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xXxD4r7hP4ndaxXx
xXxD4r7hP4ndaxXx
Member since:
May, 2012
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Posts: 91
6x^5+3x^4-2x^2+3x-8 I really need help with this one. I tried the Rational Zero Theorem but, either my understanding is flawed or it does not work with this polynomial. So, I figured it might be easiest to factor it. But that came up with: x(3x^3-1)(2x+1)(2x-3)-8  Which does not appear to be proper form to apply Factor Theorem: c is a zero if (x-c) is a factor of P(x). I definitely have binomial factors here but I think that the -8 negates the theorem. Does anyone know if I'm wrong or have a solution? Edit: I figured this out before it got approved (nice one mods). Solution: The Rational Zero Theorem works on this polynomial. My confusion came from a misinterpretation of the theorem. So, remember: The Rational Zero Theorem will only give possible zeros. Using the theorem I got: +/-1/6, +/-1/3, +/-1/2, +/-2/3, +/-1, +/-4/3, +/-2, +/-8/3, +/-4, +/-8 I got confused when I checked some of the values and found that most of them were not zeros of the polynomial. For anyone wondering, the Rational Zero Theorem basically says to find the factors of the constant term of the polynomial, 8 (-8 yeah but we're using |8| for the theorem) and then find the factors of the leading coefficient: 6. So now we have the factors of 8: 1, 2, 4, 8 and the factors of 6: 1, 2, 3, 6 Divide the factors of 8 by those of 6 and remove the duplicates. Now the factors of 8 and (1, 2, 4, Cool/(1, 2, 3, 6) are the possible rational zeros of 6x^5+3x^4-2x^2+3x-8

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