Question

Expand the expression using the Binomial Theorem:

${\left(3x-4\right)}^4.$

Short answer

The binomial expansion of${\left(3x-4\right)}^4$ is$81x^4-432x^3+864x^2-768x+256$.

Step-by-step solution

Given information:

Consider the given expression ${\left(3x-4\right)}^4.$

Rewrite the expression ${\left(3x-4\right)}^4$as ${\left[3x+\left(-4\right)\right]}^4$.

On comparing with the general form of binomial theorem ${\left(x+a\right)}^n$, we get,

$x=3x,a=-4,n=4$.

Required Formulas:

Formula of Binomial Theorem:

${\left(x+a\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+...+\left(\begin{array}{c}n\\ n\end{array}\right)a^n$

Pascal Triangle :

Substitute the values in the binomial expansion formula:

${\left[3x+\left(-4\right)\right]}^4=\left(\begin{array}{c}4\\ 0\end{array}\right){\left(3x\right)}^4+\left(\begin{array}{c}4\\ 1\end{array}\right)\left(-4\right){\left(3x\right)}^3+\left(\begin{array}{c}4\\ 2\end{array}\right){(-4)}^2{\left(3x\right)}^2+\left(\begin{array}{c}4\\ 3\end{array}\right){(-4)}^3\left(3x\right)+\left(\begin{array}{c}4\\ 4\end{array}\right){(-4)}^4$

Simplification:

Here $n=4$ , refer 4^th row of the Pascal triangle, role="math" localid="1647057469360" $\left[\left(\begin{array}{c}4\\ 0\end{array}\right)=1,\left(\begin{array}{c}4\\ 1\end{array}\right)=4,\left(\begin{array}{c}4\\ 2\end{array}\right)=6,\left(\begin{array}{c}4\\ 3\end{array}\right)=4,\left(\begin{array}{c}4\\ 4\end{array}\right)=1\right]$

$\therefore {\left[3x+\left(-4\right)\right]}^4=\left[1\times {\left(3x\right)}^4\right]+\left[4\times \left(-4\right){\left(3x\right)}^3\right]+\left[6\times {\left(-4\right)}^2{\left(3x\right)}^2\right]+\left[4\times {(-4)}^3\left(3x\right)\right]+\left[1\times {\left(-4\right)}^4\right]$

$$\begin{gathered} =\left[81x^4\right]+\left[\left(-16\right)\times 27x^3\right]+\left[96\times 9x^2\right]+\left[\left(-256\right)\times 3x\right]+\left[256\right] \\ =81x^4-432x^3+864x^2-768x+256 \end{gathered}$$