Factorize x2 – 81

An expression that is made up of variables and constants along with algebraic operations such as addition, subtraction, etc. these Expressions are made up of terms. these are the equations when the operations such as addition, subtraction, multiplication, division, etc. are operated upon any variable. A combination of terms by the operations such as addition, subtraction, multiplication, division, etc is termed as an algebraic expression (or) a variable expression. Examples: 2x + 4y – 7, 3x – 10, etc.

The above expressions are represented with the help of unknown variables, constants, and coefficients. The combination of these three terms is termed as an expression. unlike the algebraic equation, It has no sides or ‘equals to’ sign.

Types of Algebraic expression

There are majorly three types of algebraic expressions based on the number of terms present in them. They are monomial expressions, binomial expressions, and polynomial expressions. Let’s learn about their definitions in detail,

• Monomial Expression: one-term expression is termed a Monomial expression. Examples of monomial expressions include 4x⁴, 2xy, 2x, 8y, etc.
• Binomial Expression: An expression which is having two terms and unlike is termed as a binomial expression. Examples of binomial include 2xy + 8, xyz + x², etc.
• Polynomial Expression: An expression that has more than one term with non-negative integral exponents of a variable is termed a polynomial expression. Examples of polynomial expression include ax + by + ca, x³ + 4x + 2, etc.

Some Other Types of Expression

Apart from monomial, binomial, and polynomial types of expressions which are numeric expression and variable expression. Let’s take a look at their respective definitions,

• Numeric Expression: An expression that consists of only numbers and operations does not include any variable is termed a numeric expression. Some of the examples of numeric expressions are 21 + 5, 16 ÷ 2, etc.
• Variable Expression: An expression that contains variables along with numbers and operations to define an expression is termed a variable expression. Some examples of a variable expression include 5x + y, 4ab + 33, etc.

Some algebraic formulae

1. (a + b)² = a² + 2ab + b²
2. (a – b)² = a² – 2ab + b²
3. (a + b)(a – b) = a² – b²
4. (x + a)(x + b) = x² + x(a + b) + ab
5. (a + b)³ = a³ + b³ + 3ab(a + b)
6. (a – b)³ = a³ – b³ – 3ab(a – b)
7. a³ – b³ = (a – b)(a² + ab + b²)
8. a³ + b³ = (a + b)(a² – ab + b²)

Difference of squares 

It is nothing but the subtraction of the square of one number from another number. The difference of squares formula for two values x and y can be given as follows.

 x² – y² = (x + y)(x – y)

Where,

• x is the first variable
• y is the second variable

Factorize x² – 81.

Solution:

Use the difference of squares property which shows that, 

x² – y² = (x – y)(x + y)

So now, 

= x² – 81

= x² – 9²

=  (x + 9) (x – 9)

Similar Problems

Question 1: Use the “difference of squares” rule to factor the following expression: x² – 144.

Solution: 

Use the difference of squares property which shows that,

x² – y² = (x – y)(x + y)

So now,

= x² – 144

= x² – 12²

= (x + 12) (x – 12)

Question 2:  Find out the constant from the following algebraic expressions,

1. x³ + 3x² – 4
2. 55+ 2y⁵

Answer:

Constants are the terms that do not have any variable, therefore, in the first term -4 is the constant and in the second term 55 is the constant.

Question 3: Simplify (1 – (1/x))(1 – x).

Solution:

Given expression, (1 – (1/x))(1 – x)

By simplifying, 1 – (1/x),

= (x – 1)/x

So It can be written as,

= (1 – (1/x))(1 – x)

= [(x – 1)/ x] {(1 – x)}

= {x – x² -1 + x}/x

= {-x² + 2x – 1} / x

= -{x² – 2x + 1} / x

= -(x – 1)²/x

Question 4: Factor completely. then use the difference of square, If the polynomial is prime, state this. x² + 2xy + y² – 16?

Solution:

Given Expressions,

 x² + 2xy + y² – 16

By splitting,

= x² + 2xy + y² – 16

= (x + y)² – (4)² {x² + 2xy + y² =( x + y )²}

Now, x² – y² =  (x + y)(x – y)

= (x + y + 4) (x + y – 4)

Question 5: Simplify, 16x² – 25y² using the difference of squares? 

Solution:

Given, 16x² – 25y²

= 16x² – 25y²

=  (4x)² – (5y)²

Now, x² – y² =  (x + y)(x – y)

= (4x + 5y )(4x – 5y)