## 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, 6-i, 6+i. I can't remember how to solve it though. @mukushla

1. anonymous

Hi, 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.

2. anonymous

$$\frac{p}{q}$$ must be written in the lowest term.

3. anonymous

well, you have a cubic polynomial with integer coefficients:$x^3 - 15x^2 + 73x -111=0$

4. anonymous

constant term: $$-111$$ leading coefficient: $$1$$

5. anonymous

does it have to do something with multiples of those numbers?

6. anonymous

o 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.

7. anonymous

rational 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 $$-1, 1, 3, -3$$

8. anonymous

If you test those numbers, you can see that $$x=3$$ is a solution

9. anonymous

Now try to factor out $$(x-3)$$ and the rest will be a quadratic, which will give you other two solutions.

10. anonymous

How do you factor it out of it? I'm not sure if I'm doing it right. @mukushla

11. anonymous

hint:$x^3 - 15x^2 + 73x -111=x^3-3x^2-12x^2+36x+37x-111$

12. anonymous

so it's x^2 -12x +37 and x-3 and after i plug the first equation into the quadratic formula?

13. anonymous

right

14. anonymous

i got 6 + √ -4 and 6 - √ 4 not 6 + i and 6-i

15. anonymous

i meant 6 - √ -4 and 6 + √ -4

16. anonymous

wait actually i forgot to simplify √ -4 first before i divided by 2

17. anonymous

i think i know what i did wrong

18. anonymous

aha, ok

19. anonymous

Thanks for all your help! :)

20. anonymous

no problem