Help with finding Zeros of a polynomial with 1+i as a zero

  • #1
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Homework Statement


p(x) = x^3 − x^2 + ax + b is a real polynomial with 1 + i as a zero, find a and b and find all of the real zeros of p(x).


The Attempt at a Solution


[/B]
1-i is also a zero as it is the conjugate of 1+i

so

(x-(1+i))(x-(1-i))=x^2-2x+2

let X^3-x^2+ax+b=x^2-2x+2(ax+d)

constant terms: b=2d, d=b/2

x terms: -2d+2a, a=2d

This is what I have done so far but I am unsure how to find a and b and the third zero

Any Help would be appreciated
 

Answers and Replies

  • #2
PeroK
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Homework Statement


p(x) = x^3 − x^2 + ax + b is a real polynomial with 1 + i as a zero, find a and b and find all of the real zeros of p(x).


The Attempt at a Solution


[/B]
1-i is also a zero as it is the conjugate of 1+i

so

(x-(1+i))(x-(1-i))=x^2-2x+2

let X^3-x^2+ax+b=x^2-2x+2(ax+d)
You've lost me here. That last step makes no sense.

##p(x) = (x^2 -2x + 2)(x + d)##

Surely?
 
  • #3
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You've lost me here. That last step makes no sense.

##p(x) = (x^2 -2x + 2)(x + d)##

Surely?
That is the third root which I have to find

How else would I go about solving it?
 
  • #4
PeroK
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That is the third root which I have to find

How else would I go about solving it?
What's the most obvious thing you can do with

##(x^2 -2x + 2)(x + d)##
 
  • #5
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What's the most obvious thing you can do with

##(x^2 -2x + 2)(x + d)##
I can Expand it which would give me

x^3+dx^-2x^2-2xde+2x+2d

What would I do now?

Would I sub in a (1+i) in for x which would give me d2i-2d+2id+2d
 
  • #6
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I can Expand it which would give me

x^3+dx^-2x^2-2xde+2x+2d

What would I do now?

Would I sub in a (1+i) in for x which would give me d2i-2d+2id+2d
It should say x^3+dx^2-2x^2-2xd+2x+2d
 
  • #7
Hi
I have a very easy method, and i am gonna give you a hint.

Assume the third root to be C

And then use the relation between the coeffecients of the equation and the sum taken one at a time,two at a time , three at a time.
 
  • #8
pasmith
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What's the most obvious thing you can do with

##(x^2 -2x + 2)(x + d)##
It is more natural to write [itex]x^3 - x^2 + ax + b = (x^2 - 2x + 2)(x - x_0)[/itex] since the root we seek is then [itex]x_0[/itex] rather than [itex]-d[/itex].

I can Expand it which would give me

x^3+dx^-2x^2-2xde+2x+2d

What would I do now?

Would I sub in a (1+i) in for x which would give me d2i-2d+2id+2d
It should say x^3+dx^2-2x^2-2xd+2x+2d
You have three unknowns. You need therefore three simultaneous equations which they must satisfy. You can obtain these by setting [tex]
x^3 - x^2 + ax + b = (x^2 - 2x + 2)(x + d)[/tex] and comparing coefficients of powers of [itex]x[/itex].
 
  • #9
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It is more natural to write [itex]x^3 - x^2 + ax + b = (x^2 - 2x + 2)(x - x_0)[/itex] since the root we seek is then [itex]x_0[/itex] rather than [itex]-d[/itex].





You have three unknowns. You need therefore three simultaneous equations which they must satisfy. You can obtain these by setting [tex]
x^3 - x^2 + ax + b = (x^2 - 2x + 2)(x + d)[/tex] and comparing coefficients of powers of [itex]x[/itex].
using constant term
b=2d

x terms

a=2d+2

x^2 term

-1=d-2

Would this be the equations that I need to use?
 
  • #10
using constant term
b=2d

x terms

a=2d+2

x^2 term

-1=d-2

Would this be the equations that I need to use?
did you see post #7 ? use that its much quicker.
 
  • #11
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did you see post #7 ? use that its much quicker.
I'm not sure that I understand would I just use C as a root and expand (1+i)(1-i)C
 
  • #12
I'm not sure that I understand would I just use C as a root and expand (1+i)(1-i)C
Nope.

1st root :- 1+i
2nd root:- 1-i
3rd root :- C

Sum of roots = -(coeff of x^2)/(coeff of x^3)
Sum of product of the roots taken 2 at a time= (coeff of x)/(coeff of x^3)
Sum of product of the roots taken 3 at a time = -(constant)/(coeff of x^3)

EDIT:

Capture.jpg
 
  • #13
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Nope.

1st root :- 1+i
2nd root:- 1-i
3rd root :- C

Sum of roots taken 1 at a time = -(coeff of x^2)/(coeff of x^3)
Sum of roots taken 2 at a time= (coeff of x)/(coeff of x^3)
Sum of roots taken 3 at a time = -(constant)/(coeff of x^3)
Sum of roots taken 1 at a time = -(coeff of x^2)/(coeff of x^3) =-1/1=1
Sum of roots taken 2 at a time= (coeff of x)/(coeff of x^3) = a/1=a
Sum of roots taken 3 at a time = -(constant)/(coeff of x^3) = b/1=b

What would I do now?
 
