SOLUTION: factor completely over the integers: 121-64x^2-36y^2+96xy I've tried pairing the integers in different ways and still can't solve this. Please help. Thank you.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: factor completely over the integers: 121-64x^2-36y^2+96xy I've tried pairing the integers in different ways and still can't solve this. Please help. Thank you.      Log On


   

Question 21916: factor completely over the integers:
121-64x^2-36y^2+96xy
I've tried pairing the integers in different ways and still can't solve this.
Please help.
Thank you.
Found 2 solutions by AnlytcPhil, venugopalramana:
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
factor completely over the integers:

121 - 64x2 - 36y2 + 96xy

I've tried pairing the integers in different ways and still
can't solve this.
Please help.

Thank you.
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Sometimes pairing doesn't work and you have to try grouping 
three factors:

121 - 64x2 - 36y2 + 96xy

Interchange the last two terms

121 - 64x2 + 96xy - 36y2

Factor -4 out of the last three terms

121 - 4(16x2 - 24xy + 9y2)

Factor the expression in the parentheses

121 - 4(4x-3y)(4x-3y)

This is now the difference of two perfect squares

112 - 22(4x-3y)2

[11 - 2(4x-3y)][11 + 2(4x-3y)]

Remove the parentheses inside the brackets

[11-8x+6y][11+8x-6y]

Change the brackets to parentheses

(11-8x+6y)(11+8x-6y)

Edwin
AnlytcPhil@aol.com

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
121-64x^2-36y^2+96xy
=11*11-{(8X)^2+(6Y)^2-2*(8X)(6Y)}
=11^2-(8X-6Y)^2........................USING...(A-B)^2=A^2+B^2-2AB
(11+8X-6Y){11-(8X-6Y)}.................USING.... A^2-B^2=(A+B)(A-B)
(11+8X-6Y)(11-8X+6Y)