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What is 3x-12 divided by x^2-5x Then TIMES this by 7x/4-x.... Thank You!

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3 Answers

Hi Abby;
3x-12 divided by x^2-5x Then TIMES this by 7x/4-x
[(3x-12)/(x2-5x)][(7x)/(4-x)]
 
Let's multiply the numerators together...
(3x-12)(7x)
21x2-84x
21(x2-4x)
Let's multiply the denominators together...
(x2-5x)(4-x)
Let's FOIL...
FIRST...(x2)(4)=4x2
OUTER...(x2)(-x)=-x3
INNER...(-5x)(4)=-20x
LAST...(-5x)(-x)=5x2
-x3+4x2+5x2-20x
-x3+9x2-20x
Let's try to factor this such that one of the parenthetical equations is (x2-4x)
(x2-4x)(...
FIRST must be (x2)(-x)=-x3
OUTER and INNER must add-up to 9x2.
LAST must be (-4x)(5)=-20x
(x2-4x)(-x+5)
Let's FOIL...
FIRST...(x2)(-x)=-x3
OUTER...(x2)(5)=5x2
INNER...(-4x)(-x)=4x2
LAST...(-4x)(5)=-20x
-x3+5x2+4x2-20x
-x3+9x2-20x
 
The numerator is... 21(x2-4x)
The denominator is... (x2-4x)(-x+5)
(x2-4x) cancels.  This is now...
21/(-x+5)
(3x-12)/(x^2-5x) * 7x/(4-x)
[My multiply sign is *.]
 
7x(3x-12)/((x^2-5x)(4-x))
 
21x(x-4)/(x(x-5)(-1)(x-4))
 
Notice the x/x and (x-4)/(x-4).
 
x/x = 1 for all real numbers EXCEPT 0, where it's UNDEFINED.
 
(x-4)/(x-4) = 1 everywhere EXCEPT 4, where it's UNDEFINED.
 
These are called "holes". We'll exclude them by stating x <> 0,4.
[My not-equal sign is <>.] Then we can cancel them:
 
-21/(x-5), x <> 0,4,5.
 
x <> 5 b/c it would cause division by zero.
 
In fact, x = 5 is a vertical asymptote.
The horizontal asymptote is y = 0.
3x-12  x  7x
x2-5x     4-x         
 
(3x-12)*(7x)      ------>  ____(3x)(7x) - (12)(7x)        
(x2-5x)*(4-x)             (x2)(4) + (x2)(-x) - (5x)(4) - (5x)(-x)
 
                         ------>    __21x2 - 84x      
                                     4x2 - x3 - 20x + 5x2
 
you're gonna want to clean this up a bit :)
 
  Final Answer    ------> __21x (x - 4)    
                                   -x3 + 9x2 - 20x