How do you find the rest of the zeros given one of the zero c=1 and the function f(x)=x^3-6x^2+11x-6f(x)=x3−6x2+11x−6?

1 Answer
Shwetank Mauria
Aug 22, 2016

Other zeros are 22 and 33.

Explanation:

f(x)=x^3-6x^2+11x-6f(x)=x3−6x2+11x−6 has one of the zeros as 11, (x-1)(x−1) is factor of f(x)f(x).

Dividing f(x)=x^3-6x^2+11x-6f(x)=x3−6x2+11x−6 by (x-1)(x−1), we get x^2(x-1)-5x(x-1)+6(x-1)x2(x−1)−5x(x−1)+6(x−1) or

(x-1)(x^2-5x+6)(x−1)(x2−5x+6), which can be further factorized as

(x-1)(x^2-3x-2x+6)(x−1)(x2−3x−2x+6)

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

= (x-1)(x-2)(x-3)(x−1)(x−2)(x−3)

Hence, other zeros are 22 and 33.

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