Find all complex roots of polynomial

In summary, the student is trying to solve the equation x^4 - 6x^3 + 26x^2 -46x +65 = 0 but is having difficulty because there are four roots and none of them are -65. The student uses the quadratic formula and finds the roots 1-2i and 1+2i which are correct from the book.
  • #1
DryRun
Gold Member
838
4
Homework Statement
Given 2 + 3i is one root of the equation:
x^4 - 6x^3 + 26x^2 -46x +65 = 0
Find the remaining roots.

The attempt at a solution
I am thinking about this as a possible solution although it is too long to be plausible, unless I'm wrong.
let x = a + bi and then replace in the polynomial. Then, expand and finally, divide it by 2+3i

Another possibility (i'm not sure) is this:
x (x^3 - 6x^2 + 26x -46) +65 = 0
So, i try to solve the cubic equation: x^3 - 6x^2 + 26x -46
However, note that one solution to x, is not equal to -65. So, this method is wrong, i believe.

I frankly don't know how else to go about to solve this one.
 
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  • #2
What other property about complex roots do you know? (Hint: they always come in pairs)
 
  • #3
So, 2-3i is also a root? How do i confirm that? By replacing in the polynomial, it should equal zero, right?

But what about the other roots? Since the maximum power of x is 4, so i assume that there are 4 roots in all.
 
  • #4
I still can't figure it out. Can someone please give me a hint?
 
  • #5
If you have two roots p and q, then (x-p)(x-q) is a factor. Use long division to divide out that factor and see what's left.
 
  • #6
LCKurtz said:
If you have two roots p and q, then (x-p)(x-q) is a factor. Use long division to divide out that factor and see what's left.
Notice that when p & q are complex conjugates, (x-p)(x-q) can be written as a sum times a difference which comes out quite nicely as a difference of squares.

(x - (2-3i))(x - (2+3i)) = ((x-2) + 3i)((x-2) - 3i) = ...
 
  • #7
OK, here is what i did:
The two known factors are: (2+3i) and (2-3i)
So, the product is: (2+3i)(2-3i) = x^2 - 4x + 13

Using this product as divisor and the polynomial as dividend, i get this:
x^2 - 2x + 5
OK, i got it. Thanks guys. :)
 
  • #8
So, what are the two remaining roots?
 
  • #9
Initially, i did it like this:
x(x-2) = -5
So x = -5 and x = -3
But then i don't know why but these answers are wrong (from the book).

So, i used quadratic formula
and i got these: 1-2i, 1+2i, which are correct from the book.

But again, i don't understand why i can't use the first set of simpler roots; -5 and -3.
 
  • #10
sharks said:
Initially, i did it like this:
x(x-2) = -5
So x = -5 and x = -3
But then i don't know why but these answers are wrong (from the book).

So, i used quadratic formula
and i got these: 1-2i, 1+2i, which are correct from the book.

But again, i don't understand why i can't use the first set of simpler roots; -5 and -3.
You can't use -5 & -3 because they're not solutions of x(x-2) = -5. Plug them into see that.
 
  • #11
Basic rule: If ab= 0 then either a= 0 or b= 0. But that is a special property of 0. You cannot say "if ab= -5 then either a= -5 or b= -5" which is what you were trying to do.
 
  • #12
I think i have this one wrapped up. Thanks for the help. :)
 

1. How do I find all complex roots of a polynomial?

To find all complex roots of a polynomial, you can use the quadratic formula or synthetic division to factor the polynomial into linear and quadratic terms. Then, use the complex number formula to find the roots of the quadratic terms.

2. Can all polynomials have complex roots?

Yes, all polynomials can have complex roots. Even if a polynomial has only real coefficients, it can still have complex roots. This is because the Fundamental Theorem of Algebra states that a polynomial of degree n has n complex roots, including repeated roots.

3. How do I know if a complex number is a root of a polynomial?

To determine if a complex number is a root of a polynomial, you can use synthetic division to divide the polynomial by the complex number. If the remainder is 0, then the complex number is a root of the polynomial.

4. Can I use the quadratic formula to find complex roots?

Yes, you can use the quadratic formula to find complex roots of a polynomial. However, you may need to use the complex number formula to find the roots of the quadratic terms.

5. What is the difference between real and complex roots?

Real roots are solutions to a polynomial that are real numbers, meaning they do not involve any imaginary terms. Complex roots, on the other hand, involve imaginary numbers and can be written in the form of a + bi, where a and b are real numbers and i is the imaginary unit.

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