Solving 4th Degree Polynomial with Roots 3 and 1-i

[saywhaaaat]

Homework Statement

Write a fourth degree polynomial that has roots of 3 and 1-i. There is more than one correct solution

Homework Equations

The Attempt at a Solution

I'm extremely lost as to where this problem is going, I know that to be a fourth degree its simply x^4, but how in the world do you incorporate i into a polynomial when you're given zeros?
Do you multiply 3 and 1-i, and keep foiling until there is a fourth degree? because that makes absolutely no sense. Sorry if i sound dumb guys

now see if it was more along the lines of a coefficient with the i such as 3i etc. I would get it, but 1-i throws me off completely

 

[SiddharthM]

1-i is a complex number, and there is a theorem that complex roots come in conjugate pair so a generic such polynomial is:

(x-(1-i))(x-(1+i))(x-a)(x-3)

Note that a must be real, for if it wasn't 3 would have to be a complex conjugate of a.

When given zeroes of a polynomial, a,b,c,...,z remember you can write the polynomial as:
(x-a)(x-b)...(x-z)=p(x)

I'm pretty sure this is what you are asking for.

Cheers,
Siddharth M

 

[HallsofIvy]

I know that to be a fourth degree its simply x^4

You certainly do NOT know that. I suspect you meant to say that the highest power must be x⁴ but there may be lower powers of x as well.

Well, you have learned to solve polynomial equations by factoring them haven't you?

You know that you can solve x²+ 5x+ 6= 0 by factoring: (x+ 3)(x+ 2)= 0 so x= -3 and x= -2 are solutions.

It works the other way too. The only quadratic polynomial with zeroes -3 and -2 is (x-(-3))(x-(-2))= (x+3)(x+2). If you were asked for a 4^th degree polynomial having those zeroes, you would simply multiply (x+3)(x+2)(x-a)(x-b) where a and b are any numbers you want.

If you want 3 and 1- i to be zeroes of a fourth degree polynomial, you must have something of the form (x- 3)(x-(1-i))(x-a)(x-b) for some numbers a and b. Choose any values you want for a and b- that's why "There is more than one correct solution".

there is a theorem that complex roots come in conjugate pair

No, there isn't. There is a theorem that says that if a polynomial with real coefficients has complex roots they must be in conjugate pairs.

If you want the coefficients to be real (although you don't say that), you must also include the "complex conjugate" of 1- i which is 1+i. That is you need (x- 3)(x-(1-i))(x-(1+i))(x-a) where, again, a can be any (real) number you wish.

(Strictly speaking, an equation has "roots". A polynomial or other function has "zeroes": values for which the function value is 0.)