Question

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

Answer 1

Let f(x) = $x^3-6x^2+11x-6$
The constant term in f(x) is equal to -6 and factors of -6 are $\pm 1,\pm 2,\pm 3,\pm 6.$
Putting x = 1 in f(x), we have
$f\left(1\right)=1^3-6\times 1^2+11\times 1-6$
= 1 - 6 + 11 - 6 = 0
$\therefore$ (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,
$x^3-6x^2+11x-6=k(x-1)(x-2)(x-3)$
Putting x = 0 on both sides, we get
-6 = k (0 - 1) (0 - 2) (0 - 3)
$\Rightarrow$ -6 = -6 k $\Rightarrow$ 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, $x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$