Expert: Scott A Wilson - 4/30/2010

Question
Please help me!
1) Reduce the fraction -20t^5 u^2 v^3 over 48t^7 u^4 v
2) Reduce the fraction w^2+5w+6 over w^2-w-12
3)Find the illegal values of b in the fraction 2b^2 + 3b-10 over b^2-2b-8
4) Multiply 3xyz^2 over 6y^4 by 2y over xz^4
5)Multiply r^2+7r+10 over 3 by 3r-30 over r^2-5r-50
6)Find the illegal values of c in the multiplication statement c^2-3c-10 over c^2+5c-14 times c^2-c-2 over C^2 -2c-15
7)Divide m^2 n^3 over p^3 by mp over n^2
8) Divide r+5 over r^2+5r-14 by r^2=4r-21 over r-2

Answer
1) Reduce (-20t^5 u^2 v^3)/(48t^7 u^4 v).
It is known that -20/48 =  -5/12, t^5/t^7 = 1/t^2, u^2/u^4 = 1/u^2, and v^3/v = v^2.
Multiplying these all together gives -5v²/(12t²u²).

2) Reduce (w^2+5w+6)/(w^2-w-12).
Factor at first giving [(x+3)(x+2)]/[(x-4)(x+3)].
Divide the numerator and the denominator by (x+3),
so they both disappear, and what is left is the answer.

3) Find the illegal values of b in (2b^2 + 3b-10)/(b^2-2b-8) .
The illegal values have nothing to do with the numerator in this case since the numerator is a polynomial of degree 2, and polynomials are defined for all x.    The critical values are the
values of (b²-2b-8) where it is 0.  Factoring (b²-2b-8) gives us (b-4)(b+2).  The critical values
are points where b-4 = 0 and b+2 = 0.  To get them, add or subtract the constant from both sides.

4) Multiply (3xyz^2)/(6y^4) by (2y)/(xz^4).
Reducing the 1st fraction gives (xz²)/(2y³).
Multiplying them gives (2xyz^2)/(6xy^4z^4).
Note that 2/6 = 1/3, x/x = 1, y/y^4 = 1/y^3, and z^2/z^4 = 1/z^2.
Multiplying this four expressions together again gives 1/(y^3 z^2)

5) Multiply (r^2+7r+10)/3 by (3r-30)/(r^2-5r-50).
The 1st fraction is (r+5)(r+2)/3.
The 2nd one is 3(r-10)/[(r-10)(r+5) = 3/(r+5).
Mulitplying the gets the expression [(r+5)/3][3/(r+5)] to come up, which is 1.
All that is left is r+2.

6) Find the illegal values of c in the multiplication statement
[(c^2-3c-10)/(c^2+5c-14)][(c^2-c-2)/(c^2-2c-15)].
Factoring the 1st fraction gives [(c-5)(c+2)]/[(c+7)(c-2)].
Factoring the 2nd fraction gives [(c-2)(c+1)]/[(c-5)(c+3)].
Multiplying them gives a (c-5)/(c-5), which is 1 and a [1/(c-2)]/(c-2), which is 1.
This leaves [(c+2)(c+1)]/[(c+7)(c+3)].

7) Find [(m^2 n^3)/p^3]/[mp/n^2].
To divide fraction, we invert and multiply.
This gives [(m^2 n^3)/p^3][n^2/(mp)].
These combine to (m^2 n^5)/(m p^4).

To find the answer, note that m^2/m = m,
so it is what is gotten when we take m  times n^5 / p^4.

8) Divide r+5 over r^2+5r-14 by r^2=4r-21 over r-2
Note that the 1st fraction is (r+5)/[(r+7)(r-2)].
Note that the 2nd fraction is (r+7)(r-3)/(r-2) { if that = is suppose to be a + }.
To divide, invert the 2nd fraction and multiply.
This gives {(r+5)/[(r+7)(r-2)]}{(r-2)/[(r+7)(r-3)]}.

Note that there is a [(1/(r-2)](r-2), which is 1.
This leaves (r+5)/[(r+7)²(r-3)].

{ By the way, the reason I assumed the '=' was suppose to be a '+', was I think the shift key wasn't pressed down as far as it needed to be to get the '+' when the '=' was hit. }