SOLUTION: Find the real zeroes of f(x)= 3x^3-x^2-18x=6 DO NOT approximate-use rasicals as nececssary.

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Question 192286: Find the real zeroes of f(x)= 3x^3-x^2-18x=6 DO NOT approximate-use rasicals as nececssary.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First, let's factor 3x%5E3-x%5E2-18x%2B6


3x%5E3-x%5E2-18x%2B6 Start with the given expression


%283x%5E3-x%5E2%29%2B%28-18x%2B6%29 Group like terms


x%5E2%283x-1%29-6%283x-1%29 Factor out the GCF x%5E2 out of the first group. Factor out the GCF -6 out of the second group


%28x%5E2-6%29%283x-1%29 Since we have the common term 3x-1, we can combine like terms


So 3x%5E3-x%5E2-18x%2B6 factors to %28x%5E2-6%29%283x-1%29


This means that f%28x%29=+3x%5E3-x%5E2-18x%2B6 factors to f%28x%29=%28x%5E2-6%29%283x-1%29


To find the zeros, set the right side equal to zero:

%28x%5E2-6%29%283x-1%29=0


x%5E2-6=0 or 3x-1=0 Use the zero product property


x=sqrt%286%29, x=-sqrt%286%29 or x=1%2F3 Solve for "x" in each case


So the zeros are x=sqrt%286%29, x=-sqrt%286%29 or x=1%2F3