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Curve Sketching

  • Thread starter Jet1045
  • Start date
  • #1
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Homework Statement



I am finding the x intercepts of 2(x^3)-6(x^2)-18x+7 . For some reason I am NOT getting it, even though its probably really easy.

Homework Equations





The Attempt at a Solution


I set the equation to 0 and then isolated the x's and got

-7 = 2(x^3)-6(x^2)-18x

factored out a 2x and got

-7 = 2x((x^2)-3x-9)

i tried factoring (x^2)-3x-9 and have no idea how.

can someone please explain where to go from here and how to find the zeroes, or if there is another way to do so
 

Answers and Replies

  • #2
311
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well, factorizing by isolating the x's will not help at all. You should factorize the entire expression, i.e.

try and factorize 2 x^3 - 6 x^2 - 18 x+7.

The reason your method won't help is the following:

if i have an expression of the form (x-a)(x-b) = 7, given that x is any real number, I cannot conclude anything useful about x. But if the expression was

(x-a)(x-b) = 0, I can safely say that x = a or x = b.
 
  • #3
49
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Thanks very much for the help, i didn't realise that by bring the 7 over I would have that problem, im so used to just trying to isolate x's lol

so i tried factoring the whole equation with the 7 on that side , with NO luck :(
any ideas. We have not learnt to many factoring techniques in class.
 
  • #4
symbolipoint
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You want to completely factor your polynomial expression. You might be able to find up to three places where the polynomial equals zero.
 
  • #5
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well, u can do some preliminary analysis to remove out some possibilites - the solution is definitely not an integer (can you guess why?)

but, I have a doubt about whether you have copied your question down correctly? this seems to have rather complicated roots.
 
  • #6
symbolipoint
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well, u can do some preliminary analysis to remove out some possibilites - the solution is definitely not an integer (can you guess why?)

but, I have a doubt about whether you have copied your question down correctly? this seems to have rather complicated roots.
Maybe the exercise was intended for solution method including Rational Roots Theorem.
 
  • #7
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Well after graphing it on my calculator, I know that there are in fact 3 roots, and they are very long decimals.

i guess there is a chance that I remember the question wrong. hopefully when I get back in class tomorrow it is alot easier :(
 
  • #8
symbolipoint
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The idea about the Rational Roots Theorem is that you can try various combinations of rational numbers using the leading coefficient and the constant term from the polynomial, and try dividing the polynomial by a binomial which uses the various rational numbers; and if a division renders a quotient with remainder of zero, you have found a root. You continue looking for other binomial factors this way until you have factored your original polynomial as much as possible.

Are you sure your graph is showing irrational zeros, or are the zeros actually nonrepeating decimals just maybe long?

UPDATE: I used polynomial division to check for the possible eight rational zeros, and all of them gave remainders, meaning the tested roots failed as actual roots for the given polynomial. You must be correct in having irrational roots, whatever they be.
 
Last edited:
  • #9
49
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The roots are irrational when i checked my calculator....
i have no idea how , i dont think the rational roots theorem is what he is looking for because we have never done that in class.

i am PRAYING that i copied it down wrong haha

thanks for your help! :)
 

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