Multiplication of Algebraic Expressions

πŸ†Practice variables and algebraic expressions

Multiplication of Algebraic Expressions

We already know that when encountering expressions where a number is added to or subtracted from a variable, we cannot combine them directly.
However, when we see an expression where a number multiplies a variable, we can simplify it by applying the multiplication!

Multiplying algebraic expressions is the same as multiplying conventional numbers, and therefore, the rules we apply to these will also be applied to algebraic expressions.
For example: 7Γ—X+13Γ—Y=7X+13Y\green{ 7\times X}+\red{13\times Y}=\green{7X}+\red{13Y}

Multiplying algebraic expressions involves distributing each term in one expression across all terms in another, combining like terms where possible.
This includes basic products like monomials with monomials, binomials, and more complex polynomials.
In algebraic expressions that contain variables or parentheses, it is not necessary to write the multiplication sign.

Using the multiplication of algebraic expressions enables us to simplify and even solve equations. It also opens the door to more advanced algebraic techniques, such as the Distributive Property and the FOIL Method, which help manage complex expressions and build towards solving intricate problems in algebra.

7Γ—X+13Γ—Y=7X+13Y

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Test yourself on variables and algebraic expressions!

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\( 3x+4x+7+2=\text{?} \)

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Remember that the rules of multiplication for algebraic expressions are the same as the multiplication rules for numbers.

Here are some examples

  • 8Γ—X=8X 8\times X=8X
  • XΓ—Y=XY X\times Y=XY
  • βˆ’3Γ—(5+6)=βˆ’3(5+6) -3\times\left(5+6\right)=-3\left(5+6\right)
  • 8Γ—(Xβˆ’5)=8(Xβˆ’5) 8\times\left(X-5\right)=8\left(X-5\right)
  • (5βˆ’a)Γ—(3+4)=(5βˆ’a)(3+4) \left(5-a\right)\times\left(3+4\right)=(5-a)\left(3+4\right)
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How many exercises should I practice?

Since each student has a different learning pace, the answer to this question is individual for each one.
The important thing is that you are aware of your level, and know if you need to practice the formulas more.
Anyway, to memorize the basic formula, it is recommended to do 10 exercises of basic and medium level. Β 


Examples and exercises with solutions for multiplying algebraic expressions

Exercise #1

18xβˆ’7+4xβˆ’9βˆ’8x=? 18x-7+4x-9-8x=\text{?}

Video Solution

Step-by-Step Solution

To solve the exercise, we will reorder the numbers using the substitution property.

18xβˆ’8x+4xβˆ’7βˆ’9= 18x-8x+4x-7-9=

To continue, let's remember an important rule:

1. It is impossible to add or subtract numbers with variables.

That is, we cannot subtract 7 from 8X, for example...

We solve according to the order of arithmetic operations, from left to right:

18xβˆ’8x=10x 18x-8x=10x 10x+4x=14x 10x+4x=14x βˆ’7βˆ’9=βˆ’16 -7-9=-16 Remember, these two numbers cannot be added or subtracted, so the result is:

14xβˆ’16 14x-16

Answer

14xβˆ’16 14x-16

Exercise #2

7.3β‹…4a+2.3+8a=? 7.3\cdot4a+2.3+8a=\text{?}

Video Solution

Step-by-Step Solution

It is important to remember that when we have numbers and variables, it is impossible to add or subtract them from each other.

We group the elements:

 

7.3Γ—4a+2.3+8a= 7.3Γ—4a + 2.3 + 8a =

29.2a + 2.3 + 8a = 

37.2a+2.3 37.2a + 2.3

 

And in this exercise, this is the solution!

You can continue looking for the value of a.

But in this case, there is no need.

Answer

37.2a+2.3 37.2a+2.3

Exercise #3

8x24x+3x= \frac{8x^2}{4x}+3x=

Video Solution

Step-by-Step Solution

Let's break down the fraction's numerator into an expression:

8x2=4Γ—2Γ—xΓ—x 8x^2=4\times2\times x\times x

And now the expression will be:

4Γ—2Γ—xΓ—x4x+3x= \frac{4\times2\times x\times x}{4x}+3x=

Let's reduce and get:

2x+3x=5x 2x+3x=5x

Answer

5x 5x

Exercise #4

Simplifica la expresiΓ³n:

2x3β‹…x2βˆ’3xβ‹…x4+6xβ‹…x2βˆ’7x3β‹…5= 2x^3\cdot x^2-3x\cdot x^4+6x\cdot x^2-7x^3\cdot 5=

Video Solution

Step-by-Step Solution

We'll use the law of exponents for multiplication between terms with identical bases:

amβ‹…an=am+n a^m\cdot a^n=a^{m+n} We'll apply this law to the expression in the problem:

2x3β‹…x2βˆ’3xβ‹…x4+6xβ‹…x2βˆ’7x3β‹…5=2x3+2βˆ’3x1+4+6x1+2βˆ’35x3 2x^3\cdot x^2-3x\cdot x^4+6x\cdot x^2-7x^3\cdot 5=2x^{3+2}-3x^{1+4}+6x^{1+2}-35x^3 When we apply the above law to the first three terms from the left, while remembering that any number can always be considered as that number raised to the power of 1:

a=a1 a=a^1 And in the last term we performed the numerical multiplication,

We'll continue and simplify the expression we got in the last step:

2x3+2βˆ’3x1+4+6x1+2βˆ’35x3=2x5βˆ’3x5+6x3βˆ’35x3=βˆ’x5βˆ’29x3 2x^{3+2}-3x^{1+4}+6x^{1+2}-35x^3=2x^5-3x^5+6x^3-35x^3=-x^5-29x^3 Where in the first stage we simplified the expressions in the exponents of the terms in the expression and in the second stage we combined like terms,

Therefore the correct answer is answer A.

Answer

βˆ’x5βˆ’29x3 -x^5-29x^3

Exercise #5

9m3m2Γ—3m6= \frac{9m}{3m^2}\times\frac{3m}{6}=

Video Solution

Step-by-Step Solution

According to the laws of multiplication, we must first simplify everything into one exercise:

9mΓ—3m3m2Γ—6= \frac{9m\times3m}{3m^2\times6}=

We will simplify and get:

9m2m2Γ—6= \frac{9m^2}{m^2\times6}=

We will simplify and get:

96= \frac{9}{6}=

We will factor the expression into a multiplication:

3Γ—33Γ—2= \frac{3\times3}{3\times2}=

We will simplify and get:

32=1.5 \frac{3}{2}=1.5

Answer

0.5m 0.5m

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