Question

f(x) = x4 − x3 − 6x2
(a) Find all the real zeros of the polynomial function.
(b) Determine the multiplicity of each zero and the number of turning points of the graph of the function.
(2) The number of turning points?

Answer 1

factor GCF\[f(x) = x4 − x3 − 6x2=x^2(x^2-x-6)=\[x^2(x-3)(x+2)\]\]
a) let f(x)=0\[x^2(x-3)(x+2)=0\]\[x^2=0 \cup x-3=0 \cup x+2=0\]so the zeros along with their multiplicity are:

Zero Multiplicity
0 two
-2 one
3 one

A rough scetch of the graph would bethere would be three turning points (the max possible)