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=2×(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.
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=4×4×4×4×4
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
(−4)2 and −42 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
Conversely, in the second case, one must first calculate the exponent and then deal with the negative sign. That is:
−42=−(4×4)=−16
Also remember that exponents come before four of the operations in the order of mathematical operations, but not before parentheses.
For example: 3×(4−2)2=3×(2)2=3×4=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×(X+Y)=ZX+ZY
Z×(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
\( (12+2)\times(3+5)= \)
Incorrect
Correct Answer:
112
Question 2
\( (35+4)\times(10+5)= \)
Incorrect
Correct Answer:
585
Question 3
Break down the expression into basic terms:
\( 4x^2 + 6x \)
Incorrect
Correct Answer:
\( 4\cdot x\cdot x+6\cdot x \)
Example 1: Distributive Property
6×26=6×(20+6)=6×20+6×6=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
Simplify the expression:
\( 5x^3 + 3x^2 \)
Incorrect
Correct Answer:
\( 5\cdot x\cdot x\cdot x + 3\cdot x\cdot x \)
Question 2
Solve the following problem:
\( (12-x)(x-3)= \)
Incorrect
Correct Answer:
\( 15x-36-x^2 \)
Question 3
Solve the following problem:
\( (a+15)(5+a)= \)
Incorrect
Correct Answer:
\( a^2+20a+75 \)
Example 3: Distributive Property with Variables
Clear the parentheses by applying the distributive property. 3a×(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=2×(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
Break down the expression into basic terms:
\( 2x^2 \)
Incorrect
Correct Answer:
\( 2\cdot x\cdot x \)
Question 2
\( (2x+y)(x+3)= \)
Incorrect
Correct Answer:
\( 2x^2+xy+6x+3y \)
Question 3
Rewrite using basic components:
\( 8x^2 - 4x \)
Incorrect
Correct Answer:
\( 8\cdot x\cdot x-4\cdot x \)
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
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)=
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
Solve the exercise:
\( (2y-3)(y-4)= \)
Incorrect
Correct Answer:
\( 2y^2-11y+12 \)
Question 2
Break down the expression into basic terms:
\( 3y^3 \)
Incorrect
Correct Answer:
\( 3\cdot y\cdot y \cdot y \)
Question 3
\( (a+4)(c+3)= \)
Incorrect
Correct Answer:
\( ac+3a+4c+12 \)
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)×(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
Solve the following problem:
(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
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.
(x−6)(x+8)↓(x+(−6))(x+8)Let's begin then with opening the parentheses:
(x+(−6))(x+8)x⋅x+x⋅8+(−6)⋅x+(−6)⋅8x2+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:
am⋅an=am+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+8x−6x−48x2+2x−48When 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+2x−48
Exercise #2
Solve the following problem:
(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 and x. Multiply these together to obtain x×x=x2.
Step 2: Multiply the Outer terms.
The outer terms are x and −4. Multiply these. together to obtain x×−4=−4x.
Step 3: Multiply the Inner terms.
The inner terms are 2 and x. Multiply these together to obtain 2×x=2x.
Step 4: Multiply the Last terms.
The last terms are 2 and −4. Multiply these together to obtain 2×−4=−8.
Proceed to combine all these results together:
x2−4x+2x−8
Finally, combine like terms:
Combine −4x and 2x to obtain −2x.
The expanded form of the expression is therefore:
x2−2x−8
Thus, the solution to the problem is x2−2x−8, which corresponds to choice 1.
Answer
x2−2x−8
Exercise #3
It is possible to use the distributive property to simplify the expression below?
What is its simplified form?
(ab)(cd)
Video Solution
Step-by-Step Solution
Let's remember the extended distributive property:
(a+b)(c+d)=ac+ad+bc+bdNote that the operation between the terms inside the parentheses is a multiplication operation:
(ab)(cd)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)=abcdTherefore, 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 #4
(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
Therefore, the correct answer is (a).
Answer
ac + ad+bc+bd
Exercise #5
Expand the following expression:
(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
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)x⋅x+x⋅3+4⋅x+4⋅3x2+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:
am⋅an=am+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+12In 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,