SOLUTION: find the zeros of the polynomial funciton F(x)=x^3+x^2-42x

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Question 690722: find the zeros of the polynomial funciton F(x)=x^3+x^2-42x
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Factoring F%28x%29=x%5E3%2Bx%5E2-42x would help us find the zeros.
F%28x%29=x%5E3%2Bx%5E2-42x --> F%28x%29=x%28x%5E2%2Bx-42%29 would be a first step (taking out the common factor x).
Factoring further is very easy if you have practiced factoring.
If not, here is how you would do it:
To factor x%5E2%2Bx-42=%28x%2Ba%29%28x%2Bb%29 <--> x%5E2%2Bx-42=x%5E2%2B%28a%2Bb%29x%2Bab
we look for a pair of factors that multiply to give 42,
because we need to find numbers a and b such that ab=-42 and a%2Bb=1.
We look for the smaller factor, starting at 1 and going up until we get to the larger factor:
We can find four such pairs of factors:
42=1%2A42
42=2%2A21
42=3%2A14
(4 and 5 are not factors; they do not divide 42 evenly)
42=6%2A7 (42=7%2A6 is the same product and 7 is the larger factor here)
Now we consider that for the negative product ab=-42, one of those factors must be negative.
Since 42=6%2A7 and %28-6%29%2B7=1, -6 and 7 are our numbers.
x%5E2%2Bx-42=%28x%2B7%29%28x-6%29 and
F%28x%29=x%28x%2B7%29%28x-6%29 with zeros for x values that make
x=0 <--> highlight%28x=0%29,
x%2B7=0 <--> highlight%28x=-7%29, and
x-6=0 <--> highlight%28x=6%29.