Can you help me find the third zero of this complex polynomial?

[TheFallen018]

Hey, first off, I'm not sure if this is the right section. If another section is better, please let me know and I'll be more careful next time.

So, my problem is with a degree 3 complex polynomial. I'm given one zero of the equation, but since it is a complex zero, I can use the conjugate too. So, I already have two of the zeros for the polynomial, and since according to the fundamental theorem of algebra, it should only have one more. Because complex roots also have a conjugate, this suggests that the root is real. However, I'm having a hard time pinning it down. I've come up with a number of answers, and one of my more recent attempts led me to think it was -1. None of these seem right.

So, here's the question.

View attachment 8177

Any help would be amazing. Thank you :)

 

[I like Serena]

Hey, first off, I'm not sure if this is the right section. If another section is better, please let me know and I'll be more careful next time.

So, my problem is with a degree 3 complex polynomial. I'm given one zero of the equation, but since it is a complex zero, I can use the conjugate too. So, I already have two of the zeros for the polynomial, and since according to the fundamental theorem of algebra, it should only have one more. Because complex roots also have a conjugate, this suggests that the root is real. However, I'm having a hard time pinning it down. I've come up with a number of answers, and one of my more recent attempts led me to think it was -1. None of these seem right.

So, here's the question.
Any help would be amazing. Thank you :)

Hi Fallen number 18! ;)

The fact that $-1+2i$ is a zero means that we can factorize the polynomial as $(z-(-1+i2))(z^2 +az+b)$.
When we expand that and match it to the polynomial, we can deduce the values of $a$ and $b$.
Care to try?

 

[HOI]

Hey, first off, I'm not sure if this is the right section. If another section is better, please let me know and I'll be more careful next time.

So, my problem is with a degree 3 complex polynomial. I'm given one zero of the equation, but since it is a complex zero, I can use the conjugate too.

No, you can't. The theorem you are misremembering is that "if z is a zero of a polynomial with real coefficients then so is its conjugate." But this polynomial does not have real coefficients.

So, I already have two of the zeros for the polynomial, and since according to the fundamental theorem of algebra, it should only have one more. Because complex roots also have a conjugate, this suggests that the root is real. However, I'm having a hard time pinning it down. I've come up with a number of answers, and one of my more recent attempts led me to think it was -1.

Well, yes, with z= -1, $$z^2= 1$$, and $$z^3= -1$$ so the polynomial becomes $$-1+ 3- 2i- 3+ 4i+ 1- 2i= (-1+ 3- 3+ 1)+ (-2i+ 4i- 2i)= 0$$.

None of these seem right.

So, here's the question.
Any help would be amazing. Thank you :)

The fact that -1+ 2i is a root means that x+ 1- 2i will divide into $$z^3+ (3- 2i)z^2+ (3- 4i)z+ (1- 2i)$$ evenly- with no remainder.

In fact, dividing $$z^3+ (3- 2i)z^2+ (3- 4i)z+ (1- 2i)$$ by z+ 1- 2i gives a quotient of $$z^2+ 2z+ 1= (z+ 1)^2$$. The zeros of $$z^3+ (3- 2i)z^2+ (3- 4i)z+ (1- 2i)$$ are -1 and -1+ 2i with -1 being a double zero.

 

[TheFallen018]

Hi Fallen number 18! ;)

The fact that $-1+2i$ is a zero means that we can factorize the polynomial as $(z-(-1+i2))(z^2 +az+b)$.
When we expand that and match it to the polynomial, we can deduce the values of $a$ and $b$.
Care to try?

I actually already tried this method. I must have done something wrong the first time, because I didn't get the same answer when I did it just now. It worked out to be $(z-(-1+i2))(z^2 +2z+1)$ which worked nicely in the quadratic formula and got me the right answer. Thanks!

No, you can't. The theorem you are misremembering is that "if z is a zero of a polynomial with real coefficients then so is its conjugate." But this polynomial does not have real coefficients. Well, yes, with z= -1, $$z^2= 1$$, and $$z^3= -1$$ so the polynomial becomes $$-1+ 3- 2i- 3+ 4i+ 1- 2i= (-1+ 3- 3+ 1)+ (-2i+ 4i- 2i)= 0$$.The fact that -1+ 2i is a root means that x+ 1- 2i will divide into $$z^3+ (3- 2i)z^2+ (3- 4i)z+ (1- 2i)$$ evenly- with no remainder.

In fact, dividing $$z^3+ (3- 2i)z^2+ (3- 4i)z+ (1- 2i)$$ by z+ 1- 2i gives a quotient of $$z^2+ 2z+ 1= (z+ 1)^2$$. The zeros of $$z^3+ (3- 2i)z^2+ (3- 4i)z+ (1- 2i)$$ are -1 and -1+ 2i with -1 being a double zero.

I really like this method. It hadn't occurred to me to factorise the z terms that way. It makes dividing really easy, and then it's just a matter of using the quadratic formula. Thanks for the insight on this one, I'll be using it again for sure.