Question

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expand the expression to a polynomial in standard form:
$(3x - 1)^4$

Explanation:

Step1: Recall the binomial theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n!=n\times(n - 1)\times\cdots\times1\). For \((3x-1)^{4}\), we have \(a = 3x\), \(b=- 1\) and \(n = 4\).

Step2: Calculate each term of the expansion

• For \(k = 0\):

\(\binom{4}{0}(3x)^{4-0}(-1)^{0}=\frac{4!}{0!(4 - 0)!}\times(81x^{4})\times1=1\times81x^{4}=81x^{4}\)

• For \(k = 1\):

\(\binom{4}{1}(3x)^{4 - 1}(-1)^{1}=\frac{4!}{1!(4 - 1)!}\times(27x^{3})\times(-1)=4\times27x^{3}\times(-1)=- 108x^{3}\)

• For \(k = 2\):

\(\binom{4}{2}(3x)^{4-2}(-1)^{2}=\frac{4!}{2!(4 - 2)!}\times(9x^{2})\times1 = 6\times9x^{2}=54x^{2}\)

• For \(k = 3\):

\(\binom{4}{3}(3x)^{4-3}(-1)^{3}=\frac{4!}{3!(4 - 3)!}\times(3x)\times(-1)=4\times3x\times(-1)=-12x\)

• For \(k = 4\):

\(\binom{4}{4}(3x)^{4 - 4}(-1)^{4}=\frac{4!}{4!(4 - 4)!}\times1\times1 = 1\times1=1\)

Step3: Sum up all the terms

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

Answer:

\(81x^{4}-108x^{3}+54x^{2}-12x + 1\)