Question

Factorise $ {{x}^{3}}+6{{x}^{2}}+11x+6 $ using factor theorem and long division method. \[\]

Answer 1

Hint: We use trial and error method to guess a root for the given polynomial $ {{x}^{3}}+6{{x}^{2}}+11x+6 $ . Let that root be $ a $ . We use factor theorem which states that a polynomial $ p\left( x \right) $ has a factor $ \left( x-a \right) $ if and only if $ p\left( a \right)=0 $ in other words $ a $ is a zero of $ p\left( x \right) $ . So we divide $ {{x}^{3}}+6{{x}^{2}}+11x+6 $ by $ x-a $ find the other factor as the quotient in long division method. \[\]

Complete step by step answer:
Let us denote $ p\left( x \right)={{x}^{3}}+6{{x}^{2}}+11x+6 $ . Let us try to guess a root of $ p\left( x \right) $ by trial and error. We observe that the given polynomial $ {{x}^{3}}+6{{x}^{2}}+11x+6 $ have only positive coefficients and positive constant term. So the root cannot be positive. So let us put a negative number $ x=-1 $ as root in the polynomial and see if it is a root or not. We have
 \[\begin{align}
  & p\left( -1 \right)={{\left( -1 \right)}^{3}}+6{{\left( -1 \right)}^{2}}+11\left( -1 \right)+6 \\
 & \Rightarrow p\left( -1 \right)=-1+6-11+6 \\
 & \Rightarrow p\left( -1 \right)=12-12=0 \\
\end{align}\]
So $ x=-1 $ is root of the polynomial $ p\left( x \right) $ . So by factor theorem we have one factor of $ p\left( x \right) $ as $ x-\left( -1 \right)=x+1 $ . Let us use the long division method taking $ d\left( x \right)=x+1 $ as the divisor polynomial and find the other factor of $ p\left( x \right) $ in the quotient. \[\]

We have the divisor polynomial $ d\left( x \right)=x+1 $ has first term $ x $ and the dividend $ p\left( x \right)={{x}^{3}}+6{{x}^{2}}+11x+6 $ has the first term $ {{x}^{3}} $ . So we have to multiply $ \dfrac{{{x}^{3}}}{x}={{x}^{2}} $ with $ d\left( x \right) $ to reach $ p\left( x \right) $ . We initiate the long division method. \[\]
\[\begin{align}
  & x+1\overset{{{x}^{2}}}{\overline{\left){{{x}^{3}}+6{{x}^{2}}+11x+6}\right.}} \\
 & \hspace{1.2 cm}\underline{{{x}^{3}}+{{x}^{2}}} \\
 & \hspace{1.2 cm} 5{{x}^{2}}+11x+6\text{ } \\
\end{align}\]
So our new dividend polynomial is $ 5{{x}^{2}}+11x+6 $ . So we need to multiply $ \dfrac{5{{x}^{2}}}{x}=5x $ with $ d\left( x \right)=x+1 $ to proceed.
\[\begin{align}
  & x+1\overset{{{x}^{2}}+5x}{\overline{\left){{{x}^{3}}+6{{x}^{2}}+11x+6}\right.}} \\
 & \hspace{1.2 cm}\underline{{{x}^{3}}+{{x}^{2}}} \\
 & \hspace{1.2 cm}5{{x}^{2}}+11x+6 \\
 & \hspace{1.2 cm} \underline{5{{x}^{2}}\text{+}5x}\text{ } \\
 & \hspace{1.2 cm}6x+6 \\
\end{align}\]
So our new dividend polynomial is $ 5{{x}^{2}}+11x+6 $ . So we need to multiply $ \dfrac{5{{x}^{2}}}{x}=5x $ with $ d\left( x \right) $ to proceed.
\[\begin{gathered}
  x + 1\mathop{\left){\vphantom{1{{x^3} + 6{x^2} + 11x + 6}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^3} + 6{x^2} + 11x + 6}}}}
\limits^{\displaystyle \,\,\, {{x^2} + 5x + 6}} \\
  {\hspace{-1 cm }}\underline {{x^3} + {x^2}} \\
  {\text{ }}5{x^2} + 11x + 6 \\
  {\hspace{-0.4 cm }}\underline {5{x^2}{\text{ + }}5x} {\text{ }} \\
  {\hspace{0.5 cm }}6x + 6 \\
  {\hspace{0.5 cm }}\underline {6x + 6} \\
  {\hspace{0.5 cm }\text{ 0}} \\
\end{gathered} \]

So by factor theorem the polynomial at the quotient is the other factor $ q\left( x \right)={{x}^{2}}+5x+6 $ . So we have
\[\begin{align}
  & p\left( x \right)=d\left( x \right)q\left( x \right) \\
 & \Rightarrow {{x}^{3}}+6{{x}^{2}}+11x+6=\left( x+1 \right)\left( {{x}^{2}}+5x+6 \right) \\
\end{align}\]
We again guess a root for $ q\left( x \right)={{x}^{2}}+5x+6 $ . We put $ x=-2 $ in $ q\left( x \right) $ to have
\[q\left( -2 \right)={{\left( -2 \right)}^{2}}+5\left( -2 \right)+6=4-10+6=0\]
So $ x=-2 $ is zero of $ q\left( x \right) $ and hence by factor theorem $ x-\left( -2 \right)=x+2 $ . So we find other factor of by $ q\left( x \right) $ by long division method.
\[\begin{gathered}
  x + 2\mathop{\left){\vphantom{1{{x^2} + 5x + 6}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^2} + 5x + 6}}}}
\limits^{\displaystyle \,\,\, {x + 3}} \\
  {\hspace{0.5 cm}}\underline {{x^2} + 2x} \\
  {\text{ }}3x + 6 \\
  {\text{ }}\underline {3x + 6} \\
  {\text{ 0}} \\
\end{gathered} \]
So $ x+3 $ is the other factor of $ q\left( x \right) $ . So we have $ q\left( x \right)={{x}^{2}}+5x+6=\left( x+2 \right)\left( x+3 \right) $ . So the given polynomial in factorized from is
\[\begin{align}
  & {{x}^{3}}+6{{x}^{2}}+11x+6=\left( x+1 \right)\left( {{x}^{2}}+5x+6 \right) \\
 & \Rightarrow {{x}^{3}}+6{{x}^{2}}+11x+6=\left( x+1 \right)\left( x+2 \right)\left( x+3 \right) \\
\end{align}\]

Note:
 We note that remainder theorem states that if linear polynomial $ x-a $ divides the polynomial $ p\left( x \right) $ then it will leave the remainder $ p\left( a \right) $ . When $ p\left( a \right)=0 $ , the remainder theorem is specialized as factor theorem. We note that that when we divide a divided polynomial $ p\left( x \right) $ with degree $ n $ by some divisor polynomial $ d\left( x \right) $ with degree $ m\le n $ then we get the quotient polynomial $ q\left( x \right) $ of degree $ n-m $ and the remainder polynomial as $ r\left( x \right) $ of degree either equal to $ m $ or $ m-1 $ .