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September 30, 2014

September 30, 2014

Posted by **JoAnn** on Tuesday, August 12, 2008 at 4:13pm.

x^10y^3 – 4x^9y^2 – 21x^8y

- Algebra -
**Damon**, Tuesday, August 12, 2008 at 5:00pmwe 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 -
**JoAnn**, Wednesday, August 13, 2008 at 4:51pmThanks a lot, this really helps!

- Algebra -
- Algebra -
**drwls**, Tuesday, August 12, 2008 at 5:01pmStart 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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