Question

How do you expand ${{\left( x-4 \right)}^{4}}$?

Answer 1

In this problem, we have to expand the given expression. We can use the Binomial theorem, to expand the given expression. We can first write the general Binomial theorem and we can compare the given expression and the right-hand side expression in the Binomial theorem, to expand the given expression and we can simplify the exponents for each term of the expansion.

Complete step by step solution:
We know that the Binomial theorem is,
$\Rightarrow {{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^nC_k\left( {{a}^{n-k}}{{b}^{k}} \right)}$ …… (1)
Se know that the given expression to be expanded is,
$\Rightarrow {{\left( x-4 \right)}^{4}}$……. (2)
We can now compare (2) and right-hand part of (1), we get
a = x, b = -4, n = 4.
We can substitute these values in (1), we get
$\Rightarrow {{\left( x-4 \right)}^{4}}=\sum\limits_{k=0}^{4}{\dfrac{4!}{\left( 4-k \right)!k!}\left( {{x}^{4-k}}{{\left( -4 \right)}^{k}} \right)}$
Now we can expand the above summation, we get
$\begin{aligned} & \Rightarrow \dfrac{4!}{\left( 4-0 \right)!0!}{{\left( x \right)}^{4-0}}{{\left( -4 \right)}^{0}}+\dfrac{4!}{\left( 4-1 \right)!1!}{{\left( x \right)}^{4-1}}{{\left( -4 \right)}^{1}}+\dfrac{4!}{\left( 4-2 \right)!2!}{{\left( x \right)}^{4-2}}{{\left( -4 \right)}^{2}}+\dfrac{4!}{\left( 4-3 \right)!3!}{{\left( x \right)}^{4-3}}{{\left( -4 \right)}^{3}} \\\ & +\dfrac{4!}{\left( 4-4 \right)!0!}{{\left( x \right)}^{4-4}}{{\left( -4 \right)}^{4}} \\\ \end{aligned}$
We can now simplify each term, we get
$\Rightarrow 1\times {{x}^{4}}\times 1+4{{x}^{3}}\times \left( -4 \right)+6{{x}^{2}}\left( 16 \right)+4x\left( -64 \right)+256$
We can now further simplify the above step, we get
$\Rightarrow {{x}^{4}}-16{{x}^{3}}+96{{x}^{2}}-256x+256$
Therefore, the expansion of ${{\left( x-4 \right)}^{4}}$ is ${{x}^{4}}-16{{x}^{3}}+96{{x}^{2}}-256x+256$.

Note: Students make mistakes while writing the Binomial theorem in which we should concentrate.
We can also expand this using Pascal’s triangle.
The expansion follows the rule,
${{\left( a+b \right)}^{n}}={{C}_{0}}{{a}^{n}}{{b}^{0}}+{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+...+{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}$
Where, a = x, b = -4, n = 4.
The values of the coefficient from the Pascal’s triangle are
1-4-6-4-1.
Such that,
$1{{a}^{4}}{{b}^{0}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+1{{a}^{0}}{{b}^{4}}$
We can now substitute the values,
a = x, b = -4, n = 4.
We get,
$\Rightarrow 1\times {{x}^{4}}\times 1+4{{x}^{3}}\times \left( -4 \right)+6{{x}^{2}}\left( 16 \right)+4x\left( -64 \right)+256$
We can now further simplify the above step, we get
$\Rightarrow {{x}^{4}}-16{{x}^{3}}+96{{x}^{2}}-256x+256$
Therefore, the expansion of ${{\left( x-4 \right)}^{4}}$ is ${{x}^{4}}-16{{x}^{3}}+96{{x}^{2}}-256x+256$.