Question

Nonsense/No solution = Report​

Answer 1

Answer:

Zeroes of

$x^3 −6x^2 +11x−6$

By using rational theorem, the roots can be among the factors of

$\frac{6}{1}$

=6

Let us try x=1

$⇒(1)^3 −6(1)^2 +11(1)−6= 0\\ </p><p></p><p>∴(x−1) \: is \: a \: factor \: of \: x^3 −6x^2 +11x−6.</p><p>$

Now, using synthetic division method :

(attachment)

$So, \: the \: quotient \: =x^2 −5x+6</p><p>$

Now, using common factor theorem,

$⇒x^2 −5x+6=x^2 −2x−3x+6$

=x(x−2)−3(x−2)

=(x−2)(x−3)

∴ Zeroes of the polynomial =1,2,3

So, factor of the polynomial =(x−1)(x−2)(x−3).

Hence, the answer is (x−1)(x−2)(x−3).