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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

Originally Posted by Unknown008

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

Originally Posted by Unknown008

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