Question

How do you use the binomial theorem to expand and simplify the expression $2{\left( {x - 3} \right)^4} + 5{\left( {x - 3} \right)^2}$ ?

Answer 1

Hint : The given question requires us to simplify the expression given to us. The expression given in the question involves two binomial expansions. We can find the required binomial expansions by using the Binomial theorem and then substitute their values in the expression to simplify the expression given in the problem. Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem.

Complete step-by-step answer :
So, to simplify the expression $2{\left( {x - 3} \right)^4} + 5{\left( {x - 3} \right)^2}$ . We can find the values of the binomial expansions ${\left( {x - 3} \right)^4}$ and ${\left( {x - 3} \right)^2}$ using the binomial theorem and then substitute these values into the expression given to us.
So, first we will find the value of ${\left( {x - 3} \right)^4}$ .
So, using the binomial theorem, the binomial expansion of ${\left( {x + y} \right)^n}$ is $\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}}$
So, the binomial expansion of ${\left( {x - 3} \right)^4}$ is $\sum\nolimits_{r = 0}^4 {\left( {^4{C_r}} \right){{\left( x \right)}^{4 - r}}{{\left( { - 3} \right)}^r}}$ .
Now, we have to expand the expression $\sum\nolimits_{r = 0}^4 {\left( {^4{C_r}} \right){{\left( x \right)}^{4 - r}}{{\left( { - 3} \right)}^r}}$ and we are done with the binomial expansion of ${\left( {x - 3} \right)^4}$ . So, we get,
$= \left( {^4{C_0}} \right){\left( x \right)^4}{\left( { - 3} \right)^0} + \left( {^4{C_1}} \right){\left( x \right)^3}{\left( { - 3} \right)^1} + \left( {^4{C_2}} \right){\left( x \right)^2}{\left( { - 3} \right)^2} + \left( {^4{C_3}} \right){\left( x \right)^1}{\left( { - 3} \right)^3} + \left( {^4{C_4}} \right){\left( x \right)^0}{\left( { - 3} \right)^4}$
Substituting values of combination formulae and simplifying calculations, we get,
$= \left( 1 \right){\left( x \right)^4}{\left( { - 3} \right)^0} + \left( 4 \right){\left( x \right)^3}{\left( { - 3} \right)^1} + \left( 6 \right){\left( x \right)^2}{\left( { - 3} \right)^2} + \left( 4 \right){\left( x \right)^1}{\left( { - 3} \right)^3} + \left( 1 \right){\left( x \right)^0}{\left( { - 3} \right)^4}$
Now simplifying the calculations further, we get,
$= \left( 1 \right){\left( x \right)^4}\left( 1 \right) + \left( 4 \right){\left( x \right)^3}\left( { - 3} \right) + \left( 6 \right){\left( x \right)^2}\left( 9 \right) + \left( 4 \right){\left( x \right)^1}\left( { - 27} \right) + \left( 1 \right){\left( x \right)^0}\left( {81} \right)$
$= {x^4} - 12{x^3} + 54{x^2} - 108x + 81$
Hence, we get the binomial expansion of ${\left( {x - 3} \right)^4}$ as ${x^4} - 12{x^3} + 54{x^2} - 108x + 81$ .
Now, we have to find the value of ${\left( {x - 3} \right)^2}$ .
So, the binomial expansion of ${\left( {x - 3} \right)^2}$ is $\sum\nolimits_{r = 0}^2 {\left( {^2{C_r}} \right){{\left( x \right)}^{2 - r}}{{\left( { - 3} \right)}^2}}$ . $\Rightarrow \sum\nolimits_{r = 0}^2 {\left( {^2{C_r}} \right){{\left( x \right)}^{2 - r}}{{\left( { - 3} \right)}^2}} = \left( {^2{C_0}} \right){\left( x \right)^2}{\left( { - 3} \right)^0} + \left( {^2{C_1}} \right){\left( x \right)^1}{\left( { - 3} \right)^1} + \left( {^2{C_2}} \right){\left( x \right)^0}{\left( { - 3} \right)^2}$
Substituting values of combination formulae and simplifying calculations, we get,
$= \left( 1 \right){\left( x \right)^2}{\left( { - 3} \right)^0} + \left( 2 \right){\left( x \right)^1}{\left( { - 3} \right)^1} + \left( 1 \right){\left( x \right)^0}{\left( { - 3} \right)^2}$
$= \left( 1 \right){\left( x \right)^2}\left( 1 \right) + \left( 2 \right){\left( x \right)^1}\left( { - 3} \right) + \left( 1 \right){\left( x \right)^0}\left( 9 \right)$
Simplifying the expression, we get,
$= {x^2} - 6x + 9$
Now, we can substitute the values of both the binomial expansions into the original expression $2{\left( {x - 3} \right)^4} + 5{\left( {x - 3} \right)^2}$ .
So, we get,
$\Rightarrow 2\left[ {{x^4} - 12{x^3} + 54{x^2} - 108x + 81} \right] + 5\left[ {{x^2} - 6x + 9} \right]$
Opening the brackets, we get,
$\Rightarrow 2{x^4} - 24{x^3} + 108{x^2} - 216x + 162 + 5{x^2} - 30x + 45$
Adding up the like terms, we get,
$\Rightarrow 2{x^4} - 24{x^3} + 113{x^2} - 246x + 209$
Hence, the expression $2{\left( {x - 3} \right)^4} + 5{\left( {x - 3} \right)^2}$ is simplified as $2{x^4} - 24{x^3} + 113{x^2} - 246x + 209$ using binomial theorem.
So, the correct answer is “ $2{x^4} - 24{x^3} + 113{x^2} - 246x + 209$ ”.

Note : The given problem can be solved by various methods. The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. The binomial expansion of ${\left( {x - 3} \right)^2}$ can also be calculated using the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.