How do you use synthetic substitution to evaluate f(2) for #f(x)=x^3+2x^2-4x-5#?

2 Answers
Nov 4, 2016

Answer:

#f(2)=3#

Explanation:

#f(x)=x^3+2x^2-4x-5#
When the problem asks to evaluate f(2), simply substitute in 2 for every time you see "x" in the function. So we have:
#f(2)=(2)^3+2(2)^2-4(2)-5#
Now we simplify each term:
#f(2)=8+2(4)-8-5#
#f(2)=8+8-8-5#
#f(2)=3#

Nov 4, 2016

Answer:

#f(2)=3#

Explanation:

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#