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:

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

$Z\times(X-Y)=ZX-ZY$

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:

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

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:
$2A + 4B$

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

$2A+4B=2\times(A+2B)$

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.

Detailed explanation

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Test yourself on algebraic technique!

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

What is its simplified form?

$$ (ab)(c d) $$

$$

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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:
$4^5$

$4$ is the number that is multiplied by itself. It is called the Base of the exponent.
$5$ represents the number of times the multiplication of the base is repeated and it is called the Exponent.

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

Let's remember that any number raised to the power of $1$ equals the number itself
That is:

$4^1=4$

And remember that any number raised to the power of $0$ equals $1$
$4^0=1$

Mathematical definition to the power of $0$ .

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

$(-4)^2$ and $-4^2$
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

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

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

$-4^2=-(4\times4)=-16$

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

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

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The Distributive Property

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

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

$Z \times (X - Y) = ZX - ZY$

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

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Test your knowledge

Question 1

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

$$ a(b+c) $$

Question 2

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

$(a+b)(c\cdot g)$

Question 3

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

Example 1: Distributive Property

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

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

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

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?

Question 1

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

Question 2

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

Question 3

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

Example 3: Distributive Property with Variables

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

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:

$2A + 4B$

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 $2$ .
We will write it as follows:

$2A+4B=2\times(A+2B)$

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 $2$ 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.

Check your understanding

Question 1

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

Question 2

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

If so, what is its simplest form?

$(x+c)(4+c) =\text{?}$

Question 3

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

Extended Distributive Property

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

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:

$(a+2)\times(3+a)=$

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?

Question 1

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

Question 2

Solve the following problem:

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

Question 3

Solve the following problem:

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

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.

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

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

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

What is its simplified form?

$(ab)(c d)$

Video Solution

Step-by-Step Solution

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.

Answer

No, $abcd$ .

Exercise #2

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

$a(b+c)$

Video Solution

Step-by-Step Solution

To solve the problem and apply the distributive property correctly, follow these steps:

Identify the expression, which is $a(b+c)$ .
Apply the distributive property: multiply $a$ by each term inside the parentheses.

Applying this, we get:

$a \times b = ab$
$a \times c = ac$

Combine these two products:

The simplified expression is: $ab + ac$ .

This matches with answer choice 2: Yes, the answer $ab+ac$ .

Answer

Yes, the answer $ab+ac$

Exercise #3

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

$(a+b)(c\cdot g)$

Video Solution

Step-by-Step Solution

To solve this problem, we must determine if we can apply the distributive property to simplify the expression $(a+b)(c \cdot g)$ .

The distributive property states that for any three terms, the expression $x(y+z)$ results in $xy + xz$ . Here, we have the sum $(a + b)$ and the product $(c \cdot g)$ .

We can treat $(c \cdot g)$ as a single term because it involves multiplication, which makes it like a single number or variable in terms of manipulating the expression algebraically. Therefore, using the distributive property, we distribute $(c \cdot g)$ over the terms within the parentheses:

Step 1: Distribute $c \cdot g$ to $a$ , yielding $acg$ .
Step 2: Distribute $c \cdot g$ to $b$ , yielding $bcg$ .

Hence, the simplified expression is:

$acg + bcg$ .

Therefore, the correct answer, according to the choices provided, is:

No, $acg + bcg$ .

Answer

No, $acg+\text{bcg}$

Exercise #4

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

Video Solution

Step-by-Step Solution

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).

Answer

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

Exercise #5

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

Video Solution

Step-by-Step Solution

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!

Answer

$ac+3a+4c+12$

Test your knowledge

Question 1

Resolve -

$$ (x-3)(x-6)= $$

Question 2

Solve the following problem:

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

Question 3

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

What is its simplified form?

$$ (ab)(c d) $$

$$

Start practice

Algebraic Method