Algebraic Method

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Algebraic Method

Algebraic Method is a general term for various tools and techniques that will help us solve more complex exercises in the future. It is mostly concern about using algebraic operations to isolate variables and solve equations. This approach is fundamental for solving equations in various mathematical contexts.

Distributive Property

This property helps us to clear parentheses and assists us with more complex calculations. Let's remember how it works. Generally, we will write it like this:

Z×(X+Y)=ZX+ZY Z\times(X+Y)=ZX+ZY

Z×(XY)=ZXZY Z\times(X-Y)=ZX-ZY

Extended Distributive Property

The extended distributive property is very similar to the distributive property, but it allows us to solve exercises with expressions in parentheses that are multiplied by other expressions in parentheses.
It looks like this:

(a+b)×(c+d)=ac+ad+bc+bd (a+b)\times(c+d)=ac+ad+bc+bd

Factoring

The factoring method is very important. It will help us move from an expression with several terms to one that includes only one by taking out the common factor from within the parentheses.
For example:
2A+4B2A + 4B

This expression consists of two terms. We can factor it by reducin by the greatest common factor. In this case, it's the 2 2 .
We will write it as follows:

2A+4B=2×(A+2B) 2A+4B=2\times(A+2B)

Algebraic Method

In this article, we’ll explain each of these topics in detail, But each of these topics will be explained even more in detail in their respective articles.

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

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Reiteration: Powers

Let's return to the essential points within the topic of exponents:

In fact, exponents are a shorthand way of writing the multiplication of a number by itself several times. It looks like this:
454^5

44 is the number that is multiplied by itself. It is called the Base of the exponent.
55 represents the number of times the multiplication of the base is repeated and it is called the Exponent.

That is, in our example:
45=4×4×4×4×4 4^5=4\times4\times4\times4\times4

Let's remember that any number raised to the power of 11 equals the number itself
That is:

41=44^1=4

And remember that any number raised to the power of 00 equals 11
40=14^0=1

Mathematical definition to the power of 00.

An important point to note is the difference between an exponent inside brackets and an exponent outside brackets. For example, what is the difference between

(4)2(-4)^2 and 42 -4^2
It is an important case that could be confusing. When the exponent is outside of the brackets, as in the first case, you have to raise the entire expression to the given exponent, that is

(4)2=(4)×(4)=16 (-4)^2=(-4)\times(-4)=16

Conversely, in the second case, one must first calculate the exponent and then deal with the negative sign. That is:

42=(4×4)=16 -4^2=-(4\times4)=-16

Also remember that exponents come before four of the operations in the order of mathematical operations, but not before parentheses.

For example:
3×(42)2=3×(2)2=3×4=12 3\times(4-2)^2=3\times(2)^2=3\times4=12


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The Distributive Property

We usually encounter the distributive property around the age of 12 12 . This property is useful for clearing parentheses and assists with more complex calculations. Let's remember how it works. Generally, we write it as:

Z×(X+Y)=ZX+ZY Z \times (X + Y) = ZX + ZY

Z×(XY)=ZXZY Z \times (X - Y) = ZX - ZY

Now let’s look at some examples with numbers to understand the formula.


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Example 1: Distributive Property

6×26=6×(20+6)=6×20+6×6=120+36=156 6\times26=6\times(20+6)=6\times20+6\times6=120+36=156

We used the distributive property to solve a problem that would have been more difficult to compute directly.
We can also use the distributive property with division operations.


Example 2: Distributive Property

104:4=(100+4):4=100:4+4:4=25+1=26104:4=(100+4):4= 100:4 + 4:4 = 25+1 = 26

Once again, the distributive property has helped us to simplify a problem that, if solved step by step in a straightforward manner, would have been slightly more complex.


Do you know what the answer is?

Example 3: Distributive Property with Variables

Clear the parentheses by applying the distributive property.
3a×(2b+5)= 3a\times(2b+5)=

We will pay close attention to multiplying the term outside the parentheses by each of the terms inside the parentheses according to the correct order of operations.

Example 3- Distributive property with variables


Factoring: Taking Out the Common Factor from Parentheses

The method of eliminating a common factor is very important. It will help us move from an expression with several terms to one that includes only one.
For example, let's look at the expression:

2A+4B2A + 4B

This expression is now composed of two terms. We can factorize it by eliminating the greatest common term. In this case, it's the number 22.
We will write it as follows:

2A+4B=2×(A+2B) 2A+4B=2\times(A+2B)

We will realize that we have moved from a situation in which we had two parts being added together, to a situation with multiplication. This procedure is called factorization.
We can use the distributive property we mentioned earlier to do the reverse process. Multiply the 22 by each of the terms inside the parentheses:

Factorization - Extracting the common term outside of the parentheses

In certain cases we might prefer an expression with multiplication, and in other cases one with addition.
In the article that elaborates on this topic, you can see more examples regarding this.


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Extended Distributive Property

The extended distributive property is very similar to the distributive property, but it allows us to solve exercises with expressions in parentheses that are multiplied by other expressions in parentheses.
It looks like this:

(a+b)×(c+d)=ac+ad+bc+bd (a+b)\times(c+d)=ac+ad+bc+bd

How does the extended distributive property work?

