Question 574517
Rational zeros of a polynomial are fractions {{{(m/n)}}} 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(1/2)=2(1/16)-9(1/8)+2(1/4)+21(1/2)-10=1/8-9/8+1/2+21/2-10=8/8+22/2-10=1+11-10=0}}}
{{{f(1)=2(1)-9(1)+2(1)+21(1)-10=2-9+2+21-10=25-19=6}}}
{{{f(2)=2(16)-9(8)+2(4)+21(2)-10=32-72+8+42-10=82-82=0}}}
So far we have two positive zeros: 1/2 and 2.
That means the polynomial is divisible by {{{x-1/2}}} and by {{{x-2}}}.
So it must be divisible by {{{2(x-1/2)(x-2)=2x^2-5x+2}}}
Dividing f(x) by {{{2x^2-5x+2}}}, I got
{{{x^2-2x-5}}}, so {{{f(x)=(2x^2-5x+2)(x^2-2x-5)=2(x-1/2)(x-2)(x^2-2x-5)}}}
Solving {{{x^2-2x-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 +- sqrt(6)}}}
{{{1-sqrt(6)<0}}}, so the positive zeros of f(x) are:
{{{highlight(1/2)}}}, {{{highlight(2)}}}, and {{{highlight(1+sqrt(6))}}}.