Algebraic identities are algebraic equations that are true for all the variables present in them. These identities contain constants and variables on both sides of the equation. When we factorize a polynomial, we often use these algebraic identities. A polynomial function is an algebraic expression that contains more than one term. Examples of polynomials include binomials and trinomials.

Quadratic equations are also the form of polynomial functions because they contain 3 terms. The highest exponent in a quadratic function is 2 and its graph is a parabola, unlike linear equation. Linear equations are graphically represented by a straight line. You might have learned these algebraic identities in junior classes when studying basic algebra because the uses of these identities are widespread while solving equations.

You can use the substitution method to verify the algebraic identities. The substitution method means substituting any value of the variable on both sides of the algebraic identity. If you have correctly expanded or solved an example using algebraic identities, then any value of the variable will hold true for both the left and right hand side of the equation.

In this article, we will see some standard algebraic identities and their relevant examples.

 

Square of a Binomial

(a \pm b) ^ 2 = a ^ 2 \pm 2 \cdot a \cdot b +  b ^2

The above algebraic identity is also known as a quadratic identity because after expansion and setting the equation equal to zero, we get a quadratic formula. The sign with 2ab will depend upon the addition and subtraction sign between a and b.

Using the above algebraic identity, we will solve some examples. You will see that any value of the variable satisfied both sides of the equation.

Example 1

(x + 3) ^ 2

Use the algebraic identity (a \pm b) ^ 2 = a ^ 2 \pm 2 \cdot a \cdot b +  b ^2 to open up the parentheses of this question.

= x ^ 2 + 2 \cdot x \cdot 3 + 3 ^ 2

Since, there is a positive sign between x and 3, so we will use a positive sign with 2(x)(3).

= x ^ 2 + 6 x + 9

The answer x ^ 2 + 6 x + 9 is also a quadratic equation since the standard form of a quadratic equation is a x ^2 + bx + c. In this answer, the leading coefficient a = 1 , the middle coefficient b = 6 and the last constant term c = 9. The last term of the quadratics is a constant term in which no variable is involved.

Hence, (x + 3) ^ 2 = x ^ 2 + 6 x + 9. It means that all the variables hold true for both sides of this algebraic equation.

Example 2

(2x - 3) ^ 2

Use the algebraic identity (a \pm b) ^ 2 = a ^ 2 \pm 2 \cdot a \cdot b +  b ^2 to open up the brackets of this question.

=(2x) ^ 2 - 2 \cdot 2x \cdot 3 + 3 ^ 2

Since there is a negative sign between 2x and 3, so we will expand it by subtracting the middle term 2(2x)(3):

= 4x ^ 2 - 12 x + 9

Hence, (2x - 3) ^ 2 = 4x ^ 2 - 12 x + 9. It means that all the variables hold true for both sides of the equation.

Example 3

(3x - 6) ^ 2

Use the algebraic identity (a \pm b) ^ 2 = a ^ 2 \pm 2 \cdot a \cdot b +  b ^2 to solve this equality:

=(3x) ^ 2 - 2(3x)(6) + (6) ^ 2

Since there is a negative sign between 3x and 6, so we will use a negative sign with the middle term 2(3x)(6):

=9x ^2 - 36x + 36

Hence, (3x - 6) ^ 2 = 9x ^2 - 36x + 36, which means that all the variables hold true for both sides of this equality.

Difference of Squares

(a + b) · (a − b) = a ^ 2 - b^2

It means that the multiplication of the terms (a + b) and (a - b ) will give us a ^2 - b ^2. Let us solve some examples related to this identity:

Example 1

(2x + 5) \cdot (2x - 5)

Use the algebraic identity (a + b) · (a − b) = a ^ 2 - b^2 to solve this example:

= (2x) ^ 2 − 5 ^ 2

= 4x ^ 2 - 25

Hence, (2x + 5) \cdot (2x - 5) = 4x ^ 2 - 25.

Example 2

(4x + 7) \cdot (4x - 7)

Use the algebraic identity (a + b) · (a − b) = a ^ 2 - b^2 to solve this example:

=(4x) ^ 2 - 7 ^ 2

=16x ^2 - 49

Hence, (4x + 7) \cdot (4x - 7) = 16x ^2 - 49.

