How do you evaluate #f(x)=-x^5-4x^3+6x^2-x# at x=-2 using direct substitution and synthetic division?

1 Answer
Apr 29, 2017

#f(-2)=90# for details please see below.

Explanation:

To evaluate #f(x)=-x^5-4x^3+6x^2-x# at x=-2 using direct substitution, we just put value of #x# in the polynomial.

Hence, #f(-2)=-(-2)^5-4(-2)^3+6(-2)^2-(-2)#

= #-(-32)-4(-8)+6xx4+2#

= #32+32+24+2=90#

Before we talk about synthetic division, one thing regarding remainder theorem which states that when a polynomial #f(x)# is divided by #x-a#, the remainder is #f(a)#.

Hence, to find #f(-2)#, we should divide #f(x)# by #x-(-2)=x+2#, the remainder so obtained is #f(-2)#. Hence let us divide #f(x)=-x^5-4x^3+6x^2-x# by #x+2# using synthetic division method.

One Write the coefficients of #x# in the dividend inside an upside-down division symbol. We are using #0# as coefficient of #x^4# and constant term.

#color(white)(1)|color(white)(X)-1" "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" "0#
#color(white)(1)|" "color(white)(X)#
#" "stackrel("——————————————————------)#

Two As #x+2=0# gives #x=-2# put #-2# at the left.

#-2|color(white)(X)-1" "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" "0#
#color(white)(xx)|" "color(white)(XX)#
#" "stackrel("—————————————---------)#

Three Drop the first coefficient of the dividend below the division symbol.

#-2|color(white)(X)-1" "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" "0#
#color(white)(xx)|" "color(white)(X)#
#" "stackrel("—————————————---------)#
#color(white)(xx)|color(white)(xX)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)0color(white)(XX)-4" "" "6" "" "-1" "" "0#
#color(white)(xx)|" "color(white)(xxXxx)2#
#" "stackrel("————————————---------—)#
#color(white)(xx)|color(white)(X)color(blue)-1#

Five Add down the column.

#-2|color(white)(X)1" "color(white)(X)0color(white)(XX)-4" "" "6" "" "-1" "" "0#
#color(white)(xx)|" "color(white)(xXx)2#
#" "stackrel("—————————————---------)#
#color(white)(xx)|color(white)()color(blue)-1color(white)(X1)color(red)2#

Six Repeat Steps Four and Five until you can go no farther.

#-2|color(white)(X)1" "color(white)(X)0color(white)(X)-4" "" "6" "" "-1" "" "0#
#color(white)(xx)|" "color(white)(Xxx)2color(white)(xx)-4color(white)(xxx)16color(white)(xxx)-44color(white)(xxx)90#
#" "stackrel("———--------——————————)#
#color(white)(xx)|color(white)()color(blue)-1color(white)(xxx)color(red)2color(white)(X)color(red)-8color(white)(XX)color(red)22color(white)(XX)color(red)-45color(white)(xxx)90#

Hence, Quotient is #-x^4+2x^3-8x^2+22x-45# and remainder is #90# an hence #f(-2)=90#