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

Resolve -

(x3)(x6)= (x-3)(x-6)=

Video Solution

Step-by-Step Solution

To solve this problem, we will expand the expression (x3)(x6)(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)(6)=6x(x)(-6) = -6x

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

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

  • Step 5: Combine all the products
    x26x3x+18x^2 - 6x - 3x + 18

  • Step 6: Combine like terms
    6x3x=9x-6x - 3x = -9x, so we have
    x29x+18x^2 - 9x + 18

Therefore, the expanded form of (x3)(x6)(x-3)(x-6) is x29x+18\boxed{x^2 - 9x + 18}.

Therefore, the solution to the problem is x29x+18x^2 - 9x + 18. This corresponds to choice 1.

Answer

x29x+18 x^2-9x+18

Exercise #2

Solve the exercise:

(2y3)(y4)= (2y-3)(y-4)=

Video Solution

Step-by-Step Solution

To solve the algebraic expression (2y3)(y4)(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: 2y×y=2y2 2y \times y = 2y^2 .
  • Step 2: Multiply the outer terms: 2y×4=8y 2y \times -4 = -8y .
  • Step 3: Multiply the inner terms: 3×y=3y -3 \times y = -3y .
  • Step 4: Multiply the last terms: 3×4=12 -3 \times -4 = 12 .

Next, we combine all these results: 2y28y3y+12 2y^2 - 8y - 3y + 12 .

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

Therefore, the expanded expression is 2y211y+12 2y^2 - 11y + 12 .

This matches choice (3): 2y211y+12 2y^2 - 11y + 12 .

Thus, the solution to the problem is 2y211y+12 2y^2 - 11y + 12 .

Answer

2y211y+12 2y^2-11y+12

Exercise #3

Solve the exercise:

(3x1)(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 (3x1)(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)+(1)(x)+(1)(2)3x(x) + 3x(2) + (-1)(x) + (-1)(2)

  • Step 2: Calculate each term:

3x2+6xx23x^2 + 6x - x - 2

  • Step 3: Combine like terms:

3x2+(6xx)2=3x2+5x23x^2 + (6x - x) - 2 = 3x^2 + 5x - 2

Thus, the expanded expression is 3x2+5x23x^2 + 5x - 2.

The correct answer choice is 3x2+5x23x^2 + 5x - 2, corresponding to choice id="4".

Answer

3x2+5x2 3x^2+5x-2

Exercise #4

Solve the exercise:

(5x2)(3+x)= (5x-2)(3+x)=

Video Solution

Step-by-Step Solution

To solve the problem (5x2)(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: 5x×3=15x 5x \times 3 = 15x .
  • Step 2: Multiply the Outer terms: 5x×x=5x2 5x \times x = 5x^2 .
  • Step 3: Multiply the Inner terms: 2×3=6 -2 \times 3 = -6 .
  • Step 4: Multiply the Last terms: 2×x=2x -2 \times x = -2x .

Now combine all these products together:

5x2+15x2x6 5x^2 + 15x - 2x - 6

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

5x2+(15x2x)6=5x2+13x6 5x^2 + (15x - 2x) - 6 = 5x^2 + 13x - 6

Thus, the expanded form of the expression is 5x2+13x6 5x^2 + 13x - 6 .

Answer

5x2+13x6 5x^2+13x-6

Exercise #5

(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 #6

(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 #7

(35+4)×(10+5)= (35+4)\times(10+5)=

Video Solution

Step-by-Step Solution

We begin by opening the parentheses using the extended distributive property to create a long addition exercise:

We then multiply the first term of the left parenthesis by the first term of the right parenthesis.

We multiply the first term of the left parenthesis by the second term of the right parenthesis.

Now we multiply the second term of the left parenthesis by the first term of the left parenthesis.

Finally, we multiply the second term of the left parenthesis by the second term of the right parenthesis.

