Write the function in the form f(x) = (x − k)q(x) + r(x) for the given value of k. Use a graphing utility to demonstrate that f(k) = r. f(x)

x-k = x+2/3   |  k = -2/3

$$\\(15x^4+10x^3-15x^2+11):(x+\frac{2}{3})=\underbrace{15x^3-15x+10}_{q(x)}+\underbrace{\frac{13}{3}\times\frac{1}{x+\frac{2}{3}}}_{p(x)}\\\\ \frac{(15x^4+10x^3-15x^2+11)}{x+\frac{2}{3}} = q(x)+p(x)$$

$$15x^4+10x^3-15x^2+11= (x+\frac{2}{3})q(x)+\underbrace{(x+\frac{2}{3}) p(x)}_{r(x)=\frac{13}{3}}$$

k=-2/3 -> f(-2/3) = r(x) = 13/3

This appears to be a problem in synthetic division/substitution.  I'll provide the steps for the problem, then an explanation for the steps.  First we want to divide the polynomial f(x) by (x+2/3). That will look like:

-2/3     15     10     -15     0     11

                  -10      0      10   -20/3

          15     0    -15      10    13/3

Of the three lines the fist is the setup. I have stripped off the coefficients from f(x).  Please note that the zero in the first line is a placeholder for the linear term which appears to be absent in f(x) but is important to include in our calculations.  The way we do this is simply pull the 15 down to line 3, multiply by -2/3 and move it up and right one space.  That is where the -10 comes from.  We then add 10 to -10 and write the result,0, below.  This process repeats until we have a complete row 3.  With out five numbers from row three we are ready to write our factor out. 

q(x)= 15x^3 +0x^2 -15x+10    The final number, 13/3, is the remainder which for us is r. 

Therefore the final answer to this is f(x)=(x-k)q(x)+r(x)=(x+2/3)(15x^3 -15x+10)+13/3