Question

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 \(n = 4\), \(a=3x\), \(b = 1\).

Step2: Calculate each term for \(k = 0,1,2,3,4\)

• For \(k = 0\):

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

• For \(k = 1\):

\(\binom{4}{1}(3x)^{3}(1)^{1}=\frac{4!}{1!3!}\times27x^{3}\times1=4\times27x^{3} = 108x^{3}\)

• For \(k = 2\):

\(\binom{4}{2}(3x)^{2}(1)^{2}=\frac{4!}{2!2!}\times9x^{2}\times1=\frac{4\times3}{2\times1}\times9x^{2}=6\times9x^{2}=54x^{2}\)

• For \(k = 3\):

\(\binom{4}{3}(3x)^{1}(1)^{3}=\frac{4!}{3!1!}\times3x\times1 = 4\times3x=12x\)

• For \(k = 4\):

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

Step3: Sum up all the terms

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

Answer:

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