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.

Practice Algebraic Technique

Examples with solutions for Algebraic Technique

Exercise #1

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 #2

(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 #3

(x8)(x+y)= (x-8)(x+y)=

Video Solution

Step-by-Step Solution

Let's simplify the given expression, open the parentheses using the expanded 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, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first as an expression where addition operation exists between all terms:

(x8)(x+y)(x+(8))(x+y) (x-8)(x+y)\\ (\textcolor{red}{x}+\textcolor{blue}{(-8)})(x+y)\\ Let's begin then with opening the parentheses:

(x+(8))(x+y)xx+xy+(8)x+(8)yx2+xy8x8y (\textcolor{red}{x}+\textcolor{blue}{(-8)})(x+y)\\ \textcolor{red}{x}\cdot x+\textcolor{red}{x}\cdot y+\textcolor{blue}{(-8)}\cdot x +\textcolor{blue}{(-8)}\cdot y\\ x^2+xy-8x -8y

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}

Note that in the expression we got in the last stage there are four different terms, this is because there isn't even one pair of terms where the variables (different ones) have the same exponent, additionally the expression is already organized therefore the expression we got is the final and most simplified form:
x2+xy8x8y \textcolor{purple}{ x^2}\textcolor{green}{+xy}-8x \textcolor{orange}{-8y}\\ 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,

We therefore concluded that the correct answer is answer A.

Answer

x2+xy8x8y x^2+xy-8x-8y

Exercise #4

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

Video Solution

Step-by-Step Solution

Let's 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, 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, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first as an expression where addition operation exists between all terms:

(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

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, we'll define like terms 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, this is because raising any number to the zero power yields the result 1), we'll use the commutative property of addition, additionally we'll arrange the expression from highest to lowest power from left to right (we'll 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}\\ 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,

We therefore got that the correct answer is answer A.

Answer

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

Exercise #5

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

Video Solution

Step-by-Step Solution

Let's 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, 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, which we'll open using the above formula, first as an expression where addition operation exists between all terms, in this expression as it's clear, all terms have a plus sign prefix, therefore we'll proceed directly to opening the parentheses,

Let's begin then with opening the parentheses:

(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, we'll define like terms 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 since raising any number to the power of zero yields 1), we'll use the commutative property of addition, additionally 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,

We therefore got that the correct answer is answer C.

Answer

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

Exercise #6

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

Video Solution

Step-by-Step Solution

Let's simplify the given expression, open the parentheses using the extended distribution law:

(t+b)(c+d)=tc+td+bc+bd (\textcolor{red}{t}+\textcolor{blue}{b})(c+d)=\textcolor{red}{t}c+\textcolor{red}{t}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, which we'll open using the above formula, first as an expression where addition operation exists between all terms, in this expression as it's clear, all terms have a plus sign prefix, therefore we'll proceed directly to opening the parentheses,

Let's begin then with opening the parentheses:

(a+15)(5+a)a5+aa+155+15a5a+a2+75+15a (\textcolor{red}{a}+\textcolor{blue}{15})(5+a)\\ \textcolor{red}{a}\cdot 5+\textcolor{red}{a}\cdot a+\textcolor{blue}{15}\cdot 5 +\textcolor{blue}{15}\cdot a\\ 5a+a^2+75+15a

In calculating the above multiplications, we used the multiplication table and the laws of exponents for multiplication between terms with identical bases:

xmxn=xm+n x^m\cdot x^n=x^{m+n}

In the next step we'll combine like terms, we'll define like terms as terms where the variable (or variables each separately), in this case a, have identical exponents (in the absence of one of the variables from the expression, we'll consider its exponent as zero power, this is because any number raised to the power of zero equals 1), we'll use the commutative law of addition, additionally we'll arrange the expression from highest to lowest power from left to right (we'll treat the free number as power of zero):
5a+a2+75+15aa2+5a+15a+75a2+20a+75 \textcolor{purple}{5a}\textcolor{green}{+a^2}+75\textcolor{purple}{+15a}\\ \textcolor{green}{a^2}\textcolor{purple}{+5a+15a}+75\\ \textcolor{green}{a^2}\textcolor{purple}{+20a}+75\\ 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,

We therefore got that the correct answer is answer B.

Answer

a2+20a+75 a^2+20a+75

Exercise #7

(12x)(x3)= (12-x)(x-3)=

Video Solution

Step-by-Step Solution

Let's simplify the given expression, open the parentheses using the extended distribution law:

(t+k)(c+d)=tc+td+kc+kd (\textcolor{red}{t}+\textcolor{blue}{k})(c+d)=\textcolor{red}{t}c+\textcolor{red}{t}d+\textcolor{blue}{k}c+\textcolor{blue}{k}d

Note that in the formula template for the above distribution law, we take as a default that the operation between 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, and we'll also apply the rules of sign multiplication and thus we can present any expression in parentheses, which we'll open using the above formula, first as an expression where addition operation exists between all terms:

(12x)(x3)(12+(x))(x+(3)) (12-x)(x-3) \\ (\textcolor{red}{12}+\textcolor{blue}{(-x)})(x+(-3))\\ Let's begin then with opening the parentheses:

