I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 5.2 : Zeroes/Roots of Polynomials
6. \(x = r\) is a root of the following polynomial. Find the other two roots and write the polynomial in fully factored form.
\[P\left( x \right) = 3{x^3} + 16{x^2} - 33x + 14 \mbox{ ; } r = - 7\]Show All Steps Hide All Steps
Start SolutionWe know that \(x = - 7\) is a root of the polynomial and so we know that we can write the polynomial as,
\[P\left( x \right) = \left( {x + 7} \right)Q\left( x \right)\] Show Step 2To find \(Q\left( x \right)\) all we need to do is a quick synthetic division.
\[\begin{array}{*{20}{r}}{\left. {\underline {\,{ - 7} \,}}\! \right| }\\{}\\{}\end{array}\,\,\,\,\begin{array}{*{20}{r}}3&{16}&{ - 33}&{14}\\{}&{ - 21}&{35}&{ - 14}\\\hline3&{ - 5}&2&0\end{array}\]From this we see that,
\[Q\left( x \right) = 3{x^2} - 5x + 2\] Show Step 3We can now write down \(P\left( x \right)\) and it is simple enough to factor \(Q\left( x \right)\).
\[P\left( x \right) = \left( {x + 7} \right)\left( {3{x^2} - 5x + 2} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\left( {x + 7} \right)\left( {3x - 2} \right)\left( {x - 1} \right)}}\] Show Step 4Finally, from the factored for of \(P\left( x \right)\) in the previous step we can see that the full list of roots/zeroes are,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 7\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{2}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1}}\]