Expert: Ahmed Salami - 11/16/2009

Question
Given: root 3 and 5-2i are roots of the equation x^5-9x^4+16x^3+56x^2-117x-87=0.  Find the other 3 roots.
I know that 5+2i is also a root, so I multiplied (x-root 3)((x-5)-2i)((x-5)+2i) and I got x^3-10x^2+29x-root3*x^2+10root3*x - 29root3.  I know that I am supposed to divide x^5-9x^4+16x^3 +56x^2-117x-87 by the product of the roots in order to find what is left, but the root 3 complicates things a lot.  How would I divide these 2?  Or did I do something wrong?

Answer
Hi Katharine,
You havent done anything wrong (except maybe making a mistake in the presentation of the expression).
Except that i'm missing something here, but i dont think the equation
x^5 - 9x^4 + 16x^3 + 56x^2 - 117x - 87 = 0
even has a root of √3 !!!
Its real roots are -2.223, -0.6084 and 2.8608
Looking away from the supposed mistake and focusing on the method of solution which, by the way, i believe is more important.
You could find the quadratic equation whose roots are 5-2i and 5+2i. The 5th order equation can then be reduced to a cubic equation and then its easier to find the other roots.
[x - (5-2i)][x - (5+2i)] = x^2 - 10x + 29
Its now easier to divide x^5 - 9x^4 + 16x^3 + 56x^2 - 117x - 87 by x^2 - 10x + 29, but then you get a quotient of x^3 + x^2 - 3x - 3 and a remainder of 60x which seems to suggest that even 5-2i and 5+2i are not roots of the equation as well!!!
Check to see that you've written everything correctly and get back to me.

Regards.