Prove function has exactly one root

f(x) = (x^3)-7(x^2)+25x+8

--------------------------------------…
My attempt:
I can easily prove by the IVT that an arbitrary interval [-1,1] proves i have AT LEAST one root. f(-1)=-25, f(1) = 27. So, f(-1) < 0 < f(1), proves at least one c within the interval.

Now i need to restrict the the root to no more than 1. I figure i could prove that the function is either always increasing or decreasing.

Since the derivtive is ->
f'(x) = 3x^2-14x+25

3x^2 = increasing
-14x = increasing/decreasing
25 = increasing

So since not all terms are always increasing( or decreasing), i can't prove that the roots are restricted to exactly one, even though a graphing calculator proves exactly 1 root.

The discriminant of the derivative is (-14)^2 - 4(3)(25) = 196-300 = -104. The negative value means the quadratic has no real roots, so is positive everywhere. (It has a value of +25 at x=0).