Question 108149
<pre><font size = 4><b>
For the function
h(x) = x³ - x² - 17x - 15,
Use long division to determine which of the following are factors of h(x).
a) x + 5 b) x + 1 c) x + 3 

Divide x³ - x² - 17x - 15 by x + 5 to see if (a) is a factor of h(x)

            x² -  6x + 13
     --------------------
x + 5)x³ -  x² - 17x - 15
      x² + 5x²
      --------
         - 6x² - 17x
         - 6x² - 30x
         -----------
                 13x - 15
                 13x + 65
                 --------
                     - 80

No, (a) is not a factor because we get a remainder of -80,
not 0.

So, we divide x3 - x2 - 17x - 15 by x + 1 to see if (b) is a factor
of h(x)

            x² -  2x - 15
     --------------------
x + 1)x³ -  x² - 17x - 15
      x² +  x²
      --------
         - 2x² - 17x
         - 2x² -  2x
         -----------
               - 15x - 15
               - 13x - 15
                 --------
                        0

Yes, (b) is a factor of h(x) because we get a remainder of 0.


Divide x3 - x2 - 17x - 15 by x + 5 to see if (c) is a factor
of h(x).

            x² -  4x -  5
     --------------------
x + 3)x³ -  x² - 17x - 15
      x² + 3x²
      --------
         - 4x² - 17x
         - 4x² - 12x
         -----------
               -  5x - 15
               -  5x - 15
                 --------
                        0

Yes, (c) is a factor of h(x) because we get a remainder of 0.

So (b) and (c) are factors and (a) is not a factor.

------------------------------------------------------------

Problem #2
Determine the oblique asymptote of the graph of the function.

g(x)= (x² + 4x - 1)/(x + 3)

We divide the expression on the right using long division:


            x + 1
     ------------
x + 3)x² + 4x + 1
      x² + 3x
      -------
            x + 1 
            x + 3
            -----
              - 2

Now use h(x) = {{{(DIVIDEND)/(DIVISOR)}}} = {{{QUOTIENT}}} + {{{(REMAINDER)/(DIVISOR)}}}  

        h(x) = x + 1 + {{{(-2)/(x+3)}}}

The fraction on the right becomes nearer and nearer to zero as x takes
on values with larger and larger absolute values.  So therefore the graph
of h(x) becomes nearer and nearer to the graph of the right hand side
without the {{{(REMAINDER)/(DIVISOR)}}} term, so the oblique (slanted)
asymptote has the equation:

          y = x + 1 

Here is the graph of h(x) and the green line is the oblique asymptote,
whose equation is y = x + 1

[The blue line is the vertical asymptote, whose equation is x = -3
gotten by setting the denominator x+3 equal to 0.]

{{{drawing(400, 400, -10,10,-10,10,graph(400, 400, -10,10,-10,10,((x^2+4x-1)/(x+3))*sqrt(3+x)/sqrt(3+x),x+1,999(x+3)),
 graph(400, 400, -10,10,-10,10,((x^2+4x-1)/(x+3))*sqrt(-3-x)/sqrt(-3-x),x+1)

 )}}}
 
Edwin</pre>