Question

factorise x cube - 6xsqaur + 11 x - 6using factor then theorem?

Answer 1

☺☺ HERE IS YOUR SOLUTION....

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let P(x) = x^3 - 6x^2 + 11x - 6

The constant term in p(x) is -6 and factor of -6 are +-1, +-1, +-3 and +-6.

putting x = 1 in P(x) we get,

P(1) = (1)^3 - 6(1)^2 + 11(1) - 6
= 1 - 6 +11 - 6
= 12 - 12
= 0

(x-1) is a factor of P(x).

similarly (x-2) and (x-3) are factor of P(x).

P(x) is a polynomial with 3 degree.

obtain factor are equal in no. of degree.

so, P(x) = k (x-1)(x-2)(x-3)

x^3 - 6x^2 +11x - 6 = k (x-1)(x-2)(x-3)

putting x=0 on both side we get,

-6 = k(0-1)(0-2)(0-3)
- 6 = - 6k
k = 1

Now put k=1 in P(x) = k (x-1)(x-2)(x-3)

P(x) = 1(x-1)(x-2)(x-3)

Hence,

x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3)

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☺ HOPE IT HELPS YOU...

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