Divide x^3 -6x^2 +11 x - 6 x3−6x2+11x−6 by x-1x−1.
First, you let the coefficients of each degree to be used in the division (1, -6, 11, -61,−6,11,−6).
Then, dividing by x-1 x−1 implies that you use 11 in your upper left. So, put 11, then a line (I've used a double line), and then write " 1" " -6" " 11" " -6 " 1 -6 11 -6 to the right.
"1 " ||1 ∣∣ " 1" " -6" " 11" " -6 " 1 -6 11 -6
++
" " " " -----−−−−−
First, bring the first 11 down to the bottom, and multiply it by the 11. Put that 11 below the -6−6.
"1 " ||1 ∣∣ " 1" " -6" " 11" " -6 " 1 -6 11 -6
++ " "" " " 1" 1
" " " " -----−−−−−
" "" " " 1" 1
Then add -6+1 = -5−6+1=−5 Put that under the 11
"1 " ||1 ∣∣ " 1" " -6 " " 11" " -6 " 1 -6 11 -6
++ " "" " " 1" 1
" " " " -----−−−−−
" " " " " 1" " -5" 1 -5
Multiply 1 xx -5 = -51×−5=−5 and put the -5−5 under the 1111. Then add:
"1 " ||1 ∣∣ " 1" " -6 " " 11 " " -6 " 1 -6 11 -6
++ " "" " " 1 " " -5" 1 -5
" " " " --------−−−−−−−−
" " " " " 1 " " -5 " " 6" 1 -5 6
Now 1 xx 6 = 61×6=6, so we put 66 under -6−6 and add::
"1 " ||1 ∣∣ " 1" " -6 " " 11 " " -6 " 1 -6 11 -6
++ " "" "" 1 " " -5 " " 6" 1 -5 6
" " " " --------−−−−−−−−
" "" " " 1 " " -5 " " 6 " |"0 " 1 -5 6 ∣0
The bottom row ignoring the last number gives us the coefficients of the quotient.
The last number on the bottom row is the remainder (and it is also P(1)P(1)).
So the division gives us:
(x^3 -6x^2 +11 x - 6) /(x-1) = x^2-5x+6x3−6x2+11x−6x−1=x2−5x+6
You can check the answer by multiplyng:
(x-1)( x^2-5x+6)(x−1)(x2−5x+6) to make sure we get x^3 -6x^2 +11 x - 6x3−6x2+11x−6.
(I've used Synthetic Division Formatting by Truong-Son R.)
