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.
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:
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:
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=2Γ(A+2B)
In this article, weβll explain each of these topics in detail, But each of these topics will be explained even more in detail in their respective articles.
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: 45
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: 45=4Γ4Γ4Γ4Γ4
Let's remember that any number raised to the power of 1 equals the number itself That is:
41=4
And remember that any number raised to the power of 0 equals 1 40=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
(β4)2 and β42 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
(β4)2=(β4)Γ(β4)=16
Conversely, in the second case, one must first calculate the exponent and then deal with the negative sign. That is:
β42=β(4Γ4)=β16
Also remember that exponents come before four of the operations in the order of mathematical operations, but not before parentheses.
For example: 3Γ(4β2)2=3Γ(2)2=3Γ4=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:
ZΓ(X+Y)=ZX+ZY
ZΓ(XβY)=ZXβZY
Now letβs look at some examples with numbers to understand the formula.
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Test your knowledge
Question 1
Solve the exercise:
\( (2y-3)(y-4)= \)
Incorrect
Correct Answer:
\( 2y^2-11y+12 \)
Question 2
Solve the exercise:
\( (3x-1)(x+2)= \)
Incorrect
Correct Answer:
\( 3x^2+5x-2 \)
Question 3
Solve the exercise:
\( (5x-2)(3+x)= \)
Incorrect
Correct Answer:
\( 5x^2+13x-6 \)
Example 1: Distributive Property
6Γ26=6Γ(20+6)=6Γ20+6Γ6=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
\( (3+20)\times(12+4)= \)
Incorrect
Correct Answer:
368
Question 2
\( (12+2)\times(3+5)= \)
Incorrect
Correct Answer:
112
Question 3
\( (35+4)\times(10+5)= \)
Incorrect
Correct Answer:
585
Example 3: Distributive Property with Variables
Clear the parentheses by applying the distributive property. 3aΓ(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.
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=2Γ(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 the2 by each of the terms inside 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
Expand the following expression:
\( (x+4)(x+3)= \)
Incorrect
Correct Answer:
\( x^2+7x+12 \)
Question 2
\( (a+b)(c+d)= \) ?
Incorrect
Correct Answer:
\( \text{ac + ad}+bc+bd \)
Question 3
\( (2x+y)(x+3)= \)
Incorrect
Correct Answer:
\( 2x^2+xy+6x+3y \)
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
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)=
Phase 1: Let's multiply a by each of the terms in the second set of parentheses.
Do you think you will be able to solve it?
Question 1
\( (a+4)(c+3)= \)
Incorrect
Correct Answer:
\( ac+3a+4c+12 \)
Question 2
\( (x+13)(y+4)= \)
Incorrect
Correct Answer:
\( xy+4x+13y+52 \)
Question 3
Solve the following problem:
\( (x-8)(x+y)= \)
Incorrect
Correct Answer:
\( x^2+xy-8x-8y \)
Phase 2: Let's multiply the 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
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 -
(xβ3)(xβ6)=
Video Solution
Step-by-Step Solution
To solve this problem, we will expand the expression (xβ3)(xβ6) using the distributive property, which involves the following steps:
Step 1: Multiply the first terms of each binomial (x)(x)=x2
Step 2: Multiply the outer terms of the binomials (x)(β6)=β6x
Step 3: Multiply the inner terms of the binomials (β3)(x)=β3x
Step 4: Multiply the last terms of each binomial (β3)(β6)=18
Step 5: Combine all the products x2β6xβ3x+18
Step 6: Combine like terms β6xβ3x=β9x, so we have x2β9x+18
Therefore, the expanded form of (xβ3)(xβ6) is x2β9x+18β.
Therefore, the solution to the problem is x2β9x+18. This corresponds to choice 1.
Answer
x2β9x+18
Exercise #2
Solve the exercise:
(2yβ3)(yβ4)=
Video Solution
Step-by-Step Solution
To solve the algebraic expression (2yβ3)(yβ4), 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: 2yΓy=2y2.
Step 2: Multiply the outer terms: 2yΓβ4=β8y.
Step 3: Multiply the inner terms: β3Γy=β3y.
Step 4: Multiply the last terms: β3Γβ4=12.
Next, we combine all these results: 2y2β8yβ3y+12.
Then, we combine the like terms β8y and β3y to get β11y.
Therefore, the expanded expression is 2y2β11y+12.
This matches choice (3):2y2β11y+12.
Thus, the solution to the problem is 2y2β11y+12.
Answer
2y2β11y+12
Exercise #3
Solve the exercise:
(3xβ1)(x+2)=
Video Solution
Step-by-Step Solution
To solve this problem, we'll apply the distributive property to expand the expression (3xβ1)(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)+(β1)(x)+(β1)(2)
Step 2: Calculate each term:
3x2+6xβxβ2
Step 3: Combine like terms:
3x2+(6xβx)β2=3x2+5xβ2
Thus, the expanded expression is 3x2+5xβ2.
The correct answer choice is 3x2+5xβ2, corresponding to choice id="4".
Answer
3x2+5xβ2
Exercise #4
Solve the exercise:
(5xβ2)(3+x)=
Video Solution
Step-by-Step Solution
To solve the problem (5xβ2)(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: 5xΓ3=15x.
Step 2: Multiply the Outer terms: 5xΓx=5x2.
Step 3: Multiply the Inner terms: β2Γ3=β6.
Step 4: Multiply the Last terms: β2Γx=β2x.
Now combine all these products together:
5x2+15xβ2xβ6
Combine the like terms 15x and β2x:
5x2+(15xβ2x)β6=5x2+13xβ6
Thus, the expanded form of the expression is 5x2+13xβ6.
Answer
5x2+13xβ6
Exercise #5
(3+20)Γ(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: