Using factor theorem, factorize the polynomial `x^3-6x^2+11 x-6` .

Explanation

Given, that the polynomial is `x³ - 6x² +11x - 6`
We have to factorize the polynomial.
Let `p(x) = x³ - 6x² +11x - 6`
The constant term of p(x) is `-6`.
Factors of `-6 = ±1, ±2, ±3, ±6`
Let us take `x = 1`
Substitute `x = 1` in `p(x)`,
`p(1) = (1)³ - 6(1)² +11(1) - 6`
`= 1 - 6 + 11 - 6`
`= 1 + 11 - 6 - 6`
`= 12 - 12`
`= 0`
So, `x - 1` is a factor of `x³ - 6x² + 11x - 6`.
Now splitting the `x²` and `x` terms,
`x³ - 6x² - 11x - 6 = x³ - x² - 5x² + 5x + 6x - 6`
Taking `(x - 1)` as a common factor,
`= x²(x - 1) - 5x(x - 1) + 6(x - 1)`
`= (x² - 5x + 6)(x - 1)`
On factoring `x² - 5x + 6` by splitting the middle term,
`= x² - 3x - 2x + 6`
`= x(x - 3) - 2(x - 3)`
`= (x - 2)(x - 3)`
Therefore, the factors of p(x) are `(x - 2)(x - 3)(x - 1)`