SOLUTION: Find all the zeros of the function. x^3+3x^2-x+12

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Question 87279: Find all the zeros of the function.
x^3+3x^2-x+12

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
first off... find a number that makes the function zero. I am assuming that there is a nice number (ie an integer) since they have asked you to do this question... horrible number would make it virtually impossible to do.

Looking at the function: +x%5E3%2B3x%5E2-x%2B12+
to make this zero by using a positive number will not work --> the cube will get bigger and bigger and the only negative term is the x.
So we need to think about a -ve number since the cube of a negative number is still -ve.

Start with x=-1 and this is too small a number
x=-2 and x=-3 still too small

x=-4 gives:
+%28-4%29%5E3%2B3%28-4%29%5E2-%28-4%29%2B12+
+-64+%2B+3%2816%29+%2B+4+%2B+12+
+-64+%2B+48+%2B+16+
-64 + 64
0

so x=-4 is one root --> (x+4) is a factor.

You now need to divide the function by this to find the other factor and then that quadratic should factorise too, to give you nice answers like (x+4).


cheers
Jon.