How do you use synthetic division to divide #x^3- 6x^2+ 11x - 6 # by #x-1#?

2 Answers

Answer:

Synthetic division is generally used, in finding zeroes (or roots) of polynomials.

Explanation:

Well since the polynomial #x^3-6x^2+11x-6=(x-1)(x-2)(x-3)#

Thus the quotient of division with #x-1# is #(x-2)(x-3)# and the remainder is #0#

Sep 11, 2015

Answer:

See the explanation section.

Explanation:

Divide #x^3 -6x^2 +11 x - 6 # by #x-1#.

First, you let the coefficients of each degree to be used in the division (#1, -6, 11, -6#).

Then, dividing by #x-1 # implies that you use #1# in your upper left. So, put #1#, then a line (I've used a double line), and then write #" 1" " -6" " 11" " -6 "# to the right.

#"1 " ||# # " 1" " -6" " 11" " -6 "#
#+#
#" " " "##-----#

First, bring the first #1# down to the bottom, and multiply it by the #1#. Put that #1# below the #-6#.

#"1 " ||# # " 1" " -6" " 11" " -6 "#
#+# #" "" " " 1"#
#" " " "##-----#
#" "" " " 1"#

Then add #-6+1 = -5# Put that under the #1#

#"1 " ||# # " 1" " -6 " " 11" " -6 "#
#+# #" "" " " 1"#
#" " " "##-----#
#" " " " " 1" " -5"#

Multiply #1 xx -5 = -5# and put the #-5# under the #11#. Then add:

#"1 " ||# # " 1" " -6 " " 11 " " -6 "#
#+# #" "" " " 1 " " -5"#
#" " " "##--------#
#" " " " " 1 " " -5 " " 6"#

Now #1 xx 6 = 6#, so we put #6# under #-6# and add::

#"1 " ||# # " 1" " -6 " " 11 " " -6 "#
#+# #" "" "" 1 " " -5 " " 6"#
#" " " "##--------#
#" "" " " 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)#).

So the division gives us:

#(x^3 -6x^2 +11 x - 6) /(x-1) = x^2-5x+6#

You can check the answer by multiplyng:

#(x-1)( x^2-5x+6)# to make sure we get #x^3 -6x^2 +11 x - 6#.

(I've used Synthetic Division Formatting by Truong-Son R.)