How do you divide #(x^4 - x^3 - 38x^2 - 31x + 45) -: (x+5)#, using synthetic division?

1 Answer
marfre
Mar 10, 2018

#(x^4-x^3-38x^2-31x+45)/(x+5) = x^3 -6x^2-8x+9#

Explanation:

Given: Use synthetic division to divide: #(x^4-x^3-38x^2-31x+45)-:(x+5)#

In synthetic division the divisor must be a linear factor: #(x - a)#
The value used to divide is #x - a = 0; " "x = a#

In synthetic division a coefficient is required for every term. This means if there is a missing term such as #0 x^3#, a zero would be required.

First list the #x#-value, then each coefficient. Drop down the leading coefficient to the 3rd row (sum row).

#ul(-5|)" " 1" " -1" " -38" " -31" " 45#
#" "ul(+" ")#
#" "1#

Multiply the sum in the first column by the #x#-value and place in the second row. Add the first and 2nd row and place the sum in the 3rd row:
#ul(-5|)" " 1" " -1" " -38" " -31" " 45#
#" "ul(+" "-5)#
#" "1" "-6#

Multiply the sum in the 2nd column by the #x#-value and place in the 3rd row. Add the first and 2nd row and place the sum in the 3rd row:
#ul(-5|)" " 1" " -1" " -38" " -31" " 45#
#" "ul(+" "-5" "30)#
#" "1" "-6" "-8#

Continue this process:
#ul(-5|)" " 1" " -1" " -38" " -31" " 45#
#" "ul(+" "-5" "30" "40" "-45)#
#" "1" "-6" "-8" "9" "0#

#The 3rd row represents the coefficients of the quotient. They will always be one less degree than the original equation. The last number is the remainder:

#(x^4-x^3-38x^2-31x+45)/(x+5) = x^3 -6x^2-8x+9#

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