Question

How do you expand the binomial ${{\left( 3x-1 \right)}^{4}}$?

Answer 1

This type of question is based on the concept of binomials. We can solve this question with the help of the expansion of binomial, that is, ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$. And from the question, we get, a=3x, b=-1, and n=4. Then, substitute these values in the binomial expansion. Simplify the obtained equation using factorial, that is, $n!=n.\left( n-1 \right).\left( n-2 \right).........3.2.1$. Then take the common terms outside the bract, if needed.

Complete step by step answer:
It is given, the binomial is ${{\left( 3x-1 \right)}^{4}}$. And we have been asked to find the expansion of the given binomial. We know that the formula of the binomial expansion is ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$.
Now substitute, a=3x, b=-1 and n=4.
We get,
${{\left( 3x-1 \right)}^{4}}={}^{4}{{C}_{0}}{{\left( 3x \right)}^{4}}+{}^{4}{{C}_{1}}{{\left( 3x \right)}^{4-1}}\left( -1 \right)+{}^{4}{{C}_{2}}{{\left( 3x \right)}^{4-2}}{{\left( -1 \right)}^{2}}+{}^{4}{{C}_{3}}3x{{\left( -1 \right)}^{3}}+{}^{4}{{C}_{4}}{{\left( -1 \right)}^{4}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Using this formula, we have to individually find the values of ${}^{4}{{C}_{0}},{}^{4}{{C}_{1}},{}^{4}{{C}_{2}},{}^{4}{{C}_{3}}$ and ${}^{4}{{C}_{4}}$.
First, consider ${}^{4}{{C}_{0}}$.
${}^{4}{{C}_{0}}=\dfrac{4!}{0!\left( 4-0 \right)!}$
We know that, $0!=1$.
$\Rightarrow {}^{4}{{C}_{0}}=\dfrac{4!}{1\times 4!}$
Canceling out $4!$ on the numerator and the denominator, we get,
${}^{4}{{C}_{0}}=1$.
Now consider, ${}^{4}{{C}_{1}}$.
${}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!}$.
We know that $1!=1$.
$\Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1\times 3!}$.
Expanding the factorial, we get, ${}^{4}{{C}_{1}}=\dfrac{4\times 3\times 2\times 1}{3\times 2\times 1}$.
$\Rightarrow {}^{4}{{C}_{1}}=4$.
Then consider, ${}^{4}{{C}_{2}}$.
${}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!}$.
$\Rightarrow {}^{4}{{C}_{2}}=\dfrac{4!}{2!2!}$.
Expanding the factorial, we get, ${}^{4}{{C}_{2}}=\dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}$.
Canceling the common terms from numerator and denominator, we get,
${}^{4}{{C}_{2}}=3\times 2$.
$\Rightarrow {}^{4}{{C}_{2}}=6$.
Then consider, ${}^{4}{{C}_{3}}$
${}^{4}{{C}_{3}}=\dfrac{4!}{3!\left( 4-3 \right)!}$
$\Rightarrow {}^{4}{{C}_{3}}=\dfrac{4!}{3!1!}$
We know that, $1!=1$.
$\Rightarrow {}^{4}{{C}_{3}}=\dfrac{4!}{3!}$
Expanding the factorial, we get, ${}^{4}{{C}_{3}}=\dfrac{4\times 3\times 2\times 1}{3\times 2\times 1}$.
Canceling the common terms from numerator and denominator, we get,
${}^{4}{{C}_{3}}=4$.
Then consider, ${}^{4}{{C}_{4}}$.
${}^{4}{{C}_{4}}=\dfrac{4!}{4!\left( 4-4 \right)!}$
$\Rightarrow {}^{4}{{C}_{4}}=\dfrac{4!}{4!0!}$
We know that $0!=1$.
$\Rightarrow {}^{4}{{C}_{4}}=\dfrac{4!}{1\times 4!}$
Canceling out $4!$ on the numerator and the denominator, we get,
${}^{4}{{C}_{4}}=1$.
Now substitute all the obtained values in the binomial expansion of ${{\left( 3x-1 \right)}^{4}}$, that is,
${{\left( 3x-1 \right)}^{4}}={}^{4}{{C}_{0}}{{\left( 3x \right)}^{4}}+{}^{4}{{C}_{1}}{{\left( 3x \right)}^{4-1}}\left( -1 \right)+{}^{4}{{C}_{2}}{{\left( 3x \right)}^{4-2}}{{\left( -1 \right)}^{2}}+{}^{4}{{C}_{3}}3x{{\left( -1 \right)}^{3}}+{}^{4}{{C}_{4}}{{\left( -1 \right)}^{4}}$
$=1{{\left( 3x \right)}^{4}}+4{{\left( 3x \right)}^{4-1}}\left( -1 \right)+6{{\left( 3x \right)}^{4-2}}{{\left( -1 \right)}^{2}}+4\left( 3x \right){{\left( -1 \right)}^{3}}+1{{\left( -1 \right)}^{4}}$
$={{\left( 3x \right)}^{4}}+4{{\left( 3x \right)}^{3}}\left( -1 \right)+6{{\left( 3x \right)}^{2}}{{\left( -1 \right)}^{2}}+4\left( 3x \right){{\left( -1 \right)}^{3}}+{{\left( -1 \right)}^{4}}$
We know that
$\left( -1 \right)x=-x,{{\left( -1 \right)}^{2}}=1,{{\left( -1 \right)}^{3}}=-1$ and ${{\left( -1 \right)}^{4}}=1$.
We get,
${{\left( 3x-1 \right)}^{4}}={{\left( 3x \right)}^{4}}-4{{\left( 3x \right)}^{3}}+6{{\left( 3x \right)}^{2}}-4\left( 3x \right)+1$
$={{3}^{4}}{{x}^{4}}-4\times {{3}^{3}}{{x}^{3}}+6\times {{3}^{2}}{{x}^{2}}-4\left( 3x \right)+1$
Since ${{3}^{4}}=91,{{3}^{3}}=27$ and ${{3}^{2}}=9$, we get
${{\left( 3x-1 \right)}^{4}}=91{{x}^{4}}-4\times 27{{x}^{3}}+6\times 9{{x}^{2}}-4\left( 3x \right)+1$
$\Rightarrow {{\left( 3x-1 \right)}^{4}}=91{{x}^{4}}-108{{x}^{3}}+54{{x}^{2}}-12x+1$

$\therefore$The expansion of the binomial ${{\left( 3x-1 \right)}^{4}}$ is ${{\left( 3x-1 \right)}^{4}}=91{{x}^{4}}-108{{x}^{3}}+54{{x}^{2}}-12x+1$

Note: Whenever we get this type of problem, we need to make sure about the sign and the power. The factorial should be performed separately to avoid confusion. Any type of problems related to binomial can be expanded using this method. Also, we should avoid calculation mistakes to obtain accurate answers. Avoid mistakes based on sign convention. Similarly, we can expect problems to find the solution for the binomial ${{\left( x+4 \right)}^{4}}$.