Edna's Math Class for OpenStax
Factor a cubic: Factor a Cubic: Part 1
We're given the cubic polynomial #\:p(x)\:# and the linear divisor #\:d(x)\:#: \[p(x)=8\cdot x^3-53\cdot x^2-85\cdot x-24 \hspace{1.0 cm}\mbox{and}\hspace{1.0 cm}d(x) = x-8\hspace{0.05 cm}.\] The goal is to factor #\:p(x)\:# into a product of all linear terms. This problem will be accomplished in two parts. Here, in Part 1, we perform the long division in the steps outlined below, to show that #\:d(x)\:# is one of the linear factors of #\:p(x)#. #\:#
As in previous exercises, we see that determining a linear factor of a polynomial has something to do with calculating its roots. So, in the steps below, determine the quotient #q(x)# and the constant remainder #r#, so that: \[\:p(x)\:=\: \left(x-8\right)\cdot q(x) + r\] #\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)=8\cdot x^3-53\cdot x^2-85\cdot x-24\:# at the value #\:\:x=8#. #\:#So, calculate the value of the polynomial if we replace #\:x\:# with #\:8#. #\:# For the first blank, show all the steps. For the second blank, show the final, simplified result. #p(8) \:=\: # #\:=\: # #\small{}\mbox{Note. We anticipate this value will give the remainder in the next step because: }##\small{}\:p(8)\:=\: \left(8-8\right)\cdot q(8) + r\:=\:0 + r# |
| Step 2. |
Show the long division steps to divide #\:p(x)\:# by the linear term #\:d(x)=\left(x-\left(8\right) \right)#, #\:# to determine the quotient #q(x)# and the constant remainder #r#, so that: \[\frac{p(x)}{x-8}\:\:=\:\:q(x)+\frac{r}{x-8}\hspace{0.05 cm}.\] |
| 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-8\right)\cdot q(x) + r#, #\:# Put the product in the first blank and the remainder in the second blank. |
| #p(x)\:=\: 8\cdot x^3-53\cdot x^2-85\cdot x-24 \:\:=# | #\large +# |
| #\small{}\mbox{Check that} \:r\: \mbox{ agrees with the answer found in Step 1. }\:\mbox{ So, } \:p(8)=r. \: \mbox{Thus, the value }\:p(8)\: \mbox{ is the same as the }# #\small{}\mbox{remainder } \:r, \:\mbox{ when }\:p(x)\:\mbox{ is divided by }\:d(x)=x-(8)#. |
| 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-8\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) + (-8)\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 +# | #(-8)\cdot q(x)# | #\large =# | ||
| #x \cdot q(x) + (-8) \cdot q(x) + r# | #\large = # |
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