Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

constant term: \(-111\)
leading coefficient: \(1\)

Answer 5

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

Answer 6

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.

Answer 7

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 \(-1, 1, 3, -3\)

Answer 8

If you test those numbers, you can see that \(x=3\) is a solution

Answer 9

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

Answer 10

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

Answer 11

hint:\[x^3 - 15x^2 + 73x -111=x^3-3x^2-12x^2+36x+37x-111 \]

Answer 12

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

Answer 13

right

Answer 14

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

Answer 15

i meant 6 - √ -4 and 6 + √ -4

Answer 16

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

Answer 17

i think i know what i did wrong

Answer 18

aha, ok

Answer 19

Thanks for all your help! :)

Answer 20

no problem