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**Factorisation**

If g(x) and h(x)** **are two polynomials whose product is p(x).** **This can be written as p(x) = g(x) . h(x). g(x) and h(x) are called the factors of the polynomial p(x).

The process of resolving a given polynomial into factors is called factorisation.** **A non-zero constant is a factor of every polynomial.

**Algebraic Identities**

Polynomials can be factorised using algebraic identities.

A polynomial of degree two is called a quadratic polynomial. The identities used to factorise the quadratic polynomials are:

- (a + b)
^{2}= a^{2}+ 2ab + b^{2} - (a – b)
^{2}= a^{2}– 2ab + b^{2} - a
^{2}– b^{2}= (a + b)(a – b) - (x + a)(x + b) = x
^{2}+ (a + b)x + ab - (a + b + c)
^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca

A polynomial of degree three is called a cubic polynomial. The algebraic identities used in factorising a cubic polynomial are:

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

Given x
^{3}+ 1/x
^{3} =110

Recall that (x+1/x)
^{3} = x
^{3}+ 1/x
^{3} + 3(x+1/x)...

give the question properly

Now subtract 2 from both the sides, we get

Cubing on both the sides we get

Given x + y + z = 8 and xy + yz + zx = 20

Consider, x + y + z = 8

Squaring on both sides, we get

(x + y + z)
^{2}...

Given 4
^{x} - 4
^{x - 1} = 24

? 4
^{x} ? (4
^{x}/ 4
^{1}) = 24 [Since a
^{m - n}...