Find a polynomial equation with integer coefficients that has the given numbers as solutions and the indicated degree
1 - i, 7i; degree 4

a) x^4+2x^3+51x^2-98x-98=0
b) x^4-2x^3+51x^2-98x-98=0
c) x^4-2x^3+51x^2-98x+98=0
d) x^4-2x^3-51x^2-98x-98=0
e) x^4+2x^3+51x^2+98x+98=0

Question

Find a polynomial equation with integer coefficients that has the given numbers as solutions and the indicated degree
1 - i, 7i; degree 4

a) x^4+2x^3+51x^2-98x-98=0
b) x^4-2x^3+51x^2-98x-98=0
c) x^4-2x^3+51x^2-98x+98=0
d) x^4-2x^3-51x^2-98x-98=0
e) x^4+2x^3+51x^2+98x+98=0

anonymous · Answer

What exactly do you need help with? Try replacing x for the given roots 1 - i and 7i in the equations and see which one is a true equality

anonymous · Answer

yeah, I've tried that. I don't know what to do

anonymous · Answer

post your work and maybe I can help you find where you did mistakes

anonymous · Answer

I don't even have an idea how to do it

kinggeorge · Answer

Since you have roots $1-i$ and $7i$, you must also have roots $1+i$ and $-7i$. This is an important property of the complex numbers you should probably remember for your class.

Now we have 4 different roots, so we can just find what $$(x-(1-i))(x-(1+i))(x-7i)(x+7i)$$is equal to, and you should have a degree 4 polynomial with integer coefficients.

kinggeorge · Answer

The specific property says that if you have a root $a+bi$, you must also have a root $a-bi$ if your polynomial is in the real numbers.

paxpolaris · Answer

$$\implies \left( x^2-2x+2 \right)\left( x^2+49 \right)=0$$