# Standard Algebraic Identities | Class 9 Maths

An **algebraic identity** is an equality that holds for any value of its variables. They are generally used in the factorization of polynomials or simplification of algebraic calculations. Many everyday situations can be formulated in the form of mathematical equations.** Identities** help us by giving away to factorize them.

While operating a mobile phone. There are millions of chips under the screen of your phone, and they do their job perfectly so that you rarely feel any glitches. Thousands of engineers brainstorming on complex equations have made this possible. They deal with much bigger equations and problems. Algebraic identities then become small tools to help solve those problems. So maybe they don’t come in front of us directly, but somewhere behind the scenes, someone has definitely used them and making life easier for us.

## Types of Polynomials

Based on the number of terms present in an Algebraic Expression, For example, there can be 1 term, 2 terms, 3 terms, 4 terms, and so on, they are classified into different categories. The terms are separated by either a positive or a negative sign:

Types of Polynomials | Definition | Example |

Monomials | A Polynomial containing only one term | 10, 2x 4abc |

Binomials | A Polynomial containing two terms | x + y, 3p x |

Trinomials | A Polynomial containing three terms | p + q + r, x |

Quadrinomial | A Polynomial containing four terms | p + q + r + s, m |

Quintinomial | A Polynomial containing five terms | 5x^{3}y + 6x^{2 }– 9xy + 8y – 7 |

Note:

- Quadrinomials and Quintinomials are not very famous terms and are often referred as Polynomials.
- The most common Standard Identities are from Binomials and Trinomials.

**Standard Identities of Algebraic Expression**

All standard Algebraic Identities are derived from the **Binomial Theorem**. There are a number of algebraic identities but a few are standard that are listed below:

Standard Identities |

(a + b)^{2} = a^{2} + b^{2} + 2ab |

(a – b)^{2} = a^{2} + b^{2} -2ab |

a^{2} – b^{2} = (a + b)(a – b) |

(ax + b)(ax – b) = ax^{2} – b^{2} |

(x + a) (x + b) = x^{2} + (a + b)x + ab |

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) |

a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca) |

Let’s look at the different applications of these identities.

## Applications of Identities

**Identity 1: (a + b)**^{2} = a^{2} + b^{2} + 2ab

^{2}= a

^{2}+ b

^{2}+ 2ab

**Question: Find the value of (x + 6)(x + 6) using algebraic identities when x = 3. **

**Solution:**

(x+6)(x+6) can be re-written as (x + 6)

^{2}.It can be rewritten in this form,

(a + b)^{2}= a^{2}+ b^{2}+ 2ab.(x + 6)

^{2}= x^{2}+ 6^{2}+ 2(6x)= x

^{2}+ 36 + 12xGiven, x = 3.

(x + 6)

^{2}= 3^{2}+ 36 + 12(3)= 9 + 36 + 36

= 81

**Identity 2: (a – b)**^{2} = a^{2} + b^{2} -2ab

^{2}= a

^{2}+ b

^{2}-2ab

**Question: Expand (5x – 3y) ^{2}.**

**Solution:**

This is similar to expanding

(a – b)^{2}= a^{2}+ b^{2}– 2ab.where a = 5x and b = 3y,

So (5x – 3y)

^{2}= (5x)^{2}+ (3y)^{2}– 2(5x)(3y)= 25x

^{2}+ 9y^{2}– 30xy

**Identity 3: a**^{2} – b^{2} = (a + b)(a – b)

^{2}– b

^{2}= (a + b)(a – b)

**Question:** **Factorize (x ^{6} – 1) using the identities mentioned above. **

**Solution:**

(x

^{6}– 1) can be written as (x^{3})^{2}– 1^{2}.This resembles the identity a

^{2}– b^{2}= (a + b)(a – b).where a = x

^{3}, and b = 1.So, x

^{6}– 1 = (x^{3})^{2}– 1 = (x^{3}+ 1) (x^{3}– 1).

## Solved Examples on Standard Algebraic Identities

**Question 1:** **If a + b = 12 and ab = 35, what is a ^{4} + b^{4}? **

**Solution:**

a

^{4}+ b^{4}can be written as (a^{2})^{2}+ (b^{2})^{2},And we know, (x + y)

^{2}= x^{2}+ y^{2}+ 2xy⇒ x

^{2}+ y^{2}= (x + y)^{2}-2xySo, in this case, x = a

^{2}, y = b^{2};a

^{4}+ b^{4}= (a^{2}+ b^{2})^{2}– 2(a^{2})(b^{2})⇒ ((a+b)

^{2}– 2ab)^{2}– 2(a^{2})(b^{2})⇒ ((12)

^{2}– 2(35))^{2}– 2(35)^{2}⇒ 5475 – 2450

⇒ 3026

**Question 2: The identity 4(z + 7)(2z – 1) = Az ^{2 }+ Bz + C holds for all real values of z. What is A + B + C?**

**Solution:**

Multiplying out the left side of the identity, we have

4(x + 7)(2x − 1) = 8x

^{2}+ 52x − 28.This expression must be equal to the right-hand side of the identity, implying

8x

^{2 }+ 52x – 28 = Ax^{2 }+ Bx + C,So now comparing both sides of the equation.

A = 8, B = 52 ad C – 28.

A + B + C = 8 + 52 – 28 = 32

**Question 3: If a + b + c = 6, a ^{2 }+ b^{2 }+ c^{2 }= 14 and ab + bc + ca = 11, what is a^{3 }+ b^{3 }+ c^{3 }– 3abc?**

**Solution:**

We know this identity,

a

^{3}+ b^{3}+ c^{3}– 3abc = (a + b + c)(a^{2}+ b^{2}+ c^{2}– ab – bc – ca)Substituting the given values,

a

^{3}+ b^{3}+ c^{3}-3abc = (6)(14 -11)⇒ (6)(3) = 18

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