Question

using factor theorem factorize x^3- 6x^2 +3x

Answer 1

Answer:

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)