Cube of a Binomial

(a \pm b) ^ 3 = a ^ 3 \pm 3 \cdot a ^2 \cdot b + 3 \cdot a \cdot b ^ 2 \pm b ^ 3

Example 1

(x + 3) ^ 3

Use the algebraic identity (a \pm b) ^ 3 = a ^ 3 \pm 3 \cdot a ^2 \cdot b + 3 \cdot a \cdot b ^ 2 \pm b ^ 3 to solve this example:

(x + 3) ^ 3= x ^ 3 + 3 \cdot x ^ 2 \cdot 3 + 3 \cdot x \cdot 3^ 2 + 3^ 3

Since there is a positive sign between x and 3, so we will use a positive sign with 3a ^2 b and b ^3 while solving the above expression.

= x ^ 3 + 9x ^ 2 + 27 x + 27

Hence, (x + 3) ^ 3 = x ^ 3 + 9x ^ 2 + 27 x + 27.

 

Example 2

(2x - 3) ^ 3

Use the algebraic identity (a \pm b) ^ 3 = a ^ 3 \pm 3 \cdot a ^2 \cdot b + 3 \cdot a \cdot b ^ 2 \pm b ^ 3 to solve this example:

 

(2x - 3) ^ 3 = (2x) ^ 3 - 3 \cdot (2x) ^ 2 \cdot 3 + 3 \cdot 2x \cdot  3 ^ 2 - 3 ^ 3

Since there is a negative sign between 2x and 3, hence we will use a negative sign with 3a ^2 b and b ^3.

= 8x  ^ 3 - 36x ^ 2 + 54x - 27

Hence,  (2x - 3) ^ 3 = 8x  ^ 3 - 36x ^ 2 + 54x - 27.

 

Square of a Trinomial

(a + b + c) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 \cdot a \cdot b + 2 \cdot a \cdot c + 2 \cdot b \cdot c

Example 1

(x ^ 2 - x + 1) ^ 2

Use the algebraic identity (a + b + c) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 \cdot a \cdot b + 2 \cdot a \cdot c + 2 \cdot b \cdot c to solve this example:

=(x ^ 2) ^ 2 + (-x) ^ 2 + 1 ^ 2 + 2 \cdot x ^ 2 \cdot  (-x) + 2 x ^ 2 \cdot 1 + 2 \cdot (-x) \cdot 1

= x^4  + x^2 + 1 - 2x ^ 3 + 2x^2 - 2x

=x^ 4 - 2x ^ 3 + 3x^2 - 2x + 1

Hence, (x ^ 2 - x + 1) ^ 2 = x^ 4 - 2x ^ 3 + 3x^2 - 2x + 1.

Example 2

( 2x ^ 2 + 2x + 3) ^ 2

Use the algebraic identity (a + b + c) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 \cdot a \cdot b + 2 \cdot a \cdot c + 2 \cdot b \cdot c to solve this example:

= (2x ^ 2) ^2 + (2x) ^ 2 + (3) ^ 2 + 2\cdot (2x ^ 2) \cdot (2x) + 2 \cdot (2x) \cdot (3) + 2 (2x ^2) (3)

= 4x ^ 4 + 4x ^2 + 9 + 16x ^ 3 + 12x +12 x ^ 2

Hence, ( 2x ^ 2 + 2x + 3) ^ 2 = 4x ^ 4 + 4x ^2 + 9 + 16x ^ 3 + 12x +12 x ^ 2.

Sum of Cubes

a ^ 3 + b ^ 3 = (a + b) \cdot (a ^ 2 - ab + b ^ 2)

It means that multiplying ( a + b ) and (a ^ 2 - ab + b^2) will give an expression a ^3 + b ^3. Lets us consider the following examples related to this algebraic identity.

Example 1

8x ^ 3 + 27

Use the algebraic identity a ^ 3 + b ^ 3 = (a + b) \cdot (a ^ 2 - ab + b ^ 2) to solve this problem:

=(2x + 3) (4x ^ 2 - 6x + 9)

Hence, 8x ^ 3 + 27 = (2x + 3) (4x ^ 2 - 6x + 9).