(12+(x))(x+(3))12x+12(3)+(x)x+(x)(3)12x36x2+3x (\textcolor{red}{12}+\textcolor{blue}{(-x)})(x+(-3))\\ \textcolor{red}{12}\cdot x+\textcolor{red}{12}\cdot(-3)+\textcolor{blue}{(-x)}\cdot x +\textcolor{blue}{(-x)}\cdot(-3)\\ 12x-36-x^2 +3x

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, we'll define like terms 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, since raising any number to the zero power yields 1), we'll use the commutative property of addition, additionally we'll arrange the expression from highest to lowest power from left to right (we'll treat the free number as having zero power):
12x36x2+3xx2+12x+3x36x2+15x36 \textcolor{purple}{12x}\textcolor{green}{-36}-x^2\textcolor{purple}{+3x}\\ -x^2\textcolor{purple}{+12x+3x}\textcolor{green}{-36}\\ -x^2\textcolor{purple}{+15x}\textcolor{green}{-36}\\ 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,

We therefore got that the correct answer is answer A (we used the commutative property of addition to verify this).

Answer

15x36x2 15x-36-x^2

Exercise #8

(12+2)×(3+5)= (12+2)\times(3+5)=

Video Solution

Step-by-Step Solution

Simplify this expression by paying attention to the order of arithmetic operations which states that exponentiation precedes multiplication, division precedes addition and subtraction and that parentheses precede all of the above.

Thus, let's begin by simplifying the expressions within the parentheses, and following this, the multiplication between them.

(12+2)(3+5)=148=112 (12+2)\cdot(3+5)= \\ 14\cdot8=\\ 112

Therefore, the correct answer is option C.

Answer

112

Exercise #9

(3+20)×(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)(12+4)=2316=368 (3+20)\cdot(12+4)=\\ 23\cdot16=\\ 368

Therefore, the correct answer is option A.

Answer

368

Exercise #10

Break down the expression into basic terms:

3x2+2x 3x^2 + 2x

Step-by-Step Solution

The expression can be broken down as follows:

3x2+2x 3x^2 + 2x

Breaking down each term we have:

- 3x2 3x^2 becomes 3xx 3\cdot x\cdot x

- 2x 2x remains 2x 2 \cdot x

Finally, the expression is:

3xx+2x 3\cdot x\cdot x+2\cdot x

Answer

3xx+2x 3\cdot x\cdot x+2\cdot x

Exercise #11

Break down the expression into basic terms:

4x2+3x 4x^2 + 3x

Step-by-Step Solution

The expression can be broken down as follows:

4x2+3x 4x^2 + 3x

1. Notice that both terms contain a common factor of x x .

2. Factor out the common x x :

x(4x+3) x(4x + 3) .

3. Thus, breaking down each term we have:

- 4x2 4x^2 becomes 4xx 4x \cdot x after factoring out x x .

- 3x 3x remains 3x 3 \cdot x after factoring out x x .

Finally, the expression is:

4xx+3x 4x\cdot x + 3\cdot x

Answer

4xx+3x 4\cdot x\cdot x+3\cdot x

Exercise #12

Break down the expression into basic terms:

3a3 3a^3

Step-by-Step Solution

To break down the expression 3a3 3a^3 , we recognize that a3 a^3 means a×a×a a \times a \times a . Therefore, 3a3 3a^3 can be decomposed as 3aaa 3 \cdot a\cdot a\cdot a .

Answer

3aaa 3 \cdot a\cdot a\cdot a

Exercise #13

Break down the expression into basic terms:

3y2+6 3y^2 + 6

Step-by-Step Solution

To break down the expression 3y2+6 3y^2 + 6 , we need to recognize common factors or express terms in basic forms.

The term 3y2 3y^2 can be rewritten by breaking down the operations: 3yy 3\cdot y\cdot y .

The constant 6 6 remains as it is in its basic term.

Thus, the broken down expression becomes 3yy+6 3\cdot y\cdot y + 6 .

Answer

3yy+6 3\cdot y\cdot y+6

Exercise #14

Break down the expression into basic terms:

3y3 3y^3

Step-by-Step Solution

To break down the expression 3y3 3y^3 into its basic terms, we understand the components of the expression:

3is a constant multiplier 3 \, \text{is a constant multiplier}

y3 y^3 can be rewritten as yyy y \cdot y \cdot y

Thus, 3y3 3y^3 can be decomposed into 3yyy 3 \cdot y \cdot y \cdot y .

Answer

3yyy 3\cdot y\cdot y \cdot y

Exercise #15

Break down the expression into basic terms:

4a2 4a^2

Step-by-Step Solution

To break down the expression 4a2 4a^2 into basic terms, we need to look at each factor:

4is a constant multiplier 4 \, \text{is a constant multiplier}

a2 a^2 means aa a \cdot a

Hence, 4a2 4a^2 is equivalent to 4aa 4 \cdot a \cdot a .

Answer

4aa 4\cdot a\cdot a