# Expand the following algebraic expression by finding the product

$\left( {{a}^{2}}+2{{c}^{2}} \right)\left( 3a-2c \right)$.

Answer

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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.

Complete step-by-step answer:

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

Complete step-by-step answer:

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

Last updated date: 30th Sep 2023

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