Section 5.2 : Zeroes/Roots of Polynomials
5. \(x = r\) is a root of the following polynomial. Find the other two roots and write the polynomial in fully factored form.
\[P\left( x \right) = {x^3} - 7{x^2} - 6x + 72 \mbox{ ; } r = 4\]Show All Steps Hide All Steps
Start SolutionWe know that \(x = 4\) is a root of the polynomial and so we know that we can write the polynomial as,
\[P\left( x \right) = \left( {x - 4} \right)Q\left( x \right)\] Show Step 2To find \(Q\left( x \right)\) all we need to do is a quick synthetic division.
\[\begin{array}{*{20}{r}}{\left. {\underline {\,4 \,}}\! \right| }\\{}\\{}\end{array}\,\,\,\,\begin{array}{*{20}{r}}1&{ - 7}&{ - 6}&{72}\\{}&4&{ - 12}&{ - 72}\\\hline1&{ - 3}&{ - 18}&0\end{array}\]From this we see that,
\[Q\left( x \right) = {x^2} - 3x - 18\] Show Step 3We can now write down \(P\left( x \right)\) and it is simple enough to factor \(Q\left( x \right)\).
\[P\left( x \right) = \left( {x - 4} \right)\left( {{x^2} - 3x - 18} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\left( {x - 4} \right)\left( {x - 6} \right)\left( {x + 3} \right)}}\] 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = 4\,\,\,\,\,\,\,\,\,\,\,\,\,x = 6\,\,\,\,\,\,\,\,\,\,\,\,\,x = - 3}}\]