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Use the binomial expansion to fully expand (3 − 1)^4

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According to the binomial expansion, it is possible to expand (a+b)^n as:


(a+b)^n=\begin{pmatrix}n \\ 0\end{pmatrix}a^nb^0+\begin{pmatrix}n \\ 1\end{pmatrix}a^(n-1)b^1+\cdots+\begin{pmatrix}n \\ n\end{pmatrix}a^0b^n

Where the binomial coefficients are given by:


\begin{pmatrix}n \\ k\end{pmatrix}=(n!)/(k!(n-k)!)

For n=4, we have:


\begin{gathered} (a+b)^4=\begin{pmatrix}4 \\ 0\end{pmatrix}a^4+\begin{pmatrix}4 \\ 1\end{pmatrix}a^3b+\begin{pmatrix}4 \\ 2\end{pmatrix}a^2b^2+\begin{pmatrix}4 \\ 3\end{pmatrix}ab^3+\begin{pmatrix}4 \\ 4\end{pmatrix}b^4 \\ =(4!)/(0!\cdot4!)a^4+(4!)/(1!\cdot3!)a^3b+(4!)/(2!\cdot2!)a^2b^2+(4!)/(3!\cdot1!)ab^3+(4!)/(4!\cdot0!)b^4 \\ =a^4+4a^3b+6a^2b^2+4ab^3+b^4 \end{gathered}

Replace a=3x and b=-1 to fully expand (3x-1)^4:


\begin{gathered} \Rightarrow(3x-1)^4=(3x)^4+4(3x)^3(-1)+6(3x)^2(-1)^2+4(3x)(-1)^3+(-1)^4 \\ =81x^4-108x^3+54x^2-12x+1 \end{gathered}

Therefore, the full expansion of (3x-1)^4 is:


(3x-1)^4=81x^4-108x^3+54x^2-12x+1

User Almenon
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