SOLUTION: I just need this multiplied and simplified. The hashed line is supposed to be a solid line. (4c² - 100)(16c + 48) ----------- ---------- (8c² - 72) (2c – 10)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I just need this multiplied and simplified. The hashed line is supposed to be a solid line. (4c² - 100)(16c + 48) ----------- ---------- (8c² - 72) (2c – 10)       Log On


   



Question 83230: I just need this multiplied and simplified. The hashed line is supposed to be a solid line.
(4c² - 100)(16c + 48)
----------------------
(8c² - 72) (2c – 10)

Found 2 solutions by Edwin McCravy, checkley75:
Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!

I just need this multiplied and 
simplified. The hashed line is 
supposed to be a solid line. 

 (4c² - 100)(16c + 48)
-----------------------
 (8c² - 72) (2c – 10) 

You don't multiply, you just factor everything 
and cancel!


Factor 4c² - 100
       4(c² - 25)
       4(c - 5)(c + 5)

Factor 16c + 48
       16(c + 3)

Factor 8c² - 72
       8(c² - 9)
       8(c - 3)(c + 3)

Factor 2c - 10
       2(c - 5)

Now replace each parentheses by its factorization:

       4(c - 5)(c + 5)16(c + 3)
      --------------------------
       8(c - 3)(c + 3)2(c - 5)

Multiply the 4 by the 16 in the top, getting 64
Multiply the 8 by the 2 in the bottom, getting 16

       64(c - 5)(c + 5)(c + 3)
      -------------------------
       16(c - 3)(c + 3)(c - 5)

Cancel the 16 into 64 and get 4 on top:

       4    
       64(c - 5)(c + 5)(c + 3)
      -------------------------
       16(c - 3)(c + 3)(c - 5)
       1

Cancel the (c - 5)'s

       4    1   
       64(c - 5)(c + 5)(c + 3)
      -------------------------
       16(c - 3)(c + 3)(c - 5)
       1                  1

Cancel the (c + 3)'s

       4    1             1  
       64(c - 5)(c + 5)(c + 3)
      -------------------------
       16(c - 3)(c + 3)(c - 5)
       1           1      1 

Now all that's left is

        4(c + 5)
       ---------
         c - 3
Edwin

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
(4C^2-100)(16C+48)/(8C^2-72)(2C-10)
4(C^2-15)16(C+3)/8(C^2-9)2(C-5)
(4*16)/(8*2)=4 NOW MULTIPLY THE TERMS IN THE () THUS:
C^2-15
C+3
---------
C^3-15C+3 &
C^2-9
C-5
-----------
C^3-9C-5 NOW COMBINE ALL TERMS & DIVIDE IF POSSIBLE.
4(C^3-15C+3)/(C^3-9C-5)