Question 896696: How to determine the rational zeros by factoring h(x)=x^3-x^2-10x-8
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your equation is x^3 - x^2 - 10x - 8
your rational zeroes should be plus or minus factors of the constant term divided by factors of the leading coefficient.
the leading coefficient is 1 and the constant term is 8
your rational zeroes, if any, will be one or more of:
Plus or minus {8,4,2,1}}}
descartes rule of thumb is:
the number of positive zeroes of f(x) will be less then or equal to the number of sign changes of f(x) when you decrement by 2 until you get to 0.
the number of negative zeroes of f(x) will be less than or equal to the number of sign changes of f(-x) when you decrement by 2 until you get to 0.
let's see how this rule applies to your equation.
your equation is:
f(x) = x^3 - x^2 - 10x - 8
your signs are +---
there is only 1 sign change so you should only have 1 positive rational root.
f(-x) = (-x)^3 - (-x)^2 - 10(-x) - 8 which is equal to:
f(-x) = -x^3 - x^2 + 10x - 8
your signs are --+-
you have 2 sign changes.
your possible negative zeroes will be 0 or 2.
you start with 2 and decrement by 2 until you get to 0.
2 - 2 = 0 so you stop and you have 0 or 2 possible negative zeroes.
a zero is the same as a root.
so you can have 1 positive rational root and may have 0 or 2 negative rational roots.
in fact, you must have one positive rational root.
this is because the degree of your equation is 3 and you must have 3 roots in total, some of which can be real and some of which can be not real.
the not real roots come in pairs so, if you have them at all, then you must have 2 which means that there's only 1 left and it must be positive since the negative roots are either 0 or 2.
bottom line is you must have one positive rational root and can have 0 or 2 negative rational roots and can have 0 or 2 irrational or unreal roots which will alwways come in pairs.
we'll shoot for the positive root since we know it has to be there.
your equation is x^3 - x^2 - 10x - 8
your possible rational roots are +/- {8,4,2,1}
after repeated iterations of synthetic division i found that 4 is a zero of the polynomial which means that x-4 is a factor of the polynomial.
i tried 1 first and then i tried 2 but neither worked.
i hit the jackpot when i tried 4.
the remainder after the division is x^2 + 3x + 2.
this can also be factored as (x+1) * (x+2) which yields 2 negative zeroes of x = -2 and x = -1
the zeroes of the polynomial are 4, -1, -2.
descartes rule worked since we have 1 positive rational root and 2 negative rational roots.
the factors of the polynomial are (x-4), (x+1), (x+2)
multiply those factors together and you will get the original polynomial.
a good tutorial on rational zeroes and rule of signs can be found at the following link:
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut38_zero1.htm
if i set the equation equal to 0 and graph it, i should see the zeroes where the graph crosses the x-axkis.
the graph of the equation is shown below:
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