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Gaurav Seth 4 years, 3 months ago
Let p(x) = x3 - 6x2 + 11x - 6
By trial, we find that
p(1) = (1)3 - 6(1)2 + 11(1) - 6 = 0
∴ By converse of factor theorem, (x - 1) is a factor of p(x).
Now, x3 - 6x2 + 11x - 6
= x2 (x - 1)- 5x (x - 1) + 6 (x - 1)
= (x - 1) (x2 - 5x + 6)
= (x - 1) {x2 - 2x - 3x + 6}
= (x - 1) {x(x - 2)-3 (x - 2)}
= (x - 1)(x - 2)(x - 3)
1Thank You