# Algebraic Formula For Class 11th:

Either you know it or not, an amazing fact about algebra for us that there are two types of algebra one is elementary and the other one is Abstract Algebra also known as Modern Algebra. And the thing of matter is that we are reading only elementary algebra which is basics of algebra. Whereas the other part Abstract algebra is studied by algebraists who are professionals or research fellow of this category in Mathematics.

In class 11th you are dealing with complex number. There are several laws for addition, subtraction, multiplication, division of complex number. This subsection of complex number is algebra of complex number.

There it’s also stated about Fundamental theorem of algebra that “A polynomial equation has at least one root.” Thus it’s also given that “A polynomial equation of degree n has n roots.”

In the chapter of linear inequalities you have to go through “Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation”. It has two rules:

1. Equal numbers may be added to (or subtracted from) both sides of an equation.
2. Both sides of an equation may be multiplied (or divided) by the same non-zero.

## We have listed all the necessary formula and rules related to algebra from class 11th .

 Algebra Rules For Class 11 Distributive Property a(b+c) = a×b+a×c Commutative Property of Addition a+b = b+a Commutative Property of Multiplication a×b = b×a Associative Property of Addition a+(b+c) = (a+b)+c Associative Property of Multiplication a×(b×c) = (a×b)×c Additive Identity Property a+0 = a Multiplicative Identity Property a×1 = a Additive Inverse Property a+(−a) = 0 Multiplicative Inverse Property a(1/a) = 1 Zero Property of Multiplication a×(0) = 0

## For complex number related problems

• Let z1 = a + ib and z2= c + id. Then

(i) z1+ z2= (a + c) + i(b + d)

(ii) z1.z2= (ac – bd) + i(ad + bc)

• For any integer k, i4k= 1, i4k+ 1 = i, i4k+2 = (– 1), i4k + 3 = (– i)
• The conjugate of the complex number z = a + ib, denoted by z , is given by z = a – ib.

## Revise the formula which you studied earlier:

 (a+b)2 = a2+2ab+b2 (a−b)2 = a2−2ab+b2 (a+b)(a–b) = a2–b2 (x+a)(x+b) = x2+(a+b)x+ab (x+a)(x–b) = x2+(a–b)x–ab (x–a)(x+b) = x2+(b–a)x–ab (x–a)(x–b) = x2–(a+b)x+ab (a+b)3 = a3+b3+3ab(a+b) (a–b)3 = a3–b3–3ab(a–b) (x+y+z)2 = x2+y2+z2+2xy+2yz+2xz (x+y–z)2 = x2+y2+z2+2xy–2yz–2xz (x–y+z)2 = x2+y2+z2–2xy–2yz+2xz (x–y–z)2 = x2+y2+z2–2xy+2yz–2xz x3+y3+z3–3xyz = (x+y+z)(x2+y2+z2–xy–yz−xz) x2+y2 = 1/2[(x+y)2+(x–y)2] (x+a)(x+b)(x+c) = x3+(a+b+c)x2+(ab+bc+ca)x+abc x3+y3 = (x+y)(x2–xy+y2) x3–y3 = (x–y)(x2+xy+y2) x2+y2+z2−xy–yz–zx = 1/2[(x−y)2+(y−z)2+(z−x)2]