Question

expand each binomial.

$(3x + y)^4$

$\bigcirc\\ 81x^4 + 108x^3y^4 + 54x^2y^3 + 12xy^2 + y$

$\bigcirc\\ 81x^4 + y^4$

$\bigcirc\\ 81x^4 + 108x^3y + 54x^2y^2 + 12xy^3 + y^4$

$\bigcirc\\ 81x^4 - 108x^3y + 54x^2y^2 - 12xy^3 + y^4$

Explanation:

Step1: Recall 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 \(n = 4\), \(a=3x\), \(b = y\).

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

• For \(k = 0\): \(\binom{4}{0}(3x)^{4 - 0}y^{0}=\frac{4!}{0!4!}(81x^{4})(1)=81x^{4}\)
• For \(k = 1\): \(\binom{4}{1}(3x)^{4 - 1}y^{1}=\frac{4!}{1!3!}(27x^{3})(y)=4\times27x^{3}y = 108x^{3}y\)
• For \(k = 2\): \(\binom{4}{2}(3x)^{4 - 2}y^{2}=\frac{4!}{2!2!}(9x^{2})(y^{2})=6\times9x^{2}y^{2}=54x^{2}y^{2}\)
• For \(k = 3\): \(\binom{4}{3}(3x)^{4 - 3}y^{3}=\frac{4!}{3!1!}(3x)(y^{3})=4\times3xy^{3}=12xy^{3}\)
• For \(k = 4\): \(\binom{4}{4}(3x)^{4 - 4}y^{4}=\frac{4!}{4!0!}(1)(y^{4})=y^{4}\)

Step3: Sum the terms

Adding all the terms together: \(81x^{4}+108x^{3}y + 54x^{2}y^{2}+12xy^{3}+y^{4}\)

Answer:

\(81x^{4}+108x^{3}y + 54x^{2}y^{2}+12xy^{3}+y^{4}\) (the third option)