Question

Find the L.C.M of ${{x}^{3}}-2{{x}^{2}}-13x-10=0$ and ${{x}^{3}}-{{x}^{2}}-10x-8$.

Answer 1

Hint: We will be using the concept of factorization to factorize the polynomial and find its factor then we will use the concept of LCM to find the LCM.

Complete step-by-step answer:
Now, we have been given polynomial ${{x}^{3}}-2{{x}^{2}}-13x-10=0$ and ${{x}^{3}}-{{x}^{2}}-10x-8$and we have to find its LCM.
Now, we will first factorize the two polynomials as,
$f\left( x \right)={{x}^{3}}-2{{x}^{2}}-13x-10$
Now, to factorize it we have found its root by hit and trial so first we put $x=-1$.
$\begin{align}
  & f\left( -1 \right)=-1-2+13-10 \\
 & =-13+13 \\
 & =0 \\
\end{align}$
So, $x=-1$ is a root of ${{x}^{3}}-2{{x}^{2}}-13x-10$. Therefore, by factor theorem $x+1=0$ is a factor of $f\left( x \right)$. Now, we will use long division to find other factors. So, we have,
\[x+1\overset{{{x}^{2}}-3x-10}{\overline{\left){\begin{align}
  & {{x}^{3}}-2{{x}^{2}}-13x-10 \\
 & \underline{\begin{align}
  & {{x}^{3}}+{{x}^{2}} \\
 & -\ \ \ - \\
\end{align}} \\
 & \ \ \ \ \ \ -3{{x}^{2}}-13x \\
 & \ \ \ \ \ \ \,\underline{\begin{align}
  & -3{{x}^{2}}-3x \\
 & +\ \ \ \ \ \ + \\
\end{align}} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -10x-10 \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{-10x-10} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\
\end{align}}\right.}}\]
So, we have all the factors of $f\left( x \right)$ as,
$\begin{align}
  & f\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-5x+2x-10 \right) \\
 & =\left( x+1 \right)\left( x\left( x-5 \right)+2\left( x-5 \right) \right) \\
 & f\left( x \right)=\left( x+1 \right)\left( x+2 \right)\left( x-5 \right)..........\left( 1 \right) \\
\end{align}$
Now, we have to do similarly for,
$g\left( x \right)={{x}^{3}}-{{x}^{2}}-10x-8$
Now, we put $x=-1$. So, we have,
$\begin{align}
  & g\left( -1 \right)=-1-1+10-8 \\
 & =-10+10 \\
 & =0 \\
\end{align}$
So, we have $x+1=0$ a factor of $g\left( x \right)$.
Now, an long division we have,
\[x+1\overset{{{x}^{2}}-2x-8}{\overline{\left){\begin{align}
  & {{x}^{3}}-{{x}^{2}}-10x-8 \\
 & \underline{\begin{align}
  & {{x}^{3}}+{{x}^{2}} \\
 & -\ \ \ - \\
\end{align}} \\
 & \ \ \ \ \ \ -2{{x}^{2}}-10x \\
 & \ \ \ \ \ \ \,\underline{\begin{align}
  & -2{{x}^{2}}-2x \\
 & +\ \ \ \ \ \ + \\
\end{align}} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -8x-8 \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{-8x-8} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\
\end{align}}\right.}}\]
So, we have,
$\begin{align}
  & g\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-2x-8 \right) \\
 & =\left( x+1 \right)\left( {{x}^{2}}-4x+2x-8 \right) \\
 & =\left( x+1 \right)\left( x\left( x-4 \right)+2\left( x-4 \right) \right) \\
 & g\left( x \right)=\left( x+1 \right)\left( x+2 \right)\left( x-4 \right) \\
\end{align}$
Now, we know that LCM is the least divisible by both $f\left( x \right)$ and $g\left( x \right)$. Therefore, the LCM will have the highest power of all factors in $f\left( x \right)$ and $g\left( x \right)$.
$LCM\left( f\left( x \right),g\left( x \right) \right)=\left( x+1 \right)\left( x+2 \right)\left( x-4 \right)\left( x-5 \right)$

Note: To solve these types of questions it is important to note the way we have factorized the cubic equation. Also, it is important to note that the LCM has all the factors of both the numbers and if some factors are common then we take the highest power in both of the factors.