# Expand the following algebraic expression by finding the product

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

Last updated date: 16th Mar 2023

•

Total views: 303.3k

•

Views today: 5.83k

Answer

Verified

303.3k+ views

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

Recently Updated Pages

If abc are pthqth and rth terms of a GP then left fraccb class 11 maths JEE_Main

If the pthqth and rth term of a GP are abc respectively class 11 maths JEE_Main

If abcdare any four consecutive coefficients of any class 11 maths JEE_Main

If A1A2 are the two AMs between two numbers a and b class 11 maths JEE_Main

If pthqthrth and sth terms of an AP be in GP then p class 11 maths JEE_Main

One root of the equation cos x x + frac12 0 lies in class 11 maths JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?