Question

How do you expand ${(2x + 3y)^4}?$

Answer 1

To expand these types of expression with greater degrees, we should use binomial theorem by which the expression could easily expand. Binomial theorem is given as

${(a + b)^n} = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}

n \\

r

\end{array}} \right)} {a^{n - r}}{b^r};{\text{where}};\left( {\begin{array}{*{20}{c}}

n \\

r

\end{array}} \right)$ is the coefficient of respective binomial terms.

Coefficient of binomial terms can be calculated as $\left( {\begin{array}{*{20}{c}}

n \\

r

\end{array}} \right) = {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$

Complete step by step solution:
In order to expand ${(2x + 3y)^4}$ we need to use binomial theorem which is given as

${(a + b)^n} = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}

n \\

r

\end{array}} \right)} {a^{n - r}}{b^r};{\text{where}};\left( {\begin{array}{*{20}{c}}

n \\

r

\end{array}} \right)$ is the coefficient of respective binomial terms

Firstly expanding ${(2x + 3y)^4}$ with help of binomial theorem
$ \Rightarrow {(2x + 3y)^4} = \sum\limits_{r = 0}^4 {\left( {\begin{array}{*{20}{c}}

4 \\

r
\end{array}} \right)} {(2x)^{4 - r}}{(3y)^r}$

Putting value of $r$ from $0\;{\text{to}}\;4$ in the summation function, we will get
$

\Rightarrow {(2x + 3y)^4} = \left( {\begin{array}{*{20}{c}}

4 \\

0

\end{array}} \right){(2x)^{4 - 0}}{(3y)^0} + \left( {\begin{array}{*{20}{c}}

4 \\

1

\end{array}} \right){(2x)^{4 - 1}}{(3y)^1} + \left( {\begin{array}{*{20}{c}}

4 \\

2

\end{array}} \right){(2x)^{4 - 2}}{(3y)^2} + \left( {\begin{array}{*{20}{c}}

4 \\

3

\end{array}} \right){(2x)^{4 - 3}}{(3y)^3} + \left( {\begin{array}{*{20}{c}}

4 \\

4

\end{array}} \right){(2x)^{4 - 4}}{(3y)^4} \\

\Rightarrow {(2x + 3y)^4} = \left( {\begin{array}{*{20}{c}}
4 \\

0

\end{array}} \right){(2x)^4}{(3y)^0} + \left( {\begin{array}{*{20}{c}}

4 \\

1

\end{array}} \right){(2x)^3}{(3y)^1} + \left( {\begin{array}{*{20}{c}}

4 \\

2

\end{array}} \right){(2x)^2}{(3y)^2} + \left( {\begin{array}{*{20}{c}}

4 \\

3

\end{array}} \right){(2x)^1}{(3y)^3} + \left( {\begin{array}{*{20}{c}}

4 \\

4

\end{array}} \right){(2x)^0}{(3y)^4} \\

\Rightarrow {(2x + 3y)^4} = \left( {\begin{array}{*{20}{c}}

4 \\

0

\end{array}} \right)16{x^4} \times 1 + \left( {\begin{array}{*{20}{c}}

4 \\

1

\end{array}} \right)8{x^3} \times 3y + \left( {\begin{array}{*{20}{c}}

4 \\

2

\end{array}} \right)4{x^2} \times 9{y^2} + \left( {\begin{array}{*{20}{c}}

4 \\

3

\end{array}} \right)2x \times 27{y^3} + \left( {\begin{array}{*{20}{c}}

4 \\

4

\end{array}} \right)1 \times 81{y^4} \\

\Rightarrow {(2x + 3y)^4} = \left( {\begin{array}{*{20}{c}}

4 \\

0

\end{array}} \right)16{x^4} + \left( {\begin{array}{*{20}{c}}

4 \\

1

\end{array}} \right)24{x^3}y + \left( {\begin{array}{*{20}{c}}

4 \\

2

\end{array}} \right)36{x^2}{y^2} + \left( {\begin{array}{*{20}{c}}

4 \\

3

\end{array}} \right)54x{y^3} + \left( {\begin{array}{*{20}{c}}

4 \\

4

\end{array}} \right)81{y^4} \\
$

Now we will find the values of respective coefficient of binomial terms $\left(

{{\text{i}}{\text{.e}}{\text{.}};\left( {\begin{array}{*{20}{c}}

4 \\

0

\end{array}} \right),;\left( {\begin{array}{*{20}{c}}

4 \\

1

\end{array}} \right),;\left( {\begin{array}{*{20}{c}}

4 \\

2

\end{array}} \right),;\left( {\begin{array}{*{20}{c}}

4 \\

3

\end{array}} \right),;\left( {\begin{array}{*{20}{c}}

4 \\

4

\end{array}} \right)} \right)$

If $$\left( {\begin{array}{*{20}{c}}

n \\

r

\end{array}} \right)$$ is the binomial coefficient then its value can be calculated as follows

