akar akar bulat dari x^3-6x^2+11x-6=0 adalah

[tex] x^3-6x^2+11x-6=0 [/tex]

[tex] x^2 - 6x = \frac{6}{x} - 11 [/tex]

Let [tex] f(x) = x^2 - 6x [/tex] and [tex] g(x) =\frac{6}{x} - 11 [/tex]

f(x) and g(x) intersect at some points

We find that x can't be minus because if it is, f(x) => positive and g(x) => negative

We also see that [tex] x<6 [/tex] because if x>=6, f(x) >=0 and g(x) negative

We narrow it to [tex] 0 < x < 6 [/tex]

And so by trying 1 until 5 we get the integers are 1,2,3