-3 | 1 0 0 0 0 243 [tex]\longleftarrow[/tex] coefficients of the polynomial you're dividing
. | [tex]\longleftarrow[/tex] drop down the leading coefficient
- - - - - - - - - - - - - - - - - - -
. | 1
On the left side of the frame, we write -3 because we're dividing by [tex]x+3=x-(-3)[/tex]. (The algorithm is followed for division of a polynomial by a factor of [tex]x-c[/tex].) Since we're dividing a degree 5 polynomial by a degree 1 polynomial, we expect to get a degree 4 polynomial.
-3 | 1 0 0 0 0 243
. | -3 [tex]\longleftarrow[/tex] multiply -3 by 1, write in next column, add to 0
- - - - - - - - - - - - - - - - - - -
. | 1 -3
Repeat step for the remaining columns.
-3 | 1 0 0 0 0 243
. | -3 9
- - - - - - - - - - - - - - - - - - -
. | 1 -3 9
-3 | 1 0 0 0 0 243
. | -3 9 -27
- - - - - - - - - - - - - - - - - - -
. | 1 -3 9 -27
-3 | 1 0 0 0 0 243
. | -3 9 -27 81
- - - - - - - - - - - - - - - - - - - - -
. | 1 -3 9 -27 81
-3 | 1 0 0 0 0 243
. | -3 9 -27 81 -243
- - - - - - - - - - - - - - - - - - - - -
. | 1 -3 9 -27 81 0
which translates to
[tex]\dfrac{x^5+243}{x+3}=x^4-3x^3+9x^2-27x+81[/tex]
So the bottom row of the frame gives the coefficients of each term in the quotient by descending order. Since the last coefficient is 0, this means the remainder upon division vanishes, i.e. [tex]x^5+243[/tex] is exactly divisible by [tex]x+3[/tex].
- - -
Another way to get the same result is to use a well-known result: for [tex]a+b\neq0[/tex],
[tex]a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)\implies\dfrac{a^5+b^5}{a+b}=a^4-a^3b+a^2b^2-ab^3+b^4[/tex]
and in this case [tex]a=x[/tex] and [tex]b=3[/tex]
