Question

Find the least common multiple(LCM) of $3{{x}^{3}}-9{{x}^{2}}-12x$ and ${{x}^{2}}+5x+4$.\[\]

Answer 1

Hint: We denote the given polynomials as $f\left( x \right)=3{{x}^{3}}-9{{x}^{2}}-12x$ and $g\left( x \right)={{x}^{2}}+5x+4$. The least common multiple of the given polynomials is the $f\left( x \right)g\left( x \right)$ with common factors multiplied only once. We find the factor of $f\left( x \right)$ with factor theorem. We find the factors of quadratic polynomial $g\left( x \right)$ by splitting the middle term.\[\]

Complete step-by-step solution:
We know from the factor theorem of polynomials 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 $p\left( x \right).$Then we can find the other factor $\dfrac{p\left( x \right)}{x-a}.$\[\]
The given polynomials are $3{{x}^{3}}-9{{x}^{2}}-12x=f\left( x \right)$(we denote ) and ${{x}^{2}}+5x+4=g\left( x \right)$(we denote). We want to find the least common multiple of $f\left( x \right)$ and $g\left( x \right)$ which $f\left( x \right)g\left( x \right)$ with common factors multiplied only once. So we have to find their common factors first.\[\]
We shall use factor theorem to find factors of $f\left( x \right)$. We need to find a zero by trial and error first. So we put first $x=0$ and we get $f\left( x \right)=0$. This means $x$is zero of $f\left( x \right)$ which implies $x$ is a factor of $f\left( x \right)$. The other factor which we denote as $q\left( x \right)$will be
\[ q\left( x \right)=\dfrac{f\left( x \right)}{x}=\dfrac{3{{x}^{3}}-9{{x}^{2}}-12x}{x}=3{{x}^{2}}-9x-12\]
 Now we further factorize the quadratic polynomial $q\left( x \right)$ by splitting the middle term.
\[ \begin{align}
  & q\left( x \right)=3{{x}^{2}}-9x-12 \\
 & =3{{x}^{2}}-12x+3x-12 \\
 & =3x\left( x-4 \right)+3\left( x-4 \right) \\
 & =\left( x-4 \right)\left( 3x+3 \right) \\
 & =3\left( x+1 \right)\left( x-4 \right) \\
\end{align}\]
So we have $f\left( x \right)=3x\left( x+1 \right)\left( x-4 \right)$\[\]
We factorize the second polynomial $g\left( x \right)$ which is quadratic by splitting the middle term,
\[\begin{align}
  & g\left( x \right)={{x}^{2}}+5x+4 \\
 & ={{x}^{2}}+4x+x+4 \\
 & =x\left( x+4 \right)+1\left( x+4 \right) \\
 & =\left( x+4 \right)\left( x+1 \right) \\
\end{align}\]
So we have $g\left( x \right)=\left( x+1 \right)\left( x+4 \right)$ . We see that the common factors in both $f\left( x \right)$ and $g\left( x \right)$ is $x+1$. So the least common multiple of $f\left( x \right)$ and $g\left( x \right)$ is $f\left( x \right)g\left( x \right)$ with common factor multiplied $x+1$ only once which is $3x\left( x+1 \right)\left( x-4 \right)\left( x+4 \right)$.\[\]

Note: The factor theorem is special case of remainder theorem which 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 a factor theorem. The greatest common divisor (GCD) of two polynomials is only the common factors multiplied which in this problem is $x+1$.