document.write( "Question 574517: what are the positive zeros of f(x)=2x^4-9x^3+2x^2+21x-10? \n" ); document.write( "

Algebra.Com's Answer #369173 by KMST(5328) $\"\"$ $\"About$

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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).
\n" ); document.write( "Putting it all together, the only possible positive rational zeros are 1/2, 1, 2, 5/2, 5, and 10.
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\n" ); document.write( " $\"f%281%29=2%281%29-9%281%29%2B2%281%29%2B21%281%29-10=2-9%2B2%2B21-10=25-19=6\"$
\n" ); document.write( " $\"f%282%29=2%2816%29-9%288%29%2B2%284%29%2B21%282%29-10=32-72%2B8%2B42-10=82-82=0\"$
\n" ); document.write( "So far we have two positive zeros: 1/2 and 2.
\n" ); document.write( "That means the polynomial is divisible by $\"x-1%2F2\"$ and by $\"x-2\"$ .
\n" ); document.write( "So it must be divisible by $\"2%28x-1%2F2%29%28x-2%29=2x%5E2-5x%2B2\"$
\n" ); document.write( "Dividing f(x) by $\"2x%5E2-5x%2B2\"$ , I got
\n" ); document.write( " $\"x%5E2-2x-5\"$ , so $\"f%28x%29=%282x%5E2-5x%2B2%29%28x%5E2-2x-5%29=2%28x-1%2F2%29%28x-2%29%28x%5E2-2x-5%29\"$
\n" ); document.write( "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
\n" ); document.write( " $\"x=1+%2B-+sqrt%286%29\"$
\n" ); document.write( " $\"1-sqrt%286%29%3C0\"$ , so the positive zeros of f(x) are:
\n" ); document.write( " $\"highlight%281%2F2%29\"$ , $\"highlight%282%29\"$ , and $\"highlight%281%2Bsqrt%286%29%29\"$ . \n" ); document.write( "

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