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प्रश्न
Factorise:
x3 – 6x2 + 11x – 6
उत्तर
Let p(x) = x3 – 6x2 + 11x – 6
Constant term of p(x) = – 6
Factors of – 6 are ±1, ±2, ±3, ±6.
By trial, we find that p(1) = 0, so (x – 1) is a factor of p(x) ...[∵ (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 – 6 = 0]
Now, we see that x3 – 6x2 + 11x = 6
= x3 – x2 – 5x2 + 5x + 6x – 6
= x2(x – 1) – 5x(x – 1) + 6(x – 1)
= (x – 1)(x2 – 5x + 6) ...[Taking (x – 1) common factor]
Now, (x2 – 5x + 6) = x2 – 3x – 2x + 6 ...[By splitting the middle term]
= x(x – 3) – 2(x – 2)
= (x – 3)(x – 2)
∴ x3 – 6x2 + 11x – 6 = (x – 1)(x – 2)(x – 3)
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