How do I find the quotient and remainder using synthetic division?
1 Answer
Let's take this as an example:
#f(x) = x^3 - x^2 + x - 6#
To find factors, notice that
I pick
#color(blue)((x^3 - x^2 + x - 6)/(x-2) = q(x) + r(x)# where
#q(x)# is the quotient and#r(x)# is the remainder.
For synthetic division, we only use coefficients. So, we start with:
#color(white)([(color(black)(2),color(black)(|),color(black)(1),color(black)(-1),color(black)(1),color(black)(-6)),(color(black)(-),color(black)(""),color(black)("_"),color(black)("_"),color(black)("_"),color(black)("_")),(color(black)(""),color(black)(""),color(black)(""),color(black)(""),color(black)(""),color(black)(""))])#
If you don't have a term, use
The general process is:
- Bring down the first coefficient.
- Multiply it by the test factor (in this case,
#2# ), and store the result under the next coefficient. - Subtract the result from this coefficient.
- Repeat steps 2 and 3 until you've subtracted the result from the last coefficient.
The result is one degree lower.
#=> color(white)([(color(black)(2),color(black)(|),color(black)(1),color(black)(-1),color(black)(1),color(black)(-6)),(color(black)(-),color(black)(""),color(black)("_"),color(black)(ul(2)),color(black)("_"),color(black)("_")),(color(black)(""),color(black)(""),color(black)(1),color(black)(-3),color(black)(""),color(black)(""))])#
#=> color(white)([(color(black)(2),color(black)(|),color(black)(1),color(black)(-1),color(black)(1),color(black)(-6)),(color(black)(-),color(black)(""),color(black)("_"),color(black)(ul(2)),color(black)(ul(-6)),color(black)("_")),(color(black)(""),color(black)(""),color(black)(1),color(black)(-3),color(black)(7),color(black)(""))])#
#=> color(white)([(color(black)(2),color(black)(|),color(black)(1),color(black)(-1),color(black)(1),color(black)(-6)),(color(black)(-),color(black)(""),color(black)("_"),color(black)(ul(2)),color(black)(ul(-6)),color(black)(ul(14))),(color(black)(""),color(black)(""),color(black)(\mathbf(1)),color(black)(\mathbf(-3)),color(black)(\mathbf(7)),color(black)(\mathbf(-20)))])#
So, our result is
#\mathbf(q(x) = x^2 - 3x + 7)#
#\mathbf(r(x) = -20/(x-2))#
If