According to Remainder theorem, if a polynomial #f(x)# is divided by a monomial of degree one say #x-a#, the remainder is #f(a)#.
Hence, to evaluate #f(2)# for #x^3+2x^2-4x-5#, we should divided it by #x-2#
One Write the coefficients of #x# in the dividend inside an upside-down division symbol.
#color(white)(1)|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5#
#color(white)(1)|" "color(white)(X)#
#" "stackrel("—————————————)#
Two Put the divisor at the left.
#2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5#
#color(white)(1)|" "color(white)(X)#
#" "stackrel("—————————————)#
Three Drop the first coefficient of the dividend below the division symbol.
#2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5#
#color(white)(1)|" "color(white)(X)#
#" "stackrel("—————————————)#
#color(white)(1)|color(white)(X)color(red)1#
Four Multiply the result by the constant, and put the product in the next column.
#2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5#
#color(white)(1)|" "color(white)(XX1)2#
#" "stackrel("—————————————)#
#color(white)(1)|color(white)(X)color(blue)1#
Five Add down the column.
#2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5#
#color(white)(1)|" "color(white)(XX1)2#
#" "stackrel("—————————————)#
#color(white)(1)|color(white)(X)color(blue)1color(white)(X11)color(red)4#
Six Repeat Steps Four and Five until you can go no farther.
#2|color(white)(X)1" "color(white)(X)2color(white)(XX)-4" "" "-5#
#color(white)(1)|" "color(white)(XX1)2color(white)(XXXX)8color(white)(XXxx)8#
#" "stackrel("—————————————)#
#color(white)(1)|color(white)(X)color(blue)1color(white)(X11)color(red)4color(white)(XXXx)color(red)4color(white)(XXXX)color(red)3#
Remainder is #3#. Hence #f(2)=3#
