SOLUTION: what are the positive zeros of f(x)=2x^4-9x^3+2x^2+21x-10?

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: what are the positive zeros of f(x)=2x^4-9x^3+2x^2+21x-10?      Log On


   



Question 574517: what are the positive zeros of f(x)=2x^4-9x^3+2x^2+21x-10?
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
what are the positive zeros of f(x)=2x^4-9x^3+2x^2+21x-10?
x = 1/2 & x = 2
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Divide those out and solve the quadratic for the other 2, 1 is positive.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Rational zeros of a polynomial are fractions %28m%2Fn%29 with only certain possible numbers for numerator and denominator. The numerator (m) must be a factor/divisor of the independent term (-10 in this case), so 1, 2, 5, and 10 are the only positive possibilities. The denominator (n) must be a factor of the leading coefficient of the polynomial (2 in this case), so 1, and 2 are the only positive possibilities. (The rational zero fractions can have a plus or minus sign, but we will look for only the positive ones).
Putting it all together, the only possible positive rational zeros are 1/2, 1, 2, 5/2, 5, and 10.

f%281%29=2%281%29-9%281%29%2B2%281%29%2B21%281%29-10=2-9%2B2%2B21-10=25-19=6
f%282%29=2%2816%29-9%288%29%2B2%284%29%2B21%282%29-10=32-72%2B8%2B42-10=82-82=0
So far we have two positive zeros: 1/2 and 2.
That means the polynomial is divisible by x-1%2F2 and by x-2.
So it must be divisible by 2%28x-1%2F2%29%28x-2%29=2x%5E2-5x%2B2
Dividing f(x) by 2x%5E2-5x%2B2, I got
x%5E2-2x-5, so f%28x%29=%282x%5E2-5x%2B2%29%28x%5E2-2x-5%29=2%28x-1%2F2%29%28x-2%29%28x%5E2-2x-5%29
Solving x%5E2-2x-5=0 any which way we can (quadratic formula or completing the square), we get any remaining zeros of f(x). We get
x=1+%2B-+sqrt%286%29
1-sqrt%286%29%3C0, so the positive zeros of f(x) are:
highlight%281%2F2%29, highlight%282%29, and highlight%281%2Bsqrt%286%29%29.