SOLUTION: Find a rational zero of the polynomial function and use it to find all the zeros of the function. f(x) = 2x^3 - x^2 - 12x + 6

Algebra ->  Rational-functions -> SOLUTION: Find a rational zero of the polynomial function and use it to find all the zeros of the function. f(x) = 2x^3 - x^2 - 12x + 6      Log On


   

Question 166375: Find a rational zero of the polynomial function and use it to find all the zeros of the function.
f(x) = 2x^3 - x^2 - 12x + 6
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let's graph the function.
+graph%28+300%2C+300%2C+-5%2C+5%2C+-5%2C+5%2C+2x%5E3+-+x%5E2+-+12x+%2B+6%29+
Looks like x=0.5 is a zero.
2x%5E3+-+x%5E2+-+12x+%2B+6=2%280.5%29%5E3-%280.5%29%5E2-12%280.5%29%2B6
2x%5E3+-+x%5E2+-+12x+%2B+6=%280.5%29%5E2-%280.5%29%5E2-6%2B6
2x%5E3+-+x%5E2+-+12x+%2B+6=0
Even though x=2.5 and x=-2.5 look like zeros, they don't solve.
Let's use synthetic divsion to factor out (x-0.5) and get the quadratic equation that's left over.
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Showing synthetic division is a little difficult.
The first part will show the factor (left hand column), then the factor times the divisor,
then the next line will show the subtraction.
Repeat.
Hopefully it makes sense.
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...........___________________
(x-0.5)|2x^3-x^2-12x+6
2x^2:....2x^3-x^2
.................................-12x+6
-12:..........................-12x+6


%282x%5E3-x%5E2-12x%2B6%29%2F%28x-0.5%29=2x%5E2-12
%282x%5E3-x%5E2-12x%2B6%29=%28x-0.5%29%2A%282x%5E2-12%29
Then
2x%5E2-12=0
x%5E2=6
x=0+%2B-+sqrt%286%29
The three zeros are 1%2F2, sqrt%286%29, -sqrt%286%29.