Posted by George on Wednesday, December 31, 2008 at 5:40pm.
Given a polynomial and one of its factors, find the remaining factors of the polynomial. Some of the factors may not be binomials.
169x^31690x^29x+90; x10

Math  drwls, Wednesday, December 31, 2008 at 6:07pm
This is an exercise in polynomial long division. Divide 169x^3 1690x^2 9x +90 by x10 to get the quotient, which is 169 x^2  9. That happens to be the difference of the squares of 13x and 3. So it is possible to factor the quotient one more time.
If you need to review polynomial long division, see
http://www.sosmath.com/algebra/factor/fac01/fac01.html
This division turns out to be an easy one, with no remainder.

Math  George, Wednesday, December 31, 2008 at 6:17pm
Thanks!

Math  drwls, Thursday, January 1, 2009 at 5:51am
The complete answer is
169x^3  1690x^2  9x + 90
= (x10)(13x + 3)(13x  3)
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