You can
put this solution on YOUR website!5x^3-9x^2-17x-3
i have to solve this as follows:
(a) By using the Rational Zero Theorem, list all possible rational zeros of the given polynomial.
For A, I know that i have to do it with the rational zero theorem, but here is a sentence of a book that i don't understand.
if a rational number p/q is a zero of f(x), then p must be a factor of -4 and q must be a factor of 3. the task belonging to it is: 3x^4-11x^3+10x-4
Then it says possibility for p = +-1, +-2, +-4
possibility for q is =-1, +-3
i dont understand how to get +-4 and for q +-3. as the sentence above states. I dont get how to find the possibilities...
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In your problem p must divide -3 and q must divide 5.
p can be +/-1, +/-3
q can be +/-1, +/-5
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Then the rational roots may be 1/1, -1/1, 1/5, 1/-5, =3/1, 3/-1, 3/5, 3/-5
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(b) Find all of the zeros of the given polynomial. Be sure to show work, explaining how you have found the zeros.
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The possible rational zeros are listed in part a. You have to see
which ones really are zeros.
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f(x) = 5x^3-9x^2-17x-3
f(1) = 5-9-17-3 is not zero
f(-1) = -5-9+17-3 = 0 so x=-1 is a root
f(3) = 5*27-9*9-17*3-3 = 0 so x= 3 is a root
f(1/5)= 5(1/5)^3 - 9(1/5)^2 - 17(1/5) - 3 is not zero
f(-1/5) = 5(-1/5)^3 - 9(-1/5)^2 - 17(-1/5) -3 = 0 so x= -1/5 is a root
etc.
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Cheers,
Stan H.
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