How do you divide #(x^4-81)/(x-3)# using synthetic division?

1 Answer
Sep 28, 2016

#(x^4 -81) div (x-3) = x^3 +3x^2 +9x +27#

Explanation:

In synthetic division we only use the numerical coefficients.

Make sure that they are in descending order of the powers of x.
In this case we have #" "x^4" "x^3" "x^2" "x^1" " x^0#

Write a #0# for any missing term.

If #x-3 = 0 rarr x = 3#

#color(red)(3)# is used at the side.
#color(white)(xxxxxxxxxxxxx)" "(x^4" "x^3" "x^2" "x^1" " x^0)#

Coefficients :#rarrcolor(white)(xx)" "1" "0" "0" "0" "-81#

#color(white)(xxxxxxxxxxxx)color(red)(3)color(white)(xx)ul(darr " "color(blue)(3)" "color(purple)(9)" "color(mediumpurple)(27) " "color(darkcyan)(81)#
#color(white)(xxxxxxxxxxxxx)color(white)(xxx)1" "color(lime)(3)" "color(orange)(9)" "color(magenta)(27)" "0 larr# remainder

#color(white)(xxxxxxxxxxxxx)" "x^3" "x^2" "x^1" " x^0#

Steps: Follow the colours...
bring down the 1.
#color(red)(3)xx1 = color(blue)(3)" add " 0+color(blue)(3) = color(lime) (3)#
#color(red)(3)xxcolor(lime)(3) = color(purple)(9)" add " 0+color(purple)(9) =color(orange)(9)#
#color(red)(3)xxcolor(orange)(9) =color(mediumpurple)(27)" add "0 + color(mediumpurple)(27) = color(magenta)(27)#
#color(red)(3)xxcolor(magenta)(27) = color(darkcyan)(81)" add "-81+ color(darkcyan)(81) = 0 larr # the remainder

The numbers in the bottom row are now the numerical coefficients of the quotient.

#x^4 divx = x^3# So the first term will be #x^3#

The quotient is:

#1x^3 + color(lime)(3)x^2 + color(orange)(9)x + color(magenta)(27) " rem " 0#