Factorise : x2 – 6x2 + 11x – 6.
Let p(x) = x2 – 6x2 + 11x – 6 Here, constant term of p(x) is -6 and factors of -6 are ± 1, ± 2, ± 3 and ± 6 By putting x = 1, we have p(1) = (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 -6 = 0 ∴ (x – 1) is a factor of p(x) By putting x = 2, we have p(2) = (2)3 – 6(2)2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 0 ∴ (x – 2) is a factor of p(x) By putting x = 3, we have p(3) = (3)3 – 6(3)2 + 11(3) – 6 = 27 – 54 + 33 – 6 = 0 ∴ (x – 3) is a factor of p(x) Since p(x) is a polynomial of degree 3, so it cannot have more than three linear factors. ∴ x3 – 6x2 + 11x – 6 = k (x – 1) (x – 2) (x – 3) By putting x = 0, we obtain 0 – 0 + 0 – 6 = k (-1) (-2) (3) -6 = -6k k = 1 Hence, x3 – 6x2 + 11x – 6 = (x – 1) (x – 2)(x – 3).
