Posted by Maddison on Thursday, April 3, 2008 at 8:00pm.
I have to simplify these please help im not sure how
things in parenthesis are exponets
x(2)+8x12 over
x2
2a(3)+a(2)3a over
6a(3)+5a(2)6a
3x(2)+2x1 over
x(2)+3x+2
a(2)11a+30 over
a(2)9a+18
2y(3)12y(2)+2y over
y(2)6y+1

Algebra  Anonymous, Thursday, April 3, 2008 at 8:13pm
Use ^ and then the number to designate powers.
x^2 + 8x  12
1(x^2  8x + 12)
1(x6)(x2)
The x2 in the denominator can canceled with the x2 in the numerator and you are left with 1(x6).
(2a^3 + a^2  3a)/(6a^3 + 5a^2  6a)
Factor out a out of the numerator and denominator and you are left with
(2a^2 + a  3)/(6a^2 + 5a  6)
Factor
((2x+3)(x1))/((2x+3)(3x2))
2x+3 in the numerator and denominator cancels out and you are left with (x1)/(3x2).
All of those problems can be done with factoring of the numerator and denominator and cancel out the common factors.