$\left( {\begin{array}{*{20}{c}}
n \\

r

\end{array}} \right) = {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$, where$n!$means multiplication of numbers up to$n,;{\text{i}}{\text{.e}}{\text{.}};n! = 1 \times 2 \times 3 \times ...... \times n$

Now finding respective values for binomial coefficients of ${(2x + 3y)^4},;\left(

{{\text{i}}{\text{.e}}{\text{.}};\left( {\begin{array}{*{20}{c}}

4 \\

0

\end{array}} \right),;\left( {\begin{array}{*{20}{c}}

4 \\

1

\end{array}} \right),;\left( {\begin{array}{*{20}{c}}

4 \\

2

\end{array}} \right),;\left( {\begin{array}{*{20}{c}}

4 \\

3

\end{array}} \right),;\left( {\begin{array}{*{20}{c}}

4 \\

4

\end{array}} \right)} \right)$$

\left( {\begin{array}{*{20}{c}}

4 \\

0

\end{array}} \right) = {}^4{C_0} = \dfrac{{4!}}{{0!(4 - 0)!}} = \dfrac{{4!}}{{1 \times 4!}} = 1

\\

\left( {\begin{array}{*{20}{c}}

4 \\

1

\end{array}} \right) = {}^4{C_1} = \dfrac{{4!}}{{1!(4 - 1)!}} = \dfrac{{4!}}{{1 \times 3!}} =

\dfrac{{1

\times 2 \times 3 \times 4}}{{1 \times 2 \times 3}} = 4 \\

\left( {\begin{array}{*{20}{c}}

4 \\

2

\end{array}} \right) = {}^4{C_2} = \dfrac{{4!}}{{2!(4 - 2)!}} = \dfrac{{4!}}{{1 \times 2 \times
2!}} = \dfrac{{1 \times 2 \times 3 \times 4}}{{1 \times 2 \times (1 \times 2)}} = 6 \\

\left( {\begin{array}{*{20}{c}}

4 \\

3

\end{array}} \right) = {}^4{C_3} = \dfrac{{4!}}{{3!(4 - 3)!}} = \dfrac{{4!}}{{1 \times 2 \times 3 \times 1!}} = \dfrac{{1 \times 2 \times 3 \times 4}}{{1 \times 2 \times 3 \times 1}} = 4 \\

\left( {\begin{array}{*{20}{c}}

4 \\

4

\end{array}} \right) = {}^4{C_4} = \dfrac{{4!}}{{4!(4 - 4)!}} = \dfrac{{4!}}{{1 \times 2 \times 3
\times 4
\times 0!}} = \dfrac{{1 \times 2 \times 3 \times 4}}{{1 \times 2 \times 3 \times 4 \times 1}} = 1 \\
$

Now putting these respective values of coefficient of binomial terms in above expansion, we will get
$

\Rightarrow {(2x + 3y)^4} = \left( {\begin{array}{*{20}{c}}

4 \\

0

\end{array}} \right)16{x^4} + \left( {\begin{array}{*{20}{c}}

4 \\

1
\end{array}} \right)24{x^3}y + \left( {\begin{array}{*{20}{c}}

4 \\

2

\end{array}} \right)36{x^2}{y^2} + \left( {\begin{array}{*{20}{c}}

4 \\

3

\end{array}} \right)54x{y^3} + \left( {\begin{array}{*{20}{c}}

4 \\

4

\end{array}} \right)81{y^4} \\

\Rightarrow {(2x + 3y)^4} = 1 \times 16{x^4} + 4 \times 24{x^3}y + 6 \times 36{x^2}{y^2} + 4 \times
54x{y^3} + 1 \times 81{y^4} \\
\Rightarrow {(2x + 3y)^4} = 16{x^4} + 96{x^3}y + 216{x^2}{y^2} + 216x{y^3} + 81{y^4} \\
$

Therefore we get the expanded form of the expression ${(2x + 3y)^4}$ as follows
${(2x + 3y)^4} = 16{x^4} + 96{x^3}y + 216{x^2}{y^2} + 216x{y^3} + 81{y^4}$

Note: The value of $0!$ is always taken as $1$ and also value of \left( {\begin{array}{*{20}{c}} n \\\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\\ n \end{array}} \right) is always equals to $n$

If coefficients of binomial terms are \left( {\begin{array}{*{20}{c}} n \\\ r \end{array}} \right)\;{\text{and}}\;\left( {\begin{array}{*{20}{c}} n \\\ {n - r} \end{array}} \right) then we can write that their values are equal, that is \left( {\begin{array}{*{20}{c}} n \\\ r \end{array}} \right)\; = \;\left( {\begin{array}{*{20}{c}} n \\\ {n - r} \end{array}} \right)

We can find the values of coefficients of respective binomial terms from one more method, which is known as Pascal’s triangle. It is an infinite equilateral triangle with an array of numbers, the two sides of the triangle goes down with all “1” values, and the rest numbers are formed by the sum of the above two numbers. Try to solve this problem with the help of Pascal’s triangle by yourself and check the answer.