Polynomial Q (Need Help): one root of x3+ax2+36x-36=0 is product of other 2 roots

[Andy005]

One of the roots of the cubic x³+ax²+36x-36=0 is the product of the other two roots. Find the value of a and hence find all the roots.

 

[Deleted member 4993]

One of the roots of the cubic x³+ax²+36x-36=0 is the product of the other two roots. Find the value of a and hence find all the roots.

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

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[JeffM]

As our name implies, we are a "help" site, not an "answer" site. It is impossible to help you find errors or misunderstandings or roadblocks in your work unless you show us what you have done.

Sometimes you have done no work because you do not know where to start. We can help there as well, but only if you tell us that is your problem and, as mentioned by Subotosh, tell us what things you think MIGHT be relevant to the problem. If you never even thought about the Fundamental Theorem of Algebra, giving that as a hint will not help you much.

 

[stapel]

One of the roots of the cubic x³+ax²+36x-36=0 is the product of the other two roots. Find the value of a and hence find all the roots.

You applied the Rational Roots Test here to get a listing of possible roots. You looked for one root that was the product of two of the other roots. You plugged one of these roots into the given equation, solving for the value of "a". And... then what?

Please be complete, showing all of your reasoning and steps. Thank you!

 

[JeffM]

You applied the Rational Roots Test here to get a listing of possible roots. You looked for one root that was the product of two of the other roots. You plugged one of these roots into the given equation, solving for the value of "a". And... then what?

Please be complete, showing all of your reasoning and steps. Thank you!

That way works of course, but can find only one of the two possible answers. I wonder if we were told all the problem because one answer is neat and the other ugly. (See private message)

 

[mmm4444bot]

I wonder if we were told all the problem because one answer is neat and the other ugly.

I'm fairly certain they want the answer containing Real roots. (Maybe they haven't covered i, yet.)

I didn't try stapel's approach; I wrote a system of equations, using the four unknowns. Solving the system was easier than I had expected, but it misses the other solution. I commend you on due diligence. :cool:

 

[JeffM]

I'm fairly certain they want the answer containing Real roots. (Maybe they haven't covered i, yet.)

I didn't try stapel's approach; I wrote a system of equations, using the four unknowns. Solving the system was easier than I had expected, but it misses the other solution. I commend you on due diligence. :cool:

Thanks.

It's possible that the person who wrote the problem never considered whether there were multiple solutions because the problem was designed to exhibit the Rational Root Theorem. It's also possible that the original problem stipulated rational or integer solutions. Obviously, irrational or complex roots can never be found by the Rational Root Theorem.

Stapel responded to my PM by pointing out, as is indubitably correct, that someone who did not even try the Rational Root Theorem is not going to set up a system of non-linear equations based on the Fundamental Theorem of Algebra.

 

[HallsofIvy]

I don't see that this has anything to do with the "rational root theorem". There is no condition that the roots be rational here, just that one of the roots is the product of the other 2. Let p and q be the first two roots so that the third root is pq. Then we have \(\displaystyle (x- p)(x- q)(x- pq)= x^3+ax^2+36x-36\). Multiply on the left, then set corresponding coefficients equal to get 3 equations in p, q, and a.