SOLUTION: 1) List all of the possible roots of : 2x^3 - 5x^2 - 28x + 15 = 0. 2) Then find all the roots. I'm confused because I thought possible roots and all roots were the same thing

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 1) List all of the possible roots of : 2x^3 - 5x^2 - 28x + 15 = 0. 2) Then find all the roots. I'm confused because I thought possible roots and all roots were the same thing      Log On


   

Question 925070: 1) List all of the possible roots of : 2x^3 - 5x^2 - 28x + 15 = 0.
2) Then find all the roots.
I'm confused because I thought possible roots and all roots were the same thing? Also, are "zeros" also roots? Thank you.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, roots and zeros are synonymous. I prefer the term "root" since zero is a number and it gets muddied up a bit.

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1) The possible rational roots of f(x) are potential values that could make f(x) = 0 true.

In general, not all possible roots are actual roots. You will have more possible roots than actual roots for the majority of the cases.

To list the possible rational roots, you use the rational root theorem. The idea is to list the factors of the first and last coefficients, then divide each factor

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Factors of 15: 1, 3, 5, 15
Factor of 2: 1, 2

Divide each factor of 15 by each factor of 2 to get these fractions: 1/1, 1/2, 3/1, 3/2, 5/1, 5/2, 15/1, 15/2

Those fractions simplify to 1, 1/2, 3, 3/2, 5, 5/2, 15, 15/2

Now add on the negative counterparts: -1, -1/2, -3, -3/2, -5, -5/2, -15, -15/2

So the list of possible rational roots is shown below in blue


1, 1/2, 3, 3/2, 5, 5/2, 15, 15/2
-1, -1/2, -3, -3/2, -5, -5/2, -15, -15/2

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2) To find the actual rational roots, you need to check every possible rational root. To check each one, you plug each one (one at a time) into f(x).

If you get f(x) = 0, then that possible root is indeed an actual root.

In this case, f(x) = 2x^3 - 5x^2 - 28x + 15

I'll do the first possible root

Checking x = 1

f(x) = 2x^3 - 5x^2 - 28x + 15

f(1) = 2(1)^3 - 5(1)^2 - 28(1) + 15

f(1) = -16

The result is -16 and NOT 0. So x = 1 is NOT an actual root.

I'll let you check the other potential rational roots.

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Let me know if that helps or not. Thanks.

If you need more help, feel free to email me at jim_thompson5910@hotmail.com

My Website: http://www.freewebs.com/jimthompson5910/home.html