Algebraic Technique - Examples, Exercises and Solutions

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

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.

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

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

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 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.

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

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:

$ab+bc$

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

If we consider that b is the common factor, it can be removed from the equation:

$b(ab+bc)=$

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 into a multiplication exercise to help us simplify the calculation accordingly:$7\times a+7\times b\times2=$

We then extract the common factor 7:

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

$7(a+2b)$

Find the biggest common factor:

$12x+16y$

We begin by breaking down the coefficients 12 and 16 into multiplication exercises with a multiplying factor to eventually simplify:

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

We then 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 first break down the coefficient of 20 into a multiplication exercise. That will help us to simplify the calculation :$5\times4\times a\times b-4\times a\times c$

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

$4a(5b-c)$

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

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.

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

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

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

$10x^2-29x+21$

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 may use the parenthesis on the right hand side due to the fact that it can be simplified as follows :

(8+a)

Resulting in the following calculation:

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

$136+17a+8c+ca$

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 factoring out the greatest common factor, both from the numbers and the letters,

We must refer to the numbers and letters separately, remembering that a common factor is a factor (multiplier) common to all the 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, hence 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 can summarise as follows:

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

$5\cdot1\\ \downarrow\\ 5$Let's take the above value as a multiple outside the parenthesis and ask the question: "How many times must we multiply the common factor (including its sign) in order to obtain the terms of the original expression (including their signs)?" Using this method we can determine 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 refer both to the sign of the common factor (in black) that we extracted as a multiple outside of the parenthesis, as well as to the sign of the terms in the original expression (in colors) Note that there is no obligation to show it. Whilst the above method is described in stages it is recommended to jump directly to the broken down form in the last line whilst continuing to refer to the previous signs which are an integral part of the expression.

We can easily ensure that this decomposition is correct by opening the parentheses with the help of the distributive property. As such we can ensure that the original expression that we decomposed can be effectively recovered - Remember, this must be done emphasizing the sign of the members in the original expression as well as the sign (which is always selectable) of the common factor.

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

Therefore, the correct answer is option b.

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

Decompose the following expression into factors by removing the common factor:

$xyz+yzt+ztw+wtr$

Factor the given expression:

$xyz+yzt+ztw+wtr$
We will do this by extracting the highest common factor, both from the numbers and the letters.

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

As the given expression does not have numeric coefficients (other than 1), we will look for the highest common factor of the letters:

There are four terms in the expression:
$xyz,\hspace{4pt}yzt,\hspace{4pt}ztw,\hspace{4pt}wtr$We will notice that in each of the four members there are three different letters, but there is not one or more letters that are included (in the multiplication) in all the terms; that is, there is no common factor for the four terms and therefore it is not possible to factor this expression by extracting a common factor.

Therefore, the correct answer is option d.

It is not possible to decompose the given expression into factors by extracting the common factor.

Decompose the following expression into factors:

$4a+13b+58c$

Factor the given expression:

$4a+13b+58c$We will do this by extracting 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 (multiple) common to all terms of the expression,

Let's start with the numbers:

We first notice that the numerical coefficients of the terms in the given expression, that is, the numbers 4, 13, 58, do not have a common factor, and this is because the number 13 is a prime number and the other two numbers are not multiples of it,

Therefore, there is no common factor for the numbers hence we select the number 1, as the common factor for the numbers (Reminder: a number raised to the power zero is always given to be equal to one)

For the letters:

There are three terms in the expression:
$a,\hspace{4pt}b,\hspace{4pt}c$It is easy to see that there is no common factor for these three terms,

Hence, it is not possible to factor the given expression with the help of a common factor.

Therefore, the correct answer is option d.

It is not possible to factorize the given expression by extracting the common factor.