes place between all the terms:

$(2x-y)(4-3x)\\
(\textcolor{red}{2x}+\textcolor{blue}{(-y)})(4+(-3x))\\$**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 calculation of the multiplications above we used the multiplication table and the sign multiplication laws for multiplication between terms with identical bases:

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

In the next step ** we will enter similar terms**, we will define similar terms as

Let's pay attention 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**, so 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 thus received that the correct answer is answer D.__

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

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

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*∗2*a*∗*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

$10x^2-29x+21$

Find the common factor:

$25y-100xy^2$

First, we will decompose the coefficients of the multiplication exercise that will help us find the common factor:

$25\times y-4\times25\times x\times y\times y$

Now find the common factor 25y:

$25y(1-4xy)$

$25y(1-4xy)$

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

$(17+c)(5+a+3)$

We can use the right parenthesis since it can be simplified in the following way:

**(8+ a)**

Then we will get the exercise:

$(17+c)(8+a)=$

$136+17a+8c+ca$

Yes, $136+17a+8c+ca$

Decompose the following expression into factors:

$15a^2+10a+5$

** Factor** the given expression:

$15a^2+10a+5$We will do this by** taking out the greatest common factor, both from the numbers and the letters,**

We will refer to the numbers and letters __separately__, remembering that a common factor is a factor (multiplier) **common to all terms of the expression**,

__Let's start with the numbers__

Note that the numerical coefficients of the terms in the given expression, that is, the numbers: 5,10,15 are all multiples of the number 5:

$15=3\cdot\underline{5}\\
10=2\cdot\underline{5}\\$Therefore, the number 5 **is the greatest common factor of the numbers**,

__For the letters:__

**Note that only the first two terms on the left depend on x**, the third term is a free number that does not depend on x, therefore **there is no common factor for all three terms together for the letters** (that is, we will consider the number 1 as the common factor for the letters)

__Therefore, we summarize:__

The greatest common factor ** (for numbers and letters together)** is:

$5\cdot1\\
\downarrow\\
5$Let's take it, then, as a multiple outside the parenthesis and ask the question: **"How many times will we multiply the common factor (including its sign) obtaining each of the terms of the original expression (including its sign)?",** so we will know what is the expression inside the parenthesis that multiplied the common factor:

$\textcolor{red}{ 15a^2}\textcolor{blue}{+10a} \textcolor{green}{+5} \\ \underline{5}\cdot\textcolor{red}{3a^2}+\underline{5}\cdot\textcolor{blue}{(+2a)}+\underline{5}\cdot\textcolor{green}{(+1)}\\ \downarrow\\ \underline{5}(\textcolor{red}{3a^2}\textcolor{blue}{+2a}\textcolor{green}{+1})$In the previous expression, the operation is explained through colors and signs:

The common factor has been highlighted __with an underscore__, and the multiples inside the parenthesis are associated with the terms of the original expression with the help of colors, __note that in the detail of the decomposition above we also refer to the sign of the common factor (in black) that we extracted as a multiple outside the parenthesis and the sign of the terms in the original expression (in colors),__ there is no obligation to show it. This is in stages as described above, you can (and it is worth) jump directly to the broken down form in the last line, but you definitely should refer to the previous signs, since in each member the sign is an inseparable part of it,

**We can ensure that this decomposition is correct easily by opening the parentheses with the help of the distributive property and ensuring that the original expression that we decomposed is effectively obtained back - member, this must be done emphasizing the sign of the members in the original expression and the sign (which is always selectable) of the common factor.**

(Initially, you should use the previous colors to ensure you get all the terms in the original expression and belong to the multiple inside the parenthesis; later, it is recommended not to use the colors)

__Therefore, the correct answer is option b.__

$5(3a^2+2a+1)$