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

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) (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) (Show Source): You can put this solution on YOUR website!
.....to find the real zero of a polynomial function set
...since you have completely factored, use zero product rule to find solutions
...since , the product will be equal to zero if or
if =>
if =>
so, you have two real zeros
let see them on a graph:

RELATED QUESTIONS
I have never been more confused in my life... I tried this for the first one but I don't... (answered by funmath)
State the number of real zeros. Approximate each real zero to the nearest tenth.... (answered by richwmiller)
I am having extreme difficulty with this section of my homework. I have to find a... (answered by edjones)
f(x) = x^3 - 12x^2 + 36x - 32; 8 Given a polynomial function and a zero, find the... (answered by stanbon)
Hi, I am working on finding the zeros of a polynomial function and don't understand it. I (answered by ankor@dixie-net.com)
1. Is H(x) = 2x+1/x^2-9 a polynomial function and if so how 2. how do i find each zero... (answered by stanbon)
Write a polynomial function of minimum degree with real coefficents whose zeros include... (answered by jsmallt9)
Ok my problem is a graph and im supposed to find the quadractic function and put it in... (answered by solver91311,stanbon)
I have a problem:{{{ln(x)=5163}}}. I got the correct answer on a problem similar to this: (answered by Edwin McCravy,Alan3354)