How do you use synthetic substitution to evaluate f(-2) for #f(x)=x^4-4x^3-4x+6#?

1 Answer
Dec 16, 2015

Answer:

(see below for method of evaluation)
#f(-2)=62#

Explanation:

Evaluation of #f(x)=color(blue)(1)x^4color(blue)(-4)x^3color(blue)(+0)x^2color(blue)(-4)xcolor(blue)(+6)#
for #f(color(red)(-2))#
Notice the expansion of the expression to include the implied coefficients of #color(blue)(1)# for #x^4# and of #color(blue)(0)# for #x^2#

Initial set-up:
#{: (,"|",color(blue)(1),color(blue)(-4),color(blue)(+0),color(blue)(-4),color(blue)(+6),color(white)("XXXXXXXX")"line [1]"), (,"|",,,,,,color(white)("XXXXXXXX")"line[2]"), ("----",,"----","----","----","----","----",), (xxcolor(red)((-2)),"|",color(green)(1),,,,,color(white)("XXXXXXXX")"line [3]") :}#

For each column

  • Multiply the last number written on line [3] by #color(red)((-2))# and write the product on line [2] of the next column
  • Add the numbers in lines [1] and [2] of the next column and write the sum in line [3] of that column.

The number written in the last column of line [3] will be the value of #f(color(red)(-2))#

#{: (,"|",color(blue)(1),color(blue)(-4),color(blue)(+0),color(blue)(-4),color(blue)(+6),color(white)("XXXXXXXX")"line [1]"), (,"|",,-2,12,-24,56,color(white)("XXXXXXXX")"line[2]"), ("-----------",,"----","----","----","----","----",), (xxcolor(red)((-2)),"|",color(green)(1),-6,12,-28,color(cyan)(62),color(white)("XXXXXXXX")"line [3]") :}#