Question 1143229
consider 

{{{x^3 − kx + (k + 11) = 0}}}

Use long division to show that {{{k = x^2 + x + 1 + 12/(x-1)}}}
<pre>
When we divide polynomials we write the answer

{{{matrix(1,3,QUOTIENT, ""+"", REMAINDER/DIVISOR)}}}

So QUOTIENT = {{{x^2+x+1}}}, REMAINDER = 12, DIVISOR = x-1,

we know that we are to divide by x-1

     <u>        x² +       x +    (-k+1) </u>
x - 1) x³ + 0x² -      kx +    (k+11)
       <u>x³ -  x²</u> 
             x² -      kx
             <u>x² -       x</u> 
                  (-k+1)x +    (k+11)
                  <u>(-k+1)x -    (-k+1)</u>
                            (k+11)+(-k+1) = k+11-k+1 = 12


I get the remainder 12, but the last term in the quotient
is (-k+1), whereas the last term in the quotient you gave
was 1.  Did you leave out the -k?  Or does it mean that -k+1 = 1
and k = 0.  If k = 0, then the long division becomes


     <u>        x² +  x +  1</u>
x - 1) x³ + 0x² - 0x + 11
       <u>x³ -  x²</u> 
             x² - 0x
             <u>x² -  x</u> 
                   x + 11
                   <u>x -  1</u>
                       12

Edwin</pre>