Given:
Polynomial, f(x) = x3 - 6x2 + 11x - 6
Concept :
Factor theorem:
If f(x) is a polynomial of degree n ≥ 1 and 'a' is any real number, then, (x - a) is a factor of f(x), if f(a) = 0.
the factors of a polynomial using factor theorem:
1). If f(-a) = 0, then (x + c) is a factor of the polynomial f(x).
2). If f(d/c) = 0, then (cx - d) is a factor of the polynomial f(x).
3) If f(-d/c) = 0, then (cx + d) is a factor of the polynomial f(x).
4) If f(a) = 0 and f(b) = 0, then (x - a) and (x - b) are factors of the polynomial p(x).
Calculations:
Here f(x) is a polynomial of degree 3.
Using hit and trial method
Put x = 1 in f(x).
⇒ f(1) = 13 - 6(1)2 + 11(1) - 6
⇒ f(1) = 1 - 6 + 11 - 6
⇒ f(1) = 0
So, (x - 1) is a factor of f(x)
Similarly, For x = 2 and x = 3.
∴ (x - 1)(x - 2)(x - 3) is a factor of f(x)
Alternate method:
Using hit and trial method and division method
we get one root as 1, so (x - 1) is one of the factor of the given polynomial
and polynomial reduces to:
⇒ (x - 1)(x2 - 5x + 6)
⇒ (x - 1)(x2 - 3x - 2x + 6)
⇒ (x - 1)(x(x - 3) - 2(x - 3))
⇒ (x - 1)(x - 2)(x - 3)