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Section 5.1 : Dividing Polynomials
3. Use long division to divide \(2{x^5} + {x^4} - 6x + 9\) by \({x^2} - 3x + 1\).
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Start SolutionLet’s first perform the long division. Just remember that we keep going until the remainder has degree that is strictly less that the degree of the polynomial we’re dividing by, \({x^2} - 3x + 1\) in this case. The polynomial we’re dividing by has degree two and so, in this case, we’ll stop when the remainder is degree one or zero.
Here is the long division work for this problem.
\[\require{enclose}\begin{align*} & 2{{x}^{3}}+7{{x}^{2}}+19x+50 \\ {{x}^{2}}-3x+1 & {\enclose{longdiv}{{2{{x}^{5}}+{{x}^{4}}-6x+9}}} \\ - & \underline{\left( 2{{x}^{5}}-6{{x}^{4}}+2{{x}^{3}} \right)} \\ & \hspace{0.75in}7{{x}^{4}}-2{{x}^{3}}-6x+9 \\ & \hspace{0.5in}-\underline{\left( 7{{x}^{4}}-21{{x}^{3}}+7{{x}^{2}} \right)} \\ & \hspace{1.4in}19{{x}^{3}}-7{{x}^{2}}-6x+9 \\ & \hspace{1.05in}-\underline{\left( 19{{x}^{3}}-57{{x}^{2}}+19x \right)} \\ & \hspace{2.0in}50{{x}^{2}}-25x+9 \\ & \hspace{1.75in}-\underline{\left( 50{{x}^{2}}-150x+50 \right)} \\ & \hspace{2.75in}125x-41 \\ \end{align*}\] Show Step 2We can put the answer in either of the two following forms.
\[\frac{{2{x^5} + {x^4} - 6x + 9}}{{{x^2} - 3x + 1}} = 2{x^3} + 7{x^2} + 19x + 50 + \frac{{125x - 41}}{{{x^2} - 3x + 1}}\] \[2{x^5} + {x^4} - 6x + 9 = \left( {{x^2} - 3x + 1} \right)\left( {2{x^3} + 7{x^2} + 19x + 50} \right) + 125x - 41\]Either answer is acceptable here although one may be more useful than the other depending on the application that is being done when you need to actually do the long division.