Example 2

27 x ^ 3 + 8

Use the algebraic identity a ^ 3 + b ^ 3 = (a + b) \cdot (a ^ 2 - ab + b ^ 2) to solve this problem:

= (3x + 2) ( 3x ^ 2 - 6x + 4)

Hence, 27 x ^ 3 + 8 = (3x + 2) ( 3x ^ 2 - 6x + 4).

Difference of Cubes

a ^ 3 - b^3 = (a - b) \cdot (a^2 + ab + b^ 2)

Consider the following examples related to this algebraic identity.

Example 1

8x ^ 3 - 27

Since 8x ^3 and 27 are perfect cubes, therefore we will use the algebraic identity a ^ 3 - b^3 = (a - b) \cdot (a^2 + ab + b^ 2) to solve this problem:

= (2x - 3) (4x ^2 + 6x + 9)

Hence, 8x ^ 3 - 27 = (2x - 3) (4x ^2 + 6x + 9).

Example 2

125 y ^3 - 8

Since 125 y ^3 and 8 are perfect cubes, therefore we will use the algebraic identity a ^ 3 - b^3 = (a - b) \cdot (a^2 + ab + b^ 2) to solve this problem:

=(5y - 2) (25y ^2 + 10y + 4)

Hence, 125 y ^3 - 8 = (5y - 2) (25y ^2 + 10y + 4).

Special Algebraic Identity

(x + a) (x + b) = x ^ 2 + (a + b) x + ab

It means that whenever the question of the form (x + a) (x + b) will be multiplied together we will get x ^2 + (a + b ) x + ab. Consider the following examples related to this algebraic identity:

Example 1

(x + 2) (x + 3)

Use the algebraic identity (x + a) (x + b) = x ^ 2 + (a + b) x + ab to solve this problem:

=x ^ 2 + (2 + 3)<strong> \cdot </strong>x + 2<strong> \cdot </strong>3

= x ^ 2 + 5x + 6

Hence, (x + 2) (x + 3) = x ^ 2 + 5x + 6.

Example 2

(x + 7) (x + 4)

To simplify, use the algebraic identity (x + a) (x + b) = x ^ 2 + (a + b) x + ab to solve this problem:

= x ^ 2 +(7 + 4) x + 7 \cdot 4

= x ^2 + 11x + 28

Hence, (x + 7) (x + 4) = x ^2 + 11x + 28.

Factorization with Algebraic Identities

So far, we have seen the algebraic identities and their respective examples. Now, we will see how to solve equations by expanding or factoring the algebraic expressions using algebraic identities.

Example 1

Factorize 4x ^ 4 - 25 using standard algebraic identities.

Solution

In the above example, a = 4x ^4 and b = 25.

We will use a ^2 - b = (a - b ) ( a + b) to factorize the expression in this problem. Since, a = 2x ^2 and b =5, so we will write the expression as follows in the factored form:

= (2x ^2 + 5) (2x ^2 - 5)

Example 2

Expand (5z - 6y) ^ 3 using a standard algebraic identity.

Solution

Use (a \pm b) ^ 3 = a ^ 3 \pm 3 \cdot a ^2 \cdot b + 3 \cdot a \cdot b ^ 2 \pm b ^ 3 to expand the above question:

In this example, a = 5z and b = 6y. Hence, using the above formula, we will write this expression as:

= (5z) ^ 3 - 3 (5z) ^2 (6y) + 3 (5z) (6y) ^2 - (6y)^3

Since there was a negative sign between 5z and 6y, hence we will subtract -3 ab^2 while expanding the formula:

= 125 z ^3 - 450 z^2 y + 540zy^2 -216 y ^3

Example 3

Find the product of (y + 4) (y + 8) using standard algebraic identity.

Solution

For simplifying, use the algebraic identity (x + a) (x + b) = x ^ 2 + (a + b) x + ab to solve this problem:

=y ^2 + (4 + 8) y + (4)(8)

= y ^2 +12y +32

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

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