View Full Version : Factor Completely
usaboy94
Aug 24, 2009, 12:54 PM
3x^5 y^2 - 21x^2 y^7
Unknown008
Aug 24, 2009, 12:55 PM
What have you tried so far? I told you to try, remember? :rolleyes: tell me where you are stuck.
usaboy94
Aug 24, 2009, 01:00 PM
What have you tried so far? I told you to try, remember? :rolleyes: tell me where you are stuck.
I have tried subtracting the exponents, which didn't work. I already know the answer, it is 3x^2 y^2 (x^3 - 7y^5)
Unknown008
Aug 24, 2009, 01:08 PM
I don't quite understand what you mean by 'subtracting the exponents'...
Ok, this is how I'll teach it to you:
Factors of 3x^5y^2
3x^5y^2 = 3 \times x \times x \times x \times x \times x \times y \times y
And factors of
21x^2 y^7 = 3 \times 7 \times x \times x \times y \times y \times y \times y \times y \times y \times y
Group the common factors of 3x^5y^2 and 21x^2 y^7: You have 3, x, x, y, y, which makes 3x^2y^2
Now rewrite, with what remains:
(3x^2y^2)(x^3) - (3x^2y^2)(7y^5) = 3x^2 y^2 (x^3 - 7y^5)
If you expand (3x^2y^2)(x^3), you'll have 3x^5y^2 and if you expand (3x^2y^2)(7y^5), you'll have 21x^2 y^7.
Hope it helped! :)
usaboy94
Aug 24, 2009, 01:31 PM
I don't quite understand what you mean by 'subtracting the exponents'...
Ok, this is how I'll teach it to you:
Factors of 3x^5y^2
3x^5y^2 = 3 \times x \times x \times x \times x \times x \times y \times y
And factors of
21x^2 y^7 = 3 \times 7 \times x \times x \times y \times y \times y \times y \times y \times y \times y
Group the common factors of 3x^5y^2 and 21x^2 y^7: You have 3, x, x, y, y, which makes 3x^2y^2
Now rewrite, with what remains:
(3x^2y^2)(x^3) - (3x^2y^2)(7y^5) = 3x^2 y^2 (x^3 - 7y^5)
If you expand (3x^2y^2)(x^3), you'll have 3x^5y^2 and if you expand (3x^2y^2)(7y^5), you'll have 21x^2 y^7.
Hope it helped! :)
Thanks i understand now. :D