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Section 5.2 : Zeroes/Roots of Polynomials
4. \(x = r\) is a root of the following polynomial. Find the other two roots and write the polynomial in fully factored form.
\[P\left( x \right) = {x^3} - 6{x^2} - 16x \mbox{ ; }r = - 2\]Show All Steps Hide All Steps
Start SolutionWe know that \(x = - 2\) is a root of the polynomial and so we know that we can write the polynomial as,
\[P\left( x \right) = \left( {x + 2} \right)Q\left( x \right)\] Show Step 2To find \(Q\left( x \right)\) all we need to do is a quick synthetic division.
\[\begin{array}{*{20}{r}}{\left. {\underline {\,{ - 2} \,}}\! \right| }\\{}\\{}\end{array}\,\,\,\,\begin{array}{*{20}{r}}1&{ - 6}&{ - 16}&0\\{}&{ - 2}&{16}&0\\\hline1&{ - 8}&0&0\end{array}\]From this we see that,
\[Q\left( x \right) = {x^2} - 8x\] Show Step 3We can now write down \(P\left( x \right)\) and it is simple enough to factor \(Q\left( x \right)\).
\[P\left( x \right) = \left( {x + 2} \right)\left( {{x^2} - 8x} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{x\left( {x + 2} \right)\left( {x - 8} \right)}}\] 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0\,\,\,\,\,\,\,\,\,\,\,\,\,x = - 2\,\,\,\,\,\,\,\,\,\,\,\,\,x = 8}}\]