  • #14
Sum of roots taken 1 at a time = -(coeff of x^2)/(coeff of x^3) =-1/1=1
Sum of roots taken 2 at a time= (coeff of x)/(coeff of x^3) = a/1=a
Sum of roots taken 3 at a time = -(constant)/(coeff of x^3) = b/1=b

What would I do now?
I edited it, refresh and check.

your third equation should be= '-b' not 'b'

and if r1,r2,r3 are the roots of 3rd degree polynomial
then
sum of product of roots taken 2 at a time means
r1.r2+ r2.r3 + r3.r1

rest you can understand.
 
  • #15
PeroK
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I can Expand it which would give me

x^3+dx^-2x^2-2xde+2x+2d

What would I do now?

Would I sub in a (1+i) in for x which would give me d2i-2d+2id+2d
So, you have:

##p(x) = x^3 - x^2 +ax + b = x^3 + (d-2)x^2 + (2-2d)x + 2d##

It shouldn't be hard to solve for ##d, a, b## from there!
 
  • #16
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So, you have:

##p(x) = x^3 - x^2 +ax + b = x^3 + (d-2)x^2 + (2-2d)x + 2d##

It shouldn't be hard to solve for ##d, a, b## from there!
does this meant that

-x^2=(d-2)x^2
-1=d-2
d=1
 
  • #17
PeroK
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does this meant that

-x^2=(d-2)x^2
-1=d-2
d=1
Yes! Just compare coefficients, as was mentioned in post #8.
 
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  • #18
epenguin
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The way to find a simpler (reduced degree) equation from a polynomial equation of which you already know some roots (i.e. you know a factor of the polynomial) is to divide the overall polynomial by the polynomial that you know.

Equivalently and I think this way is easier here, eliminate between the two polynomials until you get a linear factor in this case.

In this way you should get an equation linear in x. You then identify coefficients.

If you are handy with determinants there is another, still equivalent, way to do it using them. Which I tried – but it seems I am not as handy as I thought. :redface:
 
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  • #19
SammyS
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The way to find a simpler (reduced degree) equation from a polynomial equation of which you already know some roots (i.e. you know a factor of the polynomial) is to divide the overall polynomial by the polynomial that you know.
...
It seems to me that polynomial division is the most direct way to get the complete solution. The real zero "pops out" immediately. Along with this, one simply chooses a and b to make the remainder be zero.

Unfortunately, this thread has been marked as "Solved" despite the fact that OP has not given his/her solution. Apparently OP has never come across the epenguin's signature message !
 
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  • #20
Since the thread has been marked solved :-

1+i+1-i+C=1 => C=-1

(1+i)(1-i) + (1-i)(C) + (1+i)(C) = a => 2 +2C=a => a=0.

(1+i)(1-i)(C) = -b => -b=2C => b=2.

All the roots : 1+i , 1-i , -1
a: 0
b: 2

The equation becomes:- ## x^3-x^2+2 =0##

## (x+1)(x^2-2x+2)=0##
 
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  • #21
epenguin
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It seems to me that polynomial division is the most direct way to get the complete solution. The real zero "pops out" immediately. Along with this, one simply chooses a and b to make the remainder be zero.

Unfortunately, this thread has been marked as "Solved" despite the fact that OP has not given his/her solution. Apparently OP has never come across the epenguin's signature message !
I have changed it to 'unsolved'. I don't know what was in the mind and is Greg when he created the solve the button – but I see no point calling anything without seeing a solution 'solved'. I also don't know what is in the mind of OP is who do this – they may have had an advantage from the thread, but what satisfaction or interest do they imagine there is for other participants or readers in discussing a question without ever seeing the solution?

Since the thread has been marked solved :-

1+i+1-i+C=1 => C=-1

(1+i)(1-i) + (1-i)(C) + (1+i)(C) = a => 2 +2C=a => a=0.

(1+i)(1-i)(C) = -b => -b=2C => b=2.

All the roots : 1+i , 1-i , -1
a: 0
b: 2

The equation becomes:- ## x^3-x^2+2 =0##

## (x+1)(x^2-2x+2)=0##
Yes I think it can be done this way with the formula for coefficients. But I am not following your logic at the end and I think there is a mistake at the end of the second line. What is needed is a linear factor which multiplied by (x2-2x2+2) gives the original cubic.
 
  • #22
I have changed it to 'unsolved'. I don't know what was in the mind and is Greg when he created the solve the button – but I see no point calling anything without seeing a solution 'solved'. I also don't know what is in the mind of OP is who do this – they may have had an advantage from the thread, but what satisfaction or interest do they imagine there is for other participants or readers in discussing a question without ever seeing the solution?



Yes I think it can be done this way with the formula for coefficients. But I am not following your logic at the end and I think there is a mistake at the end of the second line. What is needed is a linear factor which multiplied by (x2-2x2+2) gives the original cubic.
I think it is correct, you can just verify that 1+i & 1-i are the roots of x^2-2x+2. and by opening the last line, it gives the required cubic.
and in the second line, i think its correct too.
Please Reply. Thanks.
 
  • #23
epenguin
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I think it is correct, you can just verify that 1+i & 1-i are the roots of x^2-2x+2. and by opening the last line, it gives the required cubic.
and in the second line, i think its correct too.
Please Reply. Thanks.
S

The equation becomes:- ## x^3-x^2+2 =0##

## (x+1)(x^2-2x+2)=0##
Yes I see what you mean now and I agree (I just made some sign mistakes in my calculation).

It is however not necessary to consider the complex factors at this point, can all be done with the real quadratic factor that we know since #1
 
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