SOLUTION: Find all real zeros: {{{ f(x)=4x^4-8x^3-19x^2+23x-6 }}} Thanks for the help!

Algebra ->  Rational-functions -> SOLUTION: Find all real zeros: {{{ f(x)=4x^4-8x^3-19x^2+23x-6 }}} Thanks for the help!      Log On


   

Question 834474: Find all real zeros:
+f%28x%29=4x%5E4-8x%5E3-19x%5E2%2B23x-6+
Thanks for the help!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29=red%284%29x%5E4-8x%5E3-19x%5E2%2B23x-green%286%29
The possible rational zeros can be written as n%2Fd or -n%2Fd where n is a factor of green%286%29 and d is a factor of red%284%29.
They are:
-6,-3,-2,-1,1,2,3,6,
-3/2,-3/4,-1/2,-1/4,1/4,1/2,3/4,3/2.
Trying the easiest ones first, we find two zeros:
f%281%29=4-8-19%2B23-6=-6
f%28-1%29=4%2B8-19-23-6=-36
f%282%29=64-64-76%2B46-6=-36
f%28-2%29=64%2B64-76-46-6=0
f%283%29=324-216-171%2B69-6=0
Since f%28-2%29=0 and f%283%29=0 ,
%28x-%28-2%29%29=%28x%2B2%29 and %28x-3%29 are factors of f%28x%29 .
So, we can divide f%28x%29 by %28x%2B2%29 and by %28x-3%29 one after the other,
or we can divide f%28x%29 by %28x%2B2%29%28x-3%29=x%5E2-x-6 .
Either way, the final quotient is 4x%5E2-4x%2B1=%282x-1%29%5E2 ,
which has x=1%2F2 as a double zero.
So f%28x%29=%28x%2B2%29%28x-3%29%282x-1%29%5E2 , and the zeros of f%28x%29 are
x=-2, x=3, and x=1%2F2 .