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anonymous
 one year ago
Find all of the zeros of the function
x^3  15x^2 + 73x 111
How would you do this? The answer is 3, 6i, 6+i. I can't remember how to solve it though. @mukushla
anonymous
 one year ago
Find all of the zeros of the function x^3  15x^2 + 73x 111 How would you do this? The answer is 3, 6i, 6+i. I can't remember how to solve it though. @mukushla

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hi, If you remember rational root theorem, it says possible rational roots of a polynomial with integer coefficients are in the form \(\frac{p}{q}\) where \(p\) is a divisor of constant term and \(q\) is a divisor of leading coefficient.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\frac{p}{q}\) must be written in the lowest term.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, you have a cubic polynomial with integer coefficients:\[x^3  15x^2 + 73x 111=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0constant term: \(111\) leading coefficient: \(1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does it have to do something with multiples of those numbers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0o yeah, actually when you want to find the roots of a cubic polynomial by hand, you should look for a integer solution (a small integer number usually) for it in order for factoring the polynomial.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0rational root theorem helps you to find that root, now if you apply the theorem some of small possible roots that can be made in the form \(\frac{p}{q}\) are\[\frac{\pm1}{\pm1}, \frac{\pm3}{\pm1}, \frac{\pm37}{\pm1}, ...\]some of the smallest ones are \(1, 1, 3, 3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you test those numbers, you can see that \(x=3\) is a solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now try to factor out \((x3)\) and the rest will be a quadratic, which will give you other two solutions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do you factor it out of it? I'm not sure if I'm doing it right. @mukushla

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hint:\[x^3  15x^2 + 73x 111=x^33x^212x^2+36x+37x111 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it's x^2 12x +37 and x3 and after i plug the first equation into the quadratic formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got 6 + √ 4 and 6  √ 4 not 6 + i and 6i

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i meant 6  √ 4 and 6 + √ 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait actually i forgot to simplify √ 4 first before i divided by 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think i know what i did wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for all your help! :)
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