Question

Find the LCM of $ {{\text{m}}^{2}}-9m-22 $ , $ {{m}^{2}}-8m-33 $ , $ {{m}^{2}}+5m+6 $ .

Answer 1

Hint: In order to solve this problem, we have to factorize all three terms. We need to factorize the terms such that it is in the form of the product of two terms. Then while calculating the LCM we need all the terms together but we have to take the repeating terms only once.

Complete step-by-step answer:
We need to understand what we mean by LCM.
LCM of numbers is the smallest positive number that is multiple of two or more numbers.
Now we need to find the LCM of three numbers, which are $ {{\text{m}}^{2}}-9m-22 $ , $ {{m}^{2}}-8m-33 $ , $ {{m}^{2}}+5m+6 $ .
We need to factorize all the terms.
Let's start with the first term $ {{\text{m}}^{2}}-9m-22 $ .
 $ \begin{align}
  & {{\text{m}}^{2}}-9m-22={{\text{m}}^{2}}-11m+2m-22 \\
 & =m\left( m-11 \right)+2\left( m-11 \right)
\end{align} $
Taking $ \left( m-11 \right) $ common we get,
 $ {{\text{m}}^{2}}-9m-22=\left( m+2 \right)\left( m-11 \right)....................(i) $
Now, let's factorize the second number $ {{m}^{2}}-8m-33 $ .
 $ \begin{align}
  & {{\text{m}}^{2}}-8m-33={{\text{m}}^{2}}-11m+3m-33 \\
 & =m\left( m-11 \right)+3\left( m-11 \right)
\end{align} $
Taking $ \left( m-11 \right) $ common we get,
 $ {{\text{m}}^{2}}-8m-33=\left( m+3 \right)\left( m-11 \right)....................(ii) $
Now, let's factorize the third number $ {{m}^{2}}+5m+6 $ .
 $ \begin{align}
  & {{\text{m}}^{2}}+5m+6={{\text{m}}^{2}}+2m+3m+6 \\
 & =m\left( m+2 \right)+3\left( m+2 \right)
\end{align} $
Taking $ \left( m+2 \right) $ common we get,
 $ {{\text{m}}^{2}}+5m+6=\left( m+2 \right)\left( m+3 \right)....................(ii) $
Now, to calculate LCM we need to multiply all the values at least once.
By doing that we get,
 $ L.C.M.=\left( m+2 \right)\left( m+3 \right)\left( m-11 \right) $
Hence, the L.C.M of these three terms is $ \left( m+2 \right)\left( m+3 \right)\left( m-11 \right) $ .


Note: While factoring the easier way to do this is to find such a combination the sum of coefficients gives the linear term in x and the product of the same gives the constant term. We can show this by an example. For this, $ {{\text{m}}^{2}}-8m-33={{\text{m}}^{2}}-11m+3m-33 $ , the coefficients of linear terms selected are -11 and +3. the sum of these terms gives -11 + 3 = -8, which is the linear term and -11 x 3 = 33, which is the constant term of the quadratic equation.