Long and synthetic division

We're given the cubic polynomial #\:p(x)\:# and the linear divisor #\:d(x)\:#: \[p(x)=5\cdot x^3-9\cdot x^2-146\cdot x+120 \hspace{1.0 cm}\mbox{and}\hspace{1.0 cm}d(x) = x-6\hspace{0.05 cm}.\] In the steps outlined below, perform the long division to determine the quotient #q(x)# and the constant remainder #r#, so that: \[\:p(x)\:=\: \left(x-6\right)\cdot q(x) + r\] #\small{}\mbox{Note. One answer per blank. So, for each blank, do not type an equal sign in your answer.}#

Step 1.

Evaluate the polynomial #\:p(x)=5\cdot x^3-9\cdot x^2-146\cdot x+120\:# at the value #\:\:x=6#. #\:#So, calculate the value of the polynomial if we replace #\:x\:# with #\:6#. #\:# For the first blank, show all the steps. For the second blank, show the final, simplified result.

#p(6) \:=\: # #\:=\: #

#\small{}\mbox{Note. We anticipate this value will give the remainder in the next step because: }##\small{}\:p(6)\:=\: \left(6-6\right)\cdot q(6) + r\:=\:0 + r#

Step 2.

Show the long division steps to divide #\:p(x)\:# by the linear term #\:d(x)=\left(x-\left(6\right) \right)#, #\:# to determine the quotient #q(x)# and the constant remainder #r#, so that: \[\frac{p(x)}{x-6}\:\:=\:\:q(x)+\frac{r}{x-6}\hspace{0.05 cm}.\]


#x-6#	#5\cdot x^3-9\cdot x^2-146\cdot x+120#
#\boldsymbol{-}#

	#\boldsymbol{-}#

		#\boldsymbol{-}#

Step 3.

From Step 2, #\:#express #\:p(x)\:# in terms of the quotient #\:q(x)\:# and the remainder #\:r#, #\:# as follows: #\:p(x)\:=\: \left(x-6\right)\cdot q(x) + r#, #\:# Put the product in the first blank and the remainder in the second blank.

#p(x)\:=\: 5\cdot x^3-9\cdot x^2-146\cdot x+120 \:\:=#

#\large +#

#\small{}\mbox{Check that} \:r\: \mbox{ agrees with the answer found in Step 1. }\:\mbox{ So, } \:p(6)=r. \: \mbox{Thus, the value }\:p(6)\: \mbox{ is the same as the }# #\small{}\mbox{remainder } \:r, \:\mbox{ when }\:p(x)\:\mbox{ is divided by }\:d(x)=x-(6)#.

Step 4.

Check the answer found in Step 3. #\:# Expand the answer term-by-term and check that you do get: #\:p(x)\:=\: \left(x-6\right)\cdot q(x) + r#. #\:# So, in the table below, calculate each product. The blue entry is the sum of the three pink entries: #\:p(x) \:=\: x\cdot q(x) + (-6)\cdot q(x) +r#. #\:# Check that the resulting sum in the blue entry is the same as the polynomial #\:p(x)#.

	#x \cdot q(x)#	#\large =#
#\large +#	#(-6)\cdot q(x)#	#\large =#
#\large +#	#r#	#\large =#
	#x \cdot q(x) + (-6) \cdot q(x) + r#	#\large = #

Edna's Math Class for OpenStax:

Long and synthetic division: Divide by #\:(x-r)#