Question

plz tell summary of ch algebraic expression and identities class 8

plz answer quick

Answer 1

Identity I: (a + b)2 = a2 + 2ab + b2

Identity II: (a – b)2 = a2 – 2ab + b2

Identity III: a2 – b2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab

Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)

Identity VIII: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

A rational number is any number that can be written as abab where aa and bb are integers and b≠0b≠0.

The following are rational numbers:

Fractions with both numerator and denominator as integers

Integers

Decimal numbers that terminate

Decimal numbers that recur (repeat)

Irrational numbers are numbers that cannot be written as a fraction with the numerator and denominator as integers.

If the nthnth root of a number cannot be simplified to a rational number, it is called a surd.

If aa and bb are positive whole numbers, and a<ba<b, then a√n<b√nan<bn.

A binomial is an expression with two terms.

The product of two identical binomials is known as the square of the binomial.

We get the difference of two squares when we multiply (ax+b)(ax−b)(ax+b)(ax−b)

Factorising is the opposite process of expanding the brackets.

The product of a binomial and a trinomial is:

(A+B)(C+D+E)=A(C+D+E)+B(C+D+E)(A+B)(C+D+E)=A(C+D+E)+B(C+D+E)

Taking out a common factor is the basic factorisation method.

We often need to use grouping to factorise polynomials.

To factorise a quadratic we find the two binomials that were multiplied together to give the quadratic.

The sum of two cubes can be factorised as:

x3+y3=(x+y)(x2−xy+y2)x3+y3=(x+y)(x2−xy+y2)

The difference of two cubes can be factorised as:

x3−y3=(x−y)(x2+xy+y2)x3−y3=(x−y)(x2+xy+y2)

We can simplify fractions by incorporating the methods we have learnt to factorise expressions.

Only factors can be cancelled out in fractions, never terms.

To add or subtract fractions, the denominators of all the fractions must be the same.

Answer 2

Hey dear here is your answer:-

In this chapter nothing is there you have to just learn some identities and you should know how to do factories

So the identies of this chapter are:-
* (a+b)^2= a^2 + b^2 + 2ab

Example= (2x +5)^2= 2x^2 + 5^2 + 2×2x×5
= 4x^2 + 25 + 20x
some questions of this identity can come like:-
(101)^2
=(100+1)^2= 100^2 + 1^2 + 2×100×1
= 10000+1+200
= 10201
* (a-b)^2 = a^2 + b^2 - 2ab

Example= (x-7)^2= x^2 + 7^2 - 2×x×7
= x^2 + 49 - 14x
other questions of this identity are also there like :-
(48)^2
= (50-2)^2 = 50^2 + 2^2 + 2×50×2
= 2500 + 4 + 200
= 2704

* (a+b) (a-b) = a^2 - b^2

Example= (2x+7y) (2x-7y) = 2x^2 - 7y^2
= 4x^2 - 49y^2

Some questions of this identity also come like:-
(81)^2 -(19)^2
= (81+19)(81-19)
= 100×62
=6200

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

Example= (x+5)(x+4) = x^2 + (5+4)x + 5×4
= x^2 + 9x + 20

* (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Example= (x -2y+3z)^2 = x^2 + 2y^2 + 3z^2 +
2×x×(-2y) + 2×(-2y)×3z + 2×3z×x
= x^2 + 4y^2 + 9z^2 - 4xy - 12yz + 6zx

Now here are some factorise questions:-
=> x^4 - y^4
= (x^2)^2 - (y^2)^2
= (x^2 + y^2) (x^2 + y^2)
=[(x)^2 - (y)^2] (x^2 + y^2)
= (x-y)(x+y)(x^2+y^2) {using a^2-b^2= (a+b)(a-b)}

=> 9y^2 - 16
= (3y)^2 - (4)^2
= (3y + 4)(3y - 4)【using a^2 - b^2 = (a+b)(a-b)】

=> x^2 -11x +30
= x^2 - 6x - 5x + 30
= x(x-6)-5(x-6)
= (x-5)(x-6)

This all about the chapter
Hope my answer helps you☺
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