Complex numbers

[Axioms]

I'm not sure how to start this question. Can someone please assist in explaining how to approach the problem?

Find all the zeros of the polynomial function by using the given zero

 

p(x)= 4x^3 + 23x^2 +34x -10 Zero:-3+i

 

 

 

[timo]

Imagine you had the polynomial written in the form (x-a)(x-b)(x-c) and were supposed to find the x where it assumes the value zero. That would be rather simple. In this case you know that your polynomial can be written in the form (x+3-i)(ax^2 + b^x + c). Do so by polynomial division and the solve your problem. May I ask: Where have you encountered this problem?

[Axioms]

I have tried that. I cant get a factor of the equation without making the (x+a) part = 10. I think it makes the statement untrue because x=0 when x = -3+i =0.

 

If it is true then:

4x+11_______

x+3-i |4x^2+23x +34

4x^2 +12x -4xi

 

It will have a remainder of 1 -4xi +11i

It will mean that 4x + 11 =10 and x = -1/4

 

Not sure if it makes sense. Maybe if you elaborate a little bit more?

 

I got the question out of my linear algebra text book. Forms a section on conjugates, modulus and division of complex numbers.

 

 

Does this seem correct?

 

Find all the zeros of the polynomial function by using the given zero

 

p(x)= 4x^3 + 23x^2 +34x -10 Zero:-3+i

 

You say that x-(-3+i) and its conjugate x-(-3-i) are your factors. Multiply them together and get x^2 +6x +10

 

You use long division with this factor to find 4x - 1 and therefore x= 1/4 is another root?

 

Edited November 2, 2011 by Axioms

[timo]

Well, if your calculations are correct then that should of course be the correct solution. I somehow wonder why you think -3-i was a zero? Is that generally true for polynomials with real-valued coefficients that roots must be in some way be related via complex conjugation?

 

What I meant is that you should have found that (4x^3 + 23x^2 +34x -10) / ( x +3 - i) = 4x^2 + (11+4i)x - (3+i), meaning that you are looking for the solutions of (4x^3 + 23x^2 +34x -10) = [4x^2 + (11+4i)x - (3+i)] * [ x +3 - i ] = 0. A product is zero if one of its factors is zero. The two factors are a first and a 2nd order polynomial, respectively. For both cases you should know how to find the zeros.

[Axioms]

Well, if your calculations are correct then that should of course be the correct solution. I somehow wonder why you think -3-i was a zero? Is that generally true for polynomials with real-valued coefficients that roots must be in some way be related via complex conjugation?

 

What I meant is that you should have found that (4x^3 + 23x^2 +34x -10) / ( x +3 - i) = 4x^2 + (11+4i)x - (3+i), meaning that you are looking for the solutions of (4x^3 + 23x^2 +34x -10) = [4x^2 + (11+4i)x - (3+i)] * [ x +3 - i ] = 0. A product is zero if one of its factors is zero. The two factors are a first and a 2nd order polynomial, respectively. For both cases you should know how to find the zeros.

 

 

I think that the conjugate of one of the factors is always a factor. I could be wrong so that is why I am asking. What is true though is that a factor times is conjugate is equal to zero if and only if one of the factors is equal to zero. I know that the one factor is zero so its conjugate multiplied in must also be zero. This is then simple to do with long division because you get rid of the imaginary.

 

I think your method will work but it seems like a lot of work. I did the working out for the first term using your method and got 4x^4 + 12x^3 +4i/x^2+6x +10. I assume it simplifies at the end?

[the tree]

I did the working out for the first term using your method and got 4x^4 + 12x^3 +4i/x^2+6x +10

If you're looking a greater degree polynomial than you started with then you are definitely going wrong.

 

The two things you need to get right in this question are polynomial long division (for which there will be no remainder) and then solving a quadratic equation. That's it. Don't try to over complicate that.

[timo]

I think that the conjugate of one of the factors is always a factor.

Without additional restrictions that is definitely wrong. Take the solutions of "x-i=0" and "(x-3)*(3x-i)=0" for example.

I could be wrong so that is why I am asking.

It is pretty trivial to check if your indeed found the roots by just plugging your candidate solution into the original polynomial.

 

What is true though is that a factor times is conjugate is equal to zero if and only if one of the factors is equal to zero. I know that the one factor is zero so its conjugate multiplied in must also be zero. This is then simple to do with long division because you get rid of the imaginary.

I don't understand what you are saying there.

 

I think your method will work but it seems like a lot of work.

Five minutes by hand, twenty seconds with Wolfram Alpha.

 

I did the working out for the first term using your method and got 4x^4 + 12x^3 +4i/x^2+6x +10. I assume it simplifies at the end?

You probably calculated something like (4x^3 + 23x^2 +34x -10) * ( x +3 - i) rather than (4x^3 + 23x^2 +34x -10) / ( x +3 - i). But it seems to me that you are not understanding the reason behind doing this in the first place: Any 3rd degree polynomial can be written in the form p(x)=(x-a)(x-b)(x-c). The idea is to find this form, since then solving p(x)=0 is trivial.

[the tree]

Without additional restrictions that is definitely wrong. Take the solutions of "x-i=0" and "(x-3)*(3x-i)=0" for example.[...]Any 3rd degree polynomial can be written in the form p(x)=(x-a)(x-b)(x-c). The idea is to find this form, since then solving p(x)=0 is trivial.

To make this as clear as conceivably possible: an n^th degree polynomial has n roots. You are dealing with a 3rd degree polynomial, you have one root, you need 2 more. So you need to solve a 2nd degree polynomial, you have hopefully known how to do that for a very long time. Edited November 2, 2011 by the tree

[Axioms]

Without additional restrictions that is definitely wrong. Take the solutions of "x-i=0" and "(x-3)*(3x-i)=0" for example.

It is pretty trivial to check if your indeed found the roots by just plugging your candidate solution into the original polynomial.

 

I don't understand what you are saying there.

 

 

Five minutes by hand, twenty seconds with Wolfram Alpha.

 

You probably calculated something like (4x^3 + 23x^2 +34x -10) * ( x +3 - i) rather than (4x^3 + 23x^2 +34x -10) / ( x +3 - i). But it seems to me that you are not understanding the reason behind doing this in the first place: Any 3rd degree polynomial can be written in the form p(x)=(x-a)(x-b)(x-c). The idea is to find this form, since then solving p(x)=0 is trivial.

 

If you have i/3-i what answer do u get? I dont understand how you are doing your division?

 

 

1-3i/10

 

 

"I don't understand what you are saying there." Ok lets say z = a +bi then its conjugate is z*= a-bi. Now in z = 0 then zz* must also be equal to zero. It leaves you with a function a^2 +b^2 without the imaginary. It is still a factor because they stated that Zero: -3+i.

 

It should be true because if I substitute my answer it is equal to zero.

[the tree]

I dont understand how you are doing your division?

Okay you're not going to progress with out either learning polynomial long division or using a calculator.

 

Now just to move on, we might just have to tell you.

 

[math]\frac{4 x^3+23 x^2+34 x-10}{x-(-3+i)}=4 x^2+(11+4 i) x-(3+i)[/math]

 

To find the two remaining roots, you need to solve that quadratic equation. If that is posing a major difficulty then you are trying to learn too fast and you need to go back and learn how to do that.

 

In the particular case, the root you gave is indeed one of the remaining roots. There is one more needed to make a total of three.