Find all irrational zeros of the function. Fiind exact values whenever possible.

x^3-6x^2+7x+4

Question

Find all irrational zeros of the function.  Fiind exact values whenever possible.

x^3-6x^2+7x+4

directrix · Answer

x^3-6x^2+7x+4=0 will have 3 zeros given that it is a cubic polynomial. 

The task is to find, if possible, one root of the polynomial. Then, the given cubic polynomial can be factored into a quadratic and a linear. And, the quadratic formula will crank out the remaining two roots.

directrix · Answer

Do you know the rational root theorem?

directrix · Answer

@santosrubi_5

anonymous · Answer

OK the function is generally upward going from lower left to upper right, the derivative has zeroes at x = 1/3 where f(x) = 152/27, and x = 11/3 where f(x) = -46/27. Now the y intercept for f(x) is 4, so here's a rough sketch of the behaviour :|dw:1360301254159:dw|plus a point of inflexion at x = 2. So that gives one negative root, and two positive roots either side of 11/3.

anonymous · Answer

In  fact$$f(4)=4^{3}-6\times 4^{3}+7\times 4 + 4= 64 - 96 + 28 + 4 = 0$$
so x = 4, and $$\frac{f(x)}{(x-4)}=\frac{x^3-6x^2+7x+4}{(x-4)}=x^2-2x-1$$that has solutions $$x= 1- \sqrt{2}$$and $$x=1+\sqrt{2}$$which is verified by substitution.