How do you divide #(18x^4+9x^3+3x^2)/(3x^2+1)# using synthetic division?

1 Answer
Apr 3, 2018

Answer:

#(18x^4+9x^3+3x^2)/(3x^2+1) = 6x^2+3x-1+(-3x+1)/(3x^2+1)#

Explanation:

Synthetic division is more commonly used with linear divisors and especially with monic linear divisors. It is possible to use a form of it with divisors of degree #> 1# as in this example, but it may be easier to use in the form of long division of coefficients...

#color(white)(30000010")")underline(color(white)(001)6color(white)(000)3color(white)(00)-1color(white)(00-000-1)#
#3color(white)(00)0color(white)(00)1color(white)(0)")"color(white)(00)18color(white)(000)9color(white)(00-)3color(white)(00-)0color(white)(00-)0#
#color(white)(30000010")"00)underline(18color(white)(000)0color(white)(00-)6)#
#color(white)(30000010")"0000000)9color(white)(00)-3color(white)(00-)0#
#color(white)(30000010")"0000000)underline(9color(white)(00-)0color(white)(00-)3)#
#color(white)(30000010")"0000000900)-3color(white)(00)-3color(white)(00-)0#
#color(white)(30000010")"000000090)underline(color(white)(0)-3color(white)(00-)0color(white)(00)-1)#
#color(white)(30000010")"0000000900-300)-3color(white)(00-)1#

The first important thing to note is the inclusion of #0#'s for missing powers of #x# in both the dividend #18x^4+9x^3+3x^2# represented as #18color(white)(0)9color(white)(0)3color(white)(0)0color(white)(0)0# and the divisor #3x^2+1# represented as #3color(white)(0)0color(white)(0)1#.

The actual process of long division is similar to long division of numbers.

In our example, we arrive at a quotient #6color(white)(0)3color(white)(0)-1# representing #6x^2+3x-1# and remainder #-3color(white)(0)1# representing #-3x+1#.