SOLUTION: How do I find the real zero of a polynomial function fx=-9(x-8)(x-1). I put it in my calculator and it solved the problem for -72. But I do not know the amount of real zeros or h

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: How do I find the real zero of a polynomial function fx=-9(x-8)(x-1). I put it in my calculator and it solved the problem for -72. But I do not know the amount of real zeros or h      Log On


   

Question 756798: How do I find the real zero of a polynomial function fx=-9(x-8)(x-1).
I put it in my calculator and it solved the problem for -72. But I do not know the amount of real zeros or how to find them if there are any.
Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How do I find the real zero of a polynomial function fx=-9(x-8)(x-1).
-9(x-8)(x-1) = 0
(x-8)(x-1) = 0
Either (x-8) = 0 --> x = 8
or (x-1) = 0 --> x = 1

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
fx=-9%28x-8%29%28x-1%29 .....to find the real zero of a polynomial function set f%28x%29=0
-9%28x-8%29%28x-1%29=0...since you have f%28x%29 completely factored, use zero product rule to find solutions
-9%28x-8%29%28x-1%29=0...since -9%3C%3E0, the product will be equal to zero if %28x-8%29=0 or %28x-1%29=0
if %28x-8%29=0 => highlight%28x=8%29
if %28x-1%29=0 => highlight%28x=1%29
so, you have two real zeros
let see them on a graph: