Algebra Formulas For Class 11th

The term algebra is made up of alphabets and numbers. Numbers are always fixed, i.e their values are always known. Alphabets in algebra are used to denote the unknown quantities in the algebra formula. A combination of numbers, alphabets, factorials, matrices are used to form an algebraic equation or formula. This is significantly the methodology for algebra.

There are certain algebraic formulas that are important for the students to learn as it will help them to solve different algebraic equations. Only learning the algebraic formulas is not enough. The students must also understand the concepts behind the formula and learn to use them wisely.

In this article, we will provide a list of all the important algebraic formulas for class 11. The comprehensive list of algebraic formulas for class 11 will help the students to have a quick glance before the exams. Remember, only learning the formulas is not sufficient. You must also know how to use this formula to solve a given problem.

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List of Algebra Formulas For Class 11th

All the important algebraic expressions formulas for class 11th given below enable the students to easily search and learn them as it is listed by Vedantu in one single page. This list of algebraic expressions formulas for class 11th will help students not to miss any formula while studying for competitive or board exams.

We believe Algebra Formulas for class 11 enable the students to learn them and enhance confidence in solving algebraic problems. The list of all algebraic expressions formulas for class 11th  are given below.

• (x + θ)² = x² + θ² + 2xθ

• (x - θ)² = x² + θ² - 2xθ

• (x + θ)³ = x³ + θ³ + 3xθ(x + θ)

• (x - θ)³ = x³  + θ³  - 3xθ(x - θ)

• (x)²  - (θ)² = ( x + θ)(x - θ)

• (x)³  - (θ)³ = (x - θ)(x² + xθ + θ²)

• (x)³  + (θ)³ = (x + θ)(x² - xθ + θ²)

• (x)⁴  - (θ)⁴ = (x - θ) (x+ θ)(x² + θ²)

• (x)⁵  - (θ)⁵ = (x - θ) (x⁴ + x³θ + x²θ² + xθ³ +θ⁴)

• (x)⁵  + (θ)⁵ = (x + θ) (x⁴ - x³θ + x²θ² - xθ³ +θ⁴)

• ( x + y + θ)² =  x² + y² + θ² + 2xy +2yθ +2xθ

• ( x - y -θ)² =  x² + y² + θ² - 2xy + 2yθ - 2xθ

• x³ + y³ + θ³ - 3xθ = ( x + y +θ)( x² + y² + θ² - xy - yθ - xθ)

• (xm)( xn)= xm+n

• (xy)m = xmym

• (xm)n = xmn

• If n is even, then

xn - yn = (x - y) (xn-1 + xn-2 +  x n-3 y ² + …..+ x y n-2 + yn-1

• If n is odd, then

xn + yn = (x + y) (xn-1 - xn-2 +  x n-3 y ² - …..- xy n-2 + yn-1

Binomial Theorem Class 11 Formulas

The expansion of the binomial theorem for any positive integer n is derived by binomial theorem. The formulas of binomial theorem class 11 are used for expressing the powers of sums. The binomial theorem equation is given as 

(x + ϴ) n = nC0 an + nC1 an-1 ϴ + nC2 an-2 ϴ2 +…..+ nCn-1 x ϴn-1 + nCn ϴn

After simplifying the above equation,we get the following formulas of binomial theorem class 11

• The general term of an expression (x + ϴ) n is Tr + 1 = nCran-r ϴ r

• The general term of an expression (x - ϴ) n = (-1)r = nCran-r ϴ r

• The general term of an expression (1 + ϴ) n = nCr xr

• The general term of an expression (1 - ϴ) n =(-1)r = nCr xr

• In the expansion (x + ϴ) n , if n is even, then the middle term will be  (n/2 + 1)th term. If n is odd, then the middle terms are (n/2 + 1)th and (n + 1/2 + 1) th term.

• rth term from the end in (x + ϴ) n is equals to ( n + 2 – r) th term from the beginning.

Solved Examples

1. Expand (3p - 4q)³ by using the standard algebraic identities.

Solution:

(3p - 4q)³ is an algebraic identity where a = 3p and b = 4q. Accordingly, we have

(3p - 4q)³ = (3p)³  - (4q)³  -3 (3p)(4q)(3p - 4q) 

= 27p³  - 64q³  -108p²q + 11pq².

2. Factorize (p⁴ - 1) using the standard algebraic identities.

Solution: (p⁴ - 1) is an algebraic identity where a = p² and b = 1.

Accordingly, we have

(p⁴ - 1) = ((p²)². -1².) = (p² + 1)(p² - 1)

The factor (p² + 1) can be factored further using the similar identity where a = p and b = 1.

Accordingly,

(p⁴ - 1) = (p² + 1)((p)² -1²) = (p² + 1)(p - 1).

3. The total of the real values of y for which the middle term in the binomial expansion of (y³/y) + (3/y)⁸ equal to 5760 is?

Solution:

T5 =   8C4 × (y12/ 81) × (81/ y4) = 5670

⇒ 70y⁸ = 5670

⇒ y = ± √3

3. Find the coefficient of y9 in the expansion of ( 1+ y) (1+ y²) (1+ y³)...(1 + y100).

Solution:

y9 can be expressed in 8 ways.

i.e.  y9, y1+8 , y2+7 , y3+6 , y4+5  , y1+3 + 5, y2+3 + 4

Hence, the coefficient of y9 = 1 + 1+ 1 +……+ 8 times= 8

Quiz Time.

1. Find the Binomial coefficient of the 5th term of the expansion (x + y)⁸

1. 52

2. 70

3. 58

2.    (xp² - θ²) is a product of

1. (x² + θ²) (x² - θ²)

2.  (x² + θ²) (x² + θ²)

3.  (x² - θ²) (x² - θ²)

4. None of these