Question 189952: 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...
(b) Find all of the zeros of the given polynomial. Be sure to show work, explaining how you have found the zeros.
For B, This is what I thought!
f(x) = 5x3 – 9x2 – 17x – 3.
x(5x^2-9x-17)-3
but then with x(5x^2-9x-20) how would you find something that would result in 20 and 9
X9X- )(X+ )
I am stuck here or is it the wrong approach?
Anna
Answer by stanbon(75887)   (Show Source):
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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