using factor theorem factorize x^3- 6x^2 +3x
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Let f(x) = x3 – 6x2 + 3x + 10
Constant term = 10 Factors of 10 are ±1, ±2, ±5, ±10
Let x + 1 = 0
or x = -1 f(-1) = (-1)3 – 6(-1)2 + 3(-1) + 10
= 10 – 10 = 0 f(-1) = 0 Let x + 2 = 0
or x = -2 f(-2)
= (-2)3 – 6(-2)2 + 3(-2) + 10
= -8 – 24 – 6 + 10 = -28 f(-2) ≠ 0
Let x – 2 = 0
or x = 2 f(2) = (2)3 – 6(2)2 + 3(2) + 10
= 8 – 24 + 6 + 10
= 0 f(2) = 0
Let x – 5 = 0
or x = 5 f(5) = (5)3 – 6(5)2 + 3(5) + 10 = 1
25 – 150 + 15 + 10 = 0 f(5) = 0
Therefore, (x + 1), (x – 2) and (x - 5) are factors of f(x)
Hence f(x) = (x + 1) (x – 2) (x - 5)
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