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Let p(x) = x 3 - 6x 2 + 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, x 3 - 6x 2 + 11x - 6 = x 2 (x - 1)- 5x (x - 1) + 6 (x - 1) = (x - 1) (x 2 - 5x + 6) = (x - 1) {x 2 - 2x - 3x + 6} = (x - 1) {x(x - 2)-3 (x - 2)} = (x - 1)(x - 2)(x - 3)

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Polynomials

Short Answer Type

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2y³ + y² - 2y - 1

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128. Factorise: x³ - 6x² + 11x - 6.

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)