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

Question

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

Answer 1

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

 

 

 

 

 

Answer 2

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)

Answer 3

(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.