
Factorize the following expression completely:
$ 6{{d}^{2}}e-9{{e}^{2}} $
Answer
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Hint: First of all take the largest common term from the expression that can be possible. After taking the common term, you will find the remaining uncommon expression is in the form of $ {{a}^{2}}-{{b}^{2}} $ then factorize this expression using the identity of $ {{a}^{2}}-{{b}^{2}} $.
Complete step-by-step answer:
In the given expression, $ 6{{d}^{2}}e-9{{e}^{2}} $ first of all we will take the largest term which is common in $ 6{{d}^{2}}{{e}^{{}}} $ and $ 9{{e}^{2}} $ .
$ 3e\left( 2{{d}^{2}}-3e \right) $
You can see from the above expression that the largest common term is 3e.
Now, we can rewrite the uncommon expression i.e. $ \left( 2{{d}^{2}}-3e \right) $ as $ {{\left( \sqrt{2}d \right)}^{2}}-{{\left( \sqrt{3e} \right)}^{2}} $ .This expression is in the form of $ {{a}^{2}}-{{b}^{2}} $ . And we know that $ {{a}^{2}}-{{b}^{2}} $ is an identity and is equal to $ \left( a+b \right)\left( a-b \right) $ . So, we can write $ {{\left( \sqrt{2}d \right)}^{2}}-{{\left( \sqrt{3e} \right)}^{2}} $ as $ \left( \sqrt{2}d+\sqrt{3e} \right)\left[ \sqrt{2}d-\sqrt{3e} \right] $ . Now, we are writing whatever just said in the stepwise manner:
$ \begin{align}
& 3e\left( {{\left( \sqrt{2}d \right)}^{2}}-{{\left( \sqrt{3e} \right)}^{2}} \right) \\
& \Rightarrow 3e\left( \left( \sqrt{2}d+\sqrt{3e} \right)\left( \sqrt{2}d-\sqrt{3e} \right) \right) \\
\end{align} $
Hence, the complete factorization of $ 6{{d}^{2}}e-9{{e}^{2}} $ has given $ 3e\left( \left( \sqrt{2}d+\sqrt{3e} \right)\left( \sqrt{2}d-\sqrt{3e} \right) \right) $ .
From the above result of factorization, we can say that the factors of $ 6{{d}^{2}}e-9{{e}^{2}} $ are $ 3e,\left( \sqrt{2}d+\sqrt{3e} \right),\left( \sqrt{2}d-\sqrt{3e} \right) $ .
Note: Usually you have seen the factorization expression in a, b x, y or t. But in this problem, the factorization expression is in “d” or “e” which is not often seen so you got confused. The principle of factorization is the same here, you just have to factorize in the same manner as in the expression with variables x, y, a, b or t. If still you are facing the problem then assume “d” or “e” to some other variable in which you are comfortable like x or y and then factorize.
Complete step-by-step answer:
In the given expression, $ 6{{d}^{2}}e-9{{e}^{2}} $ first of all we will take the largest term which is common in $ 6{{d}^{2}}{{e}^{{}}} $ and $ 9{{e}^{2}} $ .
$ 3e\left( 2{{d}^{2}}-3e \right) $
You can see from the above expression that the largest common term is 3e.
Now, we can rewrite the uncommon expression i.e. $ \left( 2{{d}^{2}}-3e \right) $ as $ {{\left( \sqrt{2}d \right)}^{2}}-{{\left( \sqrt{3e} \right)}^{2}} $ .This expression is in the form of $ {{a}^{2}}-{{b}^{2}} $ . And we know that $ {{a}^{2}}-{{b}^{2}} $ is an identity and is equal to $ \left( a+b \right)\left( a-b \right) $ . So, we can write $ {{\left( \sqrt{2}d \right)}^{2}}-{{\left( \sqrt{3e} \right)}^{2}} $ as $ \left( \sqrt{2}d+\sqrt{3e} \right)\left[ \sqrt{2}d-\sqrt{3e} \right] $ . Now, we are writing whatever just said in the stepwise manner:
$ \begin{align}
& 3e\left( {{\left( \sqrt{2}d \right)}^{2}}-{{\left( \sqrt{3e} \right)}^{2}} \right) \\
& \Rightarrow 3e\left( \left( \sqrt{2}d+\sqrt{3e} \right)\left( \sqrt{2}d-\sqrt{3e} \right) \right) \\
\end{align} $
Hence, the complete factorization of $ 6{{d}^{2}}e-9{{e}^{2}} $ has given $ 3e\left( \left( \sqrt{2}d+\sqrt{3e} \right)\left( \sqrt{2}d-\sqrt{3e} \right) \right) $ .
From the above result of factorization, we can say that the factors of $ 6{{d}^{2}}e-9{{e}^{2}} $ are $ 3e,\left( \sqrt{2}d+\sqrt{3e} \right),\left( \sqrt{2}d-\sqrt{3e} \right) $ .
Note: Usually you have seen the factorization expression in a, b x, y or t. But in this problem, the factorization expression is in “d” or “e” which is not often seen so you got confused. The principle of factorization is the same here, you just have to factorize in the same manner as in the expression with variables x, y, a, b or t. If still you are facing the problem then assume “d” or “e” to some other variable in which you are comfortable like x or y and then factorize.
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