Factorise
³- 6
²+ 11x – 6
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1
According to Factor theorem, if (x - a) is a polynomial factor f(x), then f(a) = 0
Let f(x) = x^{3}-6 x^{2}+11 x-6f(x)=x
3
−6x
2
+11x−6
Let us check if (x - 1) is the factor of f(x),
Then,
f(1) = 1^{3}-6(1^{2})+11(1)-6=1-6+11-6=0f(1)=1
3
−6(1
2
)+11(1)−6=1−6+11−6=0
Therefore (x-1) is a factor of f(x)
Let us check for the other factors
Hence,
f(x)=(x-1)(x^{2}-5 x+6)f(x)=(x−1)(x
2
−5x+6)
x^{2}-5 x+6=x^{2}-2 x-3 x+6x
2
−5x+6=x
2
−2x−3x+6
=x(x-2)-3(x-2)=x(x−2)−3(x−2)
= (x - 2)(x - 3)=(x−2)(x−3)
f(x) = (x - 1)(x - 2)(x - 3)f(x)=(x−1)(x−2)(x−3)
Therefore, 1, 2, 3 are the factors of f(x)
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