Posted by shansen1@pacbell.net (66.126.108.253) on October 07, 2003 at 00:47:13:
In Reply to: division by monomials posted by on October 06, 2003 at 22:43:43:
For clarity, the expressions are restated using ASCII notation convention
* means multiply
/ means divide
^ means exponent
Problem 1)
[3(x-y)-3x(x-y)^2+4x^2(x-y)^3]/(x-y)
Note that each term in this expression has a common factor (x-y).
The associative/distributive law of algebra states that ab + ac = a(b + c) We can use this law to rewrite the expression in factored form as follows:
(x-y)[3-3x(x-y)+4x^2(x-y)^2]/(x-y)
Note that the term (x-y)/(x-y) = 1 ("cancels out") leaving:
3-3x(x-y)+4x^2(x-y)^2
Problem 2)
[a^(2x+4)*b^(z+2)-a^(2x+3)*b^(z+4)+a^(2x+2)*b^(z+6)-a^(2x+1)*b(z+8)]/a^2x*b^(z+1)
This is an application of the law(s) of exponents, to wit:
a^b/a^c = a^(b-c)
a^b*a^c = a^(b+c)
(a^b)^c = a(b*c)
In your problem, both a and b appear as factors of every term of the expression. The exponents of the a and b terms are computed by subtracting the exponent of the respective denominator term from the exponent of the numerator term. The expression then simplifies to:
a^4*b-a^3*b^3+a^2*b^5-a*b^7