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First we need to determine what the possible roots/zeros of the equation are. So we will look at p/q where p=factors of -5 and q = factors of 2. So we get these possibilities: 5/1,5/2, -5/1, -5/2, 1/1,1/2,-1/1,-1/2. Now I will choose one to divide into the polynomial using synthetic division.
First I will pick 1/2
1/2 | 2 -3 11 -5
-1 2 5
2 -2 10 0
So 1/2 is a root/zero of the polynomial! So we get: 2x^3-3x^2+11x-5 = (x-1/2)(2x^2-2x+10)
Now we can try to solve 2x^2-2x+10. Note if we set this to zero and try to solve it using the quadratic formula, it will give us an irrational answer, meaning it is not a rational root/solution.
x= 2 +- sq root ((-2)^-4(2)(10)) / 2(2)
x= 2 +- sq root (-76) / 4
When we take the square root of a negative, this gives us a complex solution.
Therefore, the only rational solution is: 1/2
Please let me know if you need any clarification. I'm always happy to answer your questions.
Dec 18th, 2015
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