  • Step 1: Multiply the first term of the first parentheses by each of the terms in the second parentheses.
  • Step 2: Multiply the second term of the first parentheses by each of the terms in the second parentheses.
  • Step 3: Combine like terms.

Example:

(a+2)×(3+a)= (a+2)\times(3+a)=


Phase 1: Let's multiply a by each of the terms in the second set of parentheses.

Phase 1- Let's multiply a by each of the terms inside the second parentheses


Do you think you will be able to solve it?

Phase 2: Let's multiply the 2 by each of the terms in the second parentheses.

Phase 2 - Multiply 2 by each of the terms in the second parentheses


Phase 3: Let's organize the terms and, if there are similar ones, let's associate them.

(a+2)×(3+a)=3a+a2+6+2a=a2+5a+6 (a+2)\times(3+a)=3a+a^2+6+2a=a^2+5a+6

In the full article about the extended distributive property, you can find detailed explanations and many more examples.


Examples and exercises with solutions for the Algebraic Method

Exercise #1

Solve the following problem:

(x6)(x+8)= (x-6)(x+8)=

Video Solution

Step-by-Step Solution

In order to simplify the given expression, open the parentheses using the extended distribution law:

(a+b)(c+d)=ac+ad+bc+bd (\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, it is a given that the operation between the terms inside of the parentheses is addition. Furthermore the sign preceding the term is of great significance and must be taken into consideration;

Proceed to apply the above formula to the expression to open out the parentheses.

(x6)(x+8)(x+(6))(x+8) (x-6)(x+8)\\ \downarrow\\ \big(\textcolor{red}{x}+\textcolor{blue}{(-6)}\big)(x+8)\\ Let's begin then with opening the parentheses:

(x+(6))(x+8)xx+x8+(6)x+(6)8x2+8x6x48 \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

To calculate the above multiplications operations we used the multiplication table as well as the laws of exponents for multiplication between terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n}

In the next step we'll combine like terms which we define as terms where the variable (or variables ), in this case x, have identical exponents . (Note that in the absence of one of the variables from the expression, we'll consider its exponent as zero power due to the fact that raising any number to the zero power yields the result 1) Apply the commutative property of addition and proceed to arrange the expression from highest to lowest power from left to right (Note: treat the free number as having zero power):
x2+8x6x48x2+2x48 \textcolor{purple}{x^2}\textcolor{green}{+8x-6x}\textcolor{orange}{-48}\\ \textcolor{purple}{x^2}\textcolor{green}{+2x}\textcolor{orange}{-48}\\ When combining like terms as shown above, we highlighted the different terms using colors, as well as treating the sign preceding the term as an inseparable part of it.

The correct answer is answer A.

Answer

x2+2x48 x^2+2x-48

Exercise #2

Solve the following problem:

(x+2)(x4)= (x+2)(x-4)=

Video Solution

Step-by-Step Solution

In order to solve the given problem, we will use the FOIL method. FOIL stands for First, Outer, Inner, Last. This helps us to systematically expand the product of the two binomials:

  • Step 1: Multiply the First terms.

The first terms of each binomial are x x and x x . Multiply these together to obtain x×x=x2 x \times x = x^2 .

  • Step 2: Multiply the Outer terms.

The outer terms are x x and 4-4. Multiply these. together to obtain x×4=4x x \times -4 = -4x .

  • Step 3: Multiply the Inner terms.

The inner terms are 2 2 and x x . Multiply these together to obtain 2×x=2x 2 \times x = 2x .

  • Step 4: Multiply the Last terms.

The last terms are 2 2 and 4-4. Multiply these together to obtain 2×4=8 2 \times -4 = -8 .

Proceed to combine all these results together:

x24x+2x8 x^2 - 4x + 2x - 8

Finally, combine like terms:

Combine 4x-4x and 2x2x to obtain 2x-2x.

The expanded form of the expression is therefore:

x22x8 x^2 - 2x - 8

Thus, the solution to the problem is x22x8 x^2 - 2x - 8 , which corresponds to choice 1.

Answer

x22x8 x^2-2x-8

Exercise #3

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

What is its simplified form?

(ab)(cd) (ab)(c d)

Video Solution

Step-by-Step Solution

Let's remember the extended distributive property:

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

(ab)(cd) (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:

(ab)(cd)=abcd (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.

Answer

No, abcd abcd .

Exercise #4

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

Video Solution

Step-by-Step Solution

Let's simplify the expression by opening the parentheses using the distributive property:

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

Answer

ac + ad+bc+bd \text{ac + ad}+bc+bd

Exercise #5

Expand the following expression:

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

Video Solution

Step-by-Step Solution

Let's simplify the given expression by opening the parentheses using the extended distribution law:

(a+b)(c+d)=ac+ad+bc+bd (\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. We'll open the parentheses using the above formula, first as an expression where an addition operation exists between all terms. In this expression it's clear that all terms have a plus sign prefix. Therefore we'll proceed directly to opening the parentheses,

Let's begin:

(x+4)(x+3)xx+x3+4x+43x2+3x+4x+12 (\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:

aman=am+n a^m\cdot a^n=a^{m+n}

In the next step we'll combine like terms, which we define 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 given that raising any number to the power of zero yields 1) We'll apply the commutative property of addition, furthermore 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):
x2+3x+4x+12x2+7x+12 \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,

Thus the correct answer is C.

Answer

x2+7x+12 x^2+7x+12

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