SOLUTION: how to find all the zeros for f(x)= 3x^4-11x^3-x^2+19x+6

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Question 805709: how to find all the zeros for
f(x)= 3x^4-11x^3-x^2+19x+6
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=+3x%5E4-11x%5E3-x%5E2%2B19x%2B6
If that polynomial has any rational zeros,
they are fractions of the form M%2Fn or -m%2Fn
with m= a factor of 6 (the independent term},
and n= a factor of 3 (the leading coefficient).
That would suggest to try
-6, -3, -2, -1, -2%2F3, -1%2F3%29%29%29%2C%0D%0A%7B%7B%7B6, 3, 2, 1, 2%2F3, and 1%2F3%29%29%29.%0D%0AThe+easiest+to+try+are+%7B%7B%7B1 and -1 .
f%281%29=+3%2A1%5E4-11%2A1%5E3-1%5E2%2B19%2A1%2B6=3-11-1%2B19%2B6=16

That means x=-1 is a zero of f%28x%29 ,
and f%28x%29 is divisible by x-%28-1%29=x%2B1 .
We divide and find
g%28x%29=f%28x%29%2F%28x%2B1%29=3x%5E3-14x%5E3%2B13x%2B6 <---> f%28x%29=%28x%2B1%29%283x%5E3-14x%5E3%2B13x%2B6%29
Next easier to try would be 2 and -2 .
WE can try to calculate g%282%29 or try to divide g%282%29%2F%28x-2%29 ,
but either way, we find that
g%282%29=0 and g%282%29%2F%28x-2%29=3x%5E2-8x-3 <--->g%28x%29=%28x-2%29%283x%5E2-8x-3%29
3x%5E2-8x-3 is a quadratic polynomial that can be easily factored as
3x%5E2-8x-3=%283x%2B1%29%28x-3%29
(If we did not figure out a factoring, we could use the quadratic formula to find the zeros of the 3x%5E2-8x-3 anyway).
In sum,
f%28x%29=%28x%2B1%29%28x-2%29%283x%2B1%29%28x-3%29 so the zeros are the values that make each of those four binomial factors equal to zero:
highlight%28x=-1%29 ,
highlight%28x=2%29 ,
highlight%28x=-1%2F3%29 ,
and highlight%28x=3%29