Complex polynomial of degree n with n+1 roots?

ausmathgenius420

ausmathgenius420

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Hi,

The textbook question is: [imath]z^3+mz+52=0[/imath] and [imath]2+3i[/imath] is a solution, find m.
Steps to solve:
  1. Conjugate theorem gives another solution (2-3i)
  2. Convert to factor of z (i.e. (z-(2+3i)) is a factor)
  3. Multiply both factors and long divide to give: the third factor [math]z+4 r. (m+3)z[/math]
  4. m must be -3 thus (so remainder =0) [imath]z^3-3z+52=0[/imath]
My question relates to step 3/4. If z+4 is a factor, then a solution is Z=-4, (factor theorem).
However both 4 and -4 satisfy the polynomial, giving a total of 4 solutions?
 
Why does 4 and -4 satisfy the polynomial?

Are you saying if x=7 satisfies a polynomial then so does x=-7? That is not true at all. Clearly x=3 satisfies x-3=0, but x=-3 does not.

This may clear things up for you. The conjugate of 2+3i is 2-3i, while the conjugate of 4 = 4+0i is 4-0i = 4, not -4.
 
I must be missing something here, but why do you need to worry about the conjugates?
Can't you just plug in 2+3i into the equation and see if you can find [imath]m[/imath] which turns both the real and the imaginary parts of the resulting expression to 0?
 
ausmathgenius420 said:
Hi,

The textbook question is: [imath]z^3+mz+52=0[/imath] and [imath]2+3i[/imath] is a solution, find m.
Steps to solve:
  1. Conjugate theorem gives another solution (2-3i)
  2. Convert to factor of z (i.e. (z-(2+3i)) is a factor)
  3. Multiply both factors and long divide to give: the third factor [math]z+4 r. (m+3)z[/math]
  4. m must be -3 thus (so remainder =0) [imath]z^3-3z+52=0[/imath]
My question relates to step 3/4. If z+4 is a factor, then a solution is Z=-4, (factor theorem).
However both 4 and -4 satisfy the polynomial, giving a total of 4 solutions?
Click to expand...
It took me a while to understand "[imath]z+4 r. (m+3)z[/imath]". That is not a factor. What you mean is that the quotient is z+4, and the remainder is (m+3)z. So, yes, the remainder has to be 0, and m has to be -3.

But why do you say that 4 satisfies the equation? It's true that -4 does: [imath](-4)^3-3(-4)+52=-64+12+52=0[/imath]; but 4 does not: [imath](4)^3-3(4)+52=64-12+52=102[/imath].
 
Dr.Peterson said:
It took me a while to understand "[imath]z+4 r. (m+3)z[/imath]". That is not a factor. What you mean is that the quotient is z+4, and the remainder is (m+3)z. So, yes, the remainder has to be 0, and m has to be -3.

But why do you say that 4 satisfies the equation? It's true that -4 does: [imath](-4)^3-3(-4)+52=-64+12+52=0[/imath]; but 4 does not: [imath](4)^3-3(4)+52=64-12+52=102[/imath].
Click to expand...
Correct. I must have made an error in my algebra. Apologies all
 
Fine you made a mistake in your algebra, but my concern is that you thought that z=4 would be a solution since z=-4 is a solution.
Yes, complex roots occurs in conjugate pairs and the conjugate of 4 is 4 and the conjugate of -4 is -4. Why? 4 = 4+ 0i and the conjugate of 4=4+0i is 4-0i=4. Do you see that??
 
Steven G said:
Fine you made a mistake in your algebra, but my concern is that you thought that z=4 would be a solution since z=-4 is a solution.
Yes, complex roots occurs in conjugate pairs and the conjugate of 4 is 4 and the conjugate of -4 is -4. Why? 4 = 4+ 0i and the conjugate of 4=4+0i is 4-0i=4. Do you see that??
Click to expand...
I thought z=4 because a) I subbed in 4 and it equaled 0 (my algebra mistake) and b) the textbook answer wrongly says z=4
 
ausmathgenius420 said:
I thought z=4 because a) I subbed in 4 and it equaled 0 (my algebra mistake) and b) the textbook answer wrongly says z=4
Click to expand...
I was just looking out for you. Now I am clear why you plugged in 4. It was because your textbook told you to, rather than you thought about the conjugate of -4 being 4.
 
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