Posted by JoAnn on Tuesday, August 12, 2008 at 4:13pm.
Factor completely:
x^10y^3 – 4x^9y^2 – 21x^8y

Algebra  Damon, Tuesday, August 12, 2008 at 5:00pm
we can immediately take out an x^8 and a y^1
x^8 y ( x^2 y^2  4 x y  21 )
21 = 3 * 7
3  7 = 4
so try
x^8 y ( x y  7 ) ( x y + 3 )

Algebra  drwls, Tuesday, August 12, 2008 at 5:01pm
Start by removing the common factor x^8 y.
x^10y^3 – 4x^9y^2 – 21x^8y
= x^8*y(x^2 y^2 4 xy 21)
The term in parentheses is a quadratic in the variable xy, and can be factored to
(xy 7)(xy + 3).
Now put it all together
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