Answer
(a) See explanations.
(b) 2 or 0 real positive zeros. 1 real negative zero. 0 or 2 complex zeros.
(c) See graph, $-0.931, 1.837,4.094$
Work Step by Step
(a) Given $f(x)=x^3-5x^2+2x+7$, we have $f(1)=(1)^3-5(1)^2+2(1)+7=5\gt0$ and $f(2)=(2)^3-5(2)^2+2(2)+7=-1\lt0$ . Based on the intermediate value theorem, function $f(x)$ has a zero between 1 and 2.
(b) Since $f(x)$ has 2 sign changes, based on the Descartes’ rule of signs, there may be 2 or 0 real positive zeros. For $f(-x)=-x^3-5x^2-2x+7$, there is only 1 sign change and there is 1 real negative zero. The total zeros is 3, thus there may be 0 or 2 complex zeros.
(c) See graph, we can find the zeros as $x=-0.931, 1.837,4.094$ (note that you can get more digits with a better calculator)
