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:

ZΓ—(X+Y)=ZX+ZY Z\times(X+Y)=ZX+ZY

ZΓ—(Xβˆ’Y)=ZXβˆ’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)Γ—(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=2Γ—(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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Resolve -

\( (x-3)(x-6)= \)

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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=4Γ—4Γ—4Γ—4Γ—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

(βˆ’4)2(-4)^2 and βˆ’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

(βˆ’4)2=(βˆ’4)Γ—(βˆ’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:

βˆ’42=βˆ’(4Γ—4)=βˆ’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:
3Γ—(4βˆ’2)2=3Γ—(2)2=3Γ—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:

ZΓ—(X+Y)=ZX+ZY Z \times (X + Y) = ZX + ZY

ZΓ—(Xβˆ’Y)=ZXβˆ’ZY 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

6Γ—26=6Γ—(20+6)=6Γ—20+6Γ—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.
3aΓ—(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=2Γ—(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)Γ—(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)Γ—(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)Γ—(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

Resolve -

(xβˆ’3)(xβˆ’6)= (x-3)(x-6)=

Video Solution

Step-by-Step Solution

To solve this problem, we will expand the expression (xβˆ’3)(xβˆ’6)(x-3)(x-6) using the distributive property, which involves the following steps:

  • Step 1: Multiply the first terms of each binomial
    (x)(x)=x2(x)(x) = x^2

  • Step 2: Multiply the outer terms of the binomials
    (x)(βˆ’6)=βˆ’6x(x)(-6) = -6x

  • Step 3: Multiply the inner terms of the binomials
    (βˆ’3)(x)=βˆ’3x(-3)(x) = -3x

  • Step 4: Multiply the last terms of each binomial
    (βˆ’3)(βˆ’6)=18(-3)(-6) = 18

  • Step 5: Combine all the products
    x2βˆ’6xβˆ’3x+18x^2 - 6x - 3x + 18

  • Step 6: Combine like terms
    βˆ’6xβˆ’3x=βˆ’9x-6x - 3x = -9x, so we have
    x2βˆ’9x+18x^2 - 9x + 18

Therefore, the expanded form of (xβˆ’3)(xβˆ’6)(x-3)(x-6) is x2βˆ’9x+18\boxed{x^2 - 9x + 18}.

Therefore, the solution to the problem is x2βˆ’9x+18x^2 - 9x + 18. This corresponds to choice 1.

Answer

x2βˆ’9x+18 x^2-9x+18

Exercise #2

Solve the exercise:

(2yβˆ’3)(yβˆ’4)= (2y-3)(y-4)=

Video Solution

Step-by-Step Solution

To solve the algebraic expression (2yβˆ’3)(yβˆ’4)(2y-3)(y-4), we will apply the distributive property, also known as the FOIL method for binomials. This involves multiplying each term in the first binomial by each term in the second binomial.

  • Step 1: Multiply the first terms: 2yΓ—y=2y2 2y \times y = 2y^2 .
  • Step 2: Multiply the outer terms: 2yΓ—βˆ’4=βˆ’8y 2y \times -4 = -8y .
  • Step 3: Multiply the inner terms: βˆ’3Γ—y=βˆ’3y -3 \times y = -3y .
  • Step 4: Multiply the last terms: βˆ’3Γ—βˆ’4=12 -3 \times -4 = 12 .

Next, we combine all these results: 2y2βˆ’8yβˆ’3y+12 2y^2 - 8y - 3y + 12 .

Then, we combine the like terms βˆ’8y-8y and βˆ’3y-3y to get βˆ’11y-11y.

Therefore, the expanded expression is 2y2βˆ’11y+12 2y^2 - 11y + 12 .

This matches choice (3): 2y2βˆ’11y+12 2y^2 - 11y + 12 .

Thus, the solution to the problem is 2y2βˆ’11y+12 2y^2 - 11y + 12 .

Answer

2y2βˆ’11y+12 2y^2-11y+12

Exercise #3

Solve the exercise:

(3xβˆ’1)(x+2)= (3x-1)(x+2)=

Video Solution

Step-by-Step Solution

To solve this problem, we'll apply the distributive property to expand the expression (3xβˆ’1)(x+2)(3x-1)(x+2). Below are the steps:

  • Step 1: Distribute each term in the first binomial to each term in the second binomial:

3x(x)+3x(2)+(βˆ’1)(x)+(βˆ’1)(2)3x(x) + 3x(2) + (-1)(x) + (-1)(2)

  • Step 2: Calculate each term:

3x2+6xβˆ’xβˆ’23x^2 + 6x - x - 2

  • Step 3: Combine like terms:

3x2+(6xβˆ’x)βˆ’2=3x2+5xβˆ’23x^2 + (6x - x) - 2 = 3x^2 + 5x - 2

Thus, the expanded expression is 3x2+5xβˆ’23x^2 + 5x - 2.

The correct answer choice is 3x2+5xβˆ’23x^2 + 5x - 2, corresponding to choice id="4".

Answer

3x2+5xβˆ’2 3x^2+5x-2

Exercise #4

Solve the exercise:

(5xβˆ’2)(3+x)= (5x-2)(3+x)=

Video Solution

Step-by-Step Solution

To solve the problem (5xβˆ’2)(3+x) (5x-2)(3+x) , we will use the distributive property, specifically the FOIL (First, Outer, Inner, Last) method, to expand the expression:

  • Step 1: Multiply the First terms: 5xΓ—3=15x 5x \times 3 = 15x .
  • Step 2: Multiply the Outer terms: 5xΓ—x=5x2 5x \times x = 5x^2 .
  • Step 3: Multiply the Inner terms: βˆ’2Γ—3=βˆ’6 -2 \times 3 = -6 .
  • Step 4: Multiply the Last terms: βˆ’2Γ—x=βˆ’2x -2 \times x = -2x .

Now combine all these products together:

5x2+15xβˆ’2xβˆ’6 5x^2 + 15x - 2x - 6

Combine the like terms 15x 15x and βˆ’2x -2x :

5x2+(15xβˆ’2x)βˆ’6=5x2+13xβˆ’6 5x^2 + (15x - 2x) - 6 = 5x^2 + 13x - 6

Thus, the expanded form of the expression is 5x2+13xβˆ’6 5x^2 + 13x - 6 .

Answer

5x2+13xβˆ’6 5x^2+13x-6

Exercise #5

(3+20)Γ—(12+4)= (3+20)\times(12+4)=

Video Solution

Step-by-Step Solution

Simplify this expression paying attention to the order of arithmetic operations. Exponentiation precedes multiplication whilst division precedes addition and subtraction. Parentheses precede all of the above.

Therefore, let's first start by simplifying the expressions within the parentheses. Then we can proceed to perform the multiplication between them:

(3+20)β‹…(12+4)=23β‹…16=368 (3+20)\cdot(12+4)=\\ 23\cdot16=\\ 368

Therefore, the correct answer is option A.

Answer

368

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