Given that 3-2i is a zero of x^4 - 6x^3 + 8x^2 + 30x - 65, find all zeroes.

[sava]

I don't know if this means there is no solution or if I'm doing something wrong

 

[topsquark]

View attachment 35285

Well, the first thing you need to do is the correct the division. If 3 - 2i is a root (it is), then your number below the -65 should be 0.

I would do it this way: you already know that 3 + 2i is another root, so multiply out (x - 3 + 2i)(x - 3 - 2i) and divide your polynomial by that and see what happens

-Dan

 

[Harry_the_cat]

Not really sure what you are doing there, but if 3-2i is a zero then so is the conjugate 3+2i.

 

[sava]

Not really sure what you are doing there, but if 3-2i is a zero then so is the conjugate 3+2i.

so i should be starting the division with 3+2i instead of what im doing?

 

[sava]

Well, the first thing you need to do is the correct the division. If 3 - 2i is a root (it is), then your number below the -65 should be 0.

I would do it this way: you already know that 3 + 2i is another root, so multiply out (x - 3 + 2i)(x - 3 - 2i) and divide your polynomial by that and see what happens

-Dan

like this?

 

[Harry_the_cat]

You've gone around in circles here.

Divide the original quartic by the quadratic \(\displaystyle x^2 - 6x +13\) to find the OTHER zeroes.

 

[Steven G]

You should do the multiplication this way:
(x-(3-2i))(x-(3+2i)) = ((x-3)+2i)((x-3)-2i) Now you have complex conjugates!
So ((x-3)+2i)((x-3)-2i) = (x-3)^2 + 2^2 = x^2 -6x +13

 

[Steven G]

You said that

I don't know if this means there is no solution or if I'm doing something wrong.​

Every fourth degree polynomial has 4 roots (including multiple roots). In fact every nth degree polynomial has n roots.