Question

Expand the following algebraic expression by finding the product$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)$.

Hint: Use distributive law for finding the product. According to distributive law, multiplication distributes over addition, i.e. $a\left( b+c \right)=ab+ac$. Let $x={{a}^{2}}+2{{c}^{2}}$, then apply distributive law. Revert to original variables and apply distributive law again to individual expressions. Use ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$ to simplify the expressions.

Let $x={{a}^{2}}+2{{c}^{2}}$, We have
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=x\left( 3a-2c \right)$
Applying distributive law, we get
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=3ax-2cx$
Reverting to original variables, we get
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=3a\left( {{a}^{2}}+2{{c}^{2}} \right)-2c\left( {{a}^{2}}+2{{c}^{2}} \right)$
Applying distributive law on the expression$3a\left( {{a}^{2}}+2{{c}^{2}} \right)$, we get
$3a\left( {{a}^{2}}+2{{c}^{2}} \right)=3a\left( {{a}^{2}} \right)+2{{c}^{2}}\left( 3a \right)$
We know that ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$.
Applying the above formula, we get
\begin{align} & 3a\left( {{a}^{2}}+2{{c}^{2}} \right)=3{{a}^{1+2}}+6a{{c}^{2}} \\ & =3{{a}^{3}}+6a{{c}^{2}} \\ \end{align}
Similarly, applying distributive law to the expression $2c\left( {{a}^{2}}+2{{c}^{2}} \right)$, we get
$2c\left( {{a}^{2}}+2{{c}^{2}} \right)=2c{{a}^{2}}+2c\left( 2{{c}^{2}} \right)$
We know that ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$.
Applying the above formula, we get
\begin{align} & 2c\left( {{a}^{2}}+2{{c}^{2}} \right)=2{{a}^{2}}c+4{{c}^{1+2}} \\ & =2{{a}^{2}}c+4{{c}^{3}} \\ \end{align}
Hence we have
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=3{{a}^{3}}+6a{{c}^{2}}-\left( 2{{a}^{2}}c+4{{c}^{3}} \right)$
Simplifying, we get
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=3{{a}^{3}}+6a{{c}^{2}}-2{{a}^{2}}c-4{{c}^{3}}$
Hence we have
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=3{{a}^{3}}+6a{{c}^{2}}-2{{a}^{2}}c-4{{c}^{3}}$

Note: [1] Although we substituted $x={{a}^{2}}+2{{c}^{2}}$ and then applied distributive law, we also could have substituted $y=3a-2c$ and then applied distributive law and arrived at the same solution.
Let $y=3a-2c$, we have
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=\left( {{a}^{2}}+2{{c}^{2}} \right)y$
Applying distributive law, we get
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)={{a}^{2}}y+2{{c}^{2}}y$
Reverting to original variables, we get
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)={{a}^{2}}\left( 3a-2c \right)+2{{c}^{2}}\left( 3a-2c \right)$
Applying distributive law in the expression ${{a}^{2}}\left( 3a-2c \right)$, we get
${{a}^{2}}\left( 3a-2c \right)={{a}^{2}}\left( 3a \right)-{{a}^{2}}\left( 2c \right)$
We know that ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$.
Applying the above formula, we get
\begin{align} & {{a}^{2}}\left( 3a-2c \right)=3{{a}^{2+1}}-2{{a}^{2}}c \\ & =3{{a}^{3}}-2{{a}^{2}}c \\ \end{align}
Similarly, applying distributive law in the expression $2{{c}^{2}}\left( 3a-2c \right)$, we get
$2{{c}^{2}}\left( 3a-2c \right)=2{{c}^{2}}\left( 3a \right)-2{{c}^{2}}\left( 2c \right)$
We know that ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$.
Applying the above formula, we get
\begin{align} & 2{{c}^{2}}\left( 3a-2c \right)=6a{{c}^{2}}-4{{c}^{2+1}} \\ & =6a{{c}^{2}}-4{{c}^{3}} \\ \end{align}
Hence we have
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=3{{a}^{3}}-2{{a}^{2}}c+6a{{c}^{2}}-4{{c}^{3}}$
Rearranging the terms, we get
$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)=3{{a}^{3}}+6a{{c}^{2}}-2{{a}^{2}}c-4{{c}^{3}}$
[2] Along with distributive property multiplication enjoys commutativity and associativity.
Commutative property of multiplication means ab = ba
Associative property of multiplication means a(bc)=(ab)c