In the following way:

(35×10)+(35×5)+(4×10)+(4×5)= (35\times10)+(35\times5)+(4\times10)+(4\times5)=

We solve each of the exercises within parentheses:

350+175+40+20= 350+175+40+20=

We solve the exercise from left to right:

350+175=525 350+175=525

525+40=565 525+40=565

565+20=585 565+20=585

Answer

585

Exercise #8

Expand the following expression:

(x+4)(x+3)= (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 (\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. 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)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, 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+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,

Thus the correct answer is C.

Answer

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

Exercise #9

(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 #10

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

Video Solution

Step-by-Step Solution

To solve this problem, we'll apply the FOIL method for multiplying binomials:

  • First: Multiply the first terms in each binomial: (2x)(x)=2x2(2x)(x) = 2x^2.
  • Outer: Multiply the outer terms in the product: (2x)(3)=6x(2x)(3) = 6x.
  • Inner: Multiply the inner terms: (y)(x)=xy(y)(x) = xy.
  • Last: Multiply the last terms: (y)(3)=3y(y)(3) = 3y.

Next, we combine these results to form the expanded expression:

2x2+6x+xy+3y 2x^2 + 6x + xy + 3y .

Since terms 6x6x and xyxy are not like terms, they cannot be combined, resulting in the final expression:

2x2+xy+6x+3y 2x^2 + xy + 6x + 3y .

Upon reviewing the multiple-choice options, the correct answer is the expanded expression, choice 4: 2x2+xy+6x+3y 2x^2 + xy + 6x + 3y .

Answer

2x2+xy+6x+3y 2x^2+xy+6x+3y

Exercise #11

(a+4)(c+3)= (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 ac+3a+4c+12

Exercise #12

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

Video Solution

Step-by-Step Solution

To solve this problem, we'll perform a step-by-step expansion of the expression (x+13)(y+4)(x+13)(y+4) using the distributive property:

  • Step 1: Multiply the first terms (xy)=xy (x \cdot y) = xy .
  • Step 2: Multiply the outer terms (x4)=4x (x \cdot 4) = 4x .
  • Step 3: Multiply the inner terms (13y)=13y (13 \cdot y) = 13y .
  • Step 4: Multiply the last terms (134)=52 (13 \cdot 4) = 52 .

After completing these steps, combine the results:

xy+4x+13y+52 xy + 4x + 13y + 52

This is the final expanded form of the expression. By comparing with the given choices, the correct answer is:

xy+4x+13y+52 xy + 4x + 13y + 52

Therefore, the correct choice is option 3.

Answer

xy+4x+13y+52 xy+4x+13y+52

Exercise #13

Solve the following problem:

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

Video Solution

Step-by-Step Solution

Let's simplify the given expression, using the expanded distribution law in order to open the parentheses :

(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 of the parentheses is addition. We must remember that the sign preceding the term is an inseparable part of it. We'll also apply the rules of sign multiplication and thus we can present any expression inside of the parentheses. We'll open the parentheses 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)\\ Proceed to open 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 that we obtained in the last stage there are four different terms, this is due to the fact that 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 that we obtain 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,

Therefore the correct answer is answer A.

Answer

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

Exercise #14

Solve the following problem:

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

Video Solution

Step-by-Step Solution

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

(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 of the parentheses is addition. Remember that the sign preceding the term is an inseparable part of it. Apply the rules of sign multiplication and we can present any expression inside of the parentheses. We'll open the parentheses by using the above formula, 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 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 zero power yields 1) Apply the commutative property of addition and proceed to 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 remained 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 #15

Solve the following problem:

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

Video Solution

Step-by-Step Solution

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

(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 of the parentheses is addition. Remember that the sign preceding the term is an inseparable part of it. Apply the rules of sign multiplication so that we can present any expression in parentheses. We'll open the parentheses using the above formula, as an expression where addition operation exists between all terms. In this expression it's clear, all terms have a plus sign prefix.

Therefore we'll proceed directly to opening the parentheses as shown below:

:

(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, which we define 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 due to the fact that any number raised to the power of zero equals 1) Apply the commutative law of addition and 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