Let f(x) = x3−6x2+11x−6
The constant term in f(x) is equal to -6 and factors of -6 are ±1,±2,±3,±6.
Putting x = 1 in f(x), we have
f(1)=13−6×12+11×1−6
= 1 - 6 + 11 - 6 = 0
∴ (x - 1) is a factor of f(x)
Similarly, x - 2 and x - 3 are factors of f(x)
Since f(x) is a polynomial of degree 3. So, it can not have more than three linear factors.
Let f(x) = k (x - 1) (x - 2) (x - 3). Then,
x3−6x2+11x−6=k(x−1)(x−2)(x−3)
Putting x = 0 on both sides, we get
-6 = k (0 - 1) (0 - 2) (0 - 3)
⇒ -6 = -6 k ⇒ k = 1
Putting k = 1 in f(x) = k (x - 1) (x - 2) (x - 3), we get
f(x) = (x - 1) (x - 2) (x - 3)
Hence, x3−6x2+11x−6=(x−1)(x−2)(x−3)