Question

Factorize the expression $25{{\left( a\text{ + 2}b\text{ }-\text{ 3}c \right)}^{2}}\text{ }-\text{ 9}{{\left( 2a\text{ }-\text{ }b\text{ }-\text{ c} \right)}^{2}}$

Answer 1

Hint: Express the terms of the expression in perfect square form. Then use the formula ${{x}^{2}}\text{ }-\text{ }{{y}^{2}}\text{ = }\left( x\text{ + }y \right)\left( x\text{ }-\text{ }y \right)$ to turn the difference of the two perfect square quantities into two factors.

Complete step-by-step answer:
In the expression $25{{\left( a\text{ + 2}b\text{ }-\text{ 3}c \right)}^{2}}\text{ }-\text{ 9}{{\left( 2a\text{ }-\text{ }b\text{ }-\text{ c} \right)}^{2}}$, first we separate the perfect square terms as,
$\begin{align}
  & 25{{\left( a\text{ + 2}b\text{ }-\text{ 3}c \right)}^{2}} \\
 & =\text{ }{{\left\{ 5\left( a\text{ + 2}b\text{ }-\text{ 3}c \right) \right\}}^{2}} \\
 & =\text{ }{{\left( 5a\text{ + 10}b\text{ }-\text{ 15c} \right)}^{2}} \\
\end{align}$
Again,
$\begin{align}
  & 9{{\left( 2a\text{ }-\text{ }b\text{ }-\text{ }c \right)}^{2}} \\
 & =\text{ }{{\left\{ 3\left( 2a\text{ }-\text{ }b\text{ }-\text{ }c \right) \right\}}^{2}} \\
 & =\text{ }{{\left( 6a\text{ }-\text{3}b\text{ }-\ 3c \right)}^{2}} \\
\end{align}$
Thus, we obtain the perfect square forms of the two terms in the expression$25{{\left( a\text{ + 2}b\text{ }-\text{ 3}c \right)}^{2}}\text{ }-\text{ 9}{{\left( 2a\text{ }-\text{ }b\text{ }-\text{ c} \right)}^{2}}$.
Now, comparing the expression $25{{\left( a\text{ + 2}b\text{ }-\text{ 3}c \right)}^{2}}\text{ }-\text{ 9}{{\left( 2a\text{ }-\text{ }b\text{ }-\text{ c} \right)}^{2}}$ with ${{x}^{2}}\text{ }-\text{ }{{y}^{2}}$, we get,
x = 5a + 10b – 15c and y = 6a – 3b – 3c
Putting these values x = 5a + 10b – 15c and y = 6a – 3b – 3c in the formula ${{x}^{2}}\text{ }-\text{ }{{y}^{2}}\text{ = }\left( x\text{ + }y \right)\left( x\text{ }-\text{ }y \right)$, we get,
\[\begin{align}
  & {{\left( 5a\text{ + 10}b\text{ }-\text{ 15c} \right)}^{2}}\text{ }-\text{ }{{\left( 6a\text{ }-\text{3}b\text{ }-\ 3c \right)}^{2}} \\
 & =\text{ }\left( 5a\text{ + 10}b\text{ }-\text{ 15c + }6a\text{ }-\text{3}b\text{ }-\ 3c \right)\cdot \left( 5a\text{ + 10}b\text{ }-\text{ 15c }-\text{ }6a\text{ +3}b\,\text{+}\ 3c \right) \\
 & =\text{ }\left( 11a\text{ + 7}b\text{ }-\text{ 18}c \right)\cdot \left( -a\text{ + 13}b\text{ }-\text{ 12}c \right) \\
\end{align}\]
Thus, the expression $25{{\left( a\text{ + 2}b\text{ }-\text{ 3}c \right)}^{2}}\text{ }-\text{ 9}{{\left( 2a\text{ }-\text{ }b\text{ }-\text{ c} \right)}^{2}}$ is factored into (11a + 7b – 18c) (- a + 13b – 12c).

Note: Using the formula ${{x}^{2}}\text{ }-\text{ }{{y}^{2}}\text{ = }\left( x\text{ + }y \right)\left( x\text{ }-\text{ }y \right)$ is much more beneficial than trying to factorize it by middle term breaking because the expression contains more than three variables. Moreover, the quantities are perfect square terms, so using the formula is a natural choice.