SOLUTION: find zeros and multiplicities a) f(x)= x^3-12x^2-55x+150

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Question 579150: find zeros and multiplicities
a) f(x)= x^3-12x^2-55x+150
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=+x%5E3-12x%5E2-55x%2B150 is a polynomial
We know that if it has a rational zero, it will be a "fraction" whose denominator is a factor of the leading coefficient and whose numerator is a factor of the constant (or independent term, whatever you call it), with a plus or minus sign in front.
The leading coefficient (the coefficient of the term of highest degree) is that invisible 1 in front of x%5E3. So the denominator is a factor of 1, which could only be 1. Any rational zero will be an integer. It has to be a factor of 150. There are 12 positive factors of 150: 1, 2, 3, 5, 6, 10, 15, 25, 30,7 5, and 150. The same numbers with a minus sign in front are also possibilities. I tried 1, and it was not a zero, but then I tried 2, and 2 was a zero.
Since 2 is a zero of f%28x%29, %28x-2%29 is a factor of f%28x%29. f%28x%29 can be evenly divided by %28x-2%29.
I did the division and the result was x%5E2-10-75. So
f%28x%29=+x%5E3-12x%5E2-55x%2B150=%28x-2%29%28x%5E2-10-75%29
Factoring x%5E2-10-75, I found that x%5E2-10-75=%28x-15%29%28x%2B10%29
So f%28x%29=+%28x-2%29%28x-15%29%28x%2B10%29 and its zeros are x=2, x=15, and x=-10.