Question

question
find the coefficient of $x^{2}y^{2}$ in the expansion of $(-x - 2y)^{4}$. give your answer as an integer.
provide your answer below:

Explanation:

Step1: Apply binomial theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\). Here \(a=-x\), \(b = - 2y\) and \(n = 4\), so \((-x-2y)^4=\sum_{k = 0}^{4}\binom{4}{k}(-x)^{4 - k}(-2y)^{k}=\sum_{k = 0}^{4}\binom{4}{k}(-1)^{4 - k}(-2)^{k}x^{4 - k}y^{k}\).

Step2: Find the value of k for \(x^{2}y^{2}\)

We want to find the coefficient of \(x^{2}y^{2}\). Set \(4 - k=2\), then \(k = 2\).

Step3: Calculate the coefficient

When \(k = 2\), the term is \(\binom{4}{2}(-1)^{4 - 2}(-2)^{2}x^{4 - 2}y^{2}\). First, \(\binom{4}{2}=\frac{4!}{2!(4 - 2)!}=\frac{4\times3\times2!}{2!\times2!}=\frac{4\times3}{2\times1}=6\). Then \((-1)^{4 - 2}=1\) and \((-2)^{2}=4\). So the coefficient is \(\binom{4}{2}(-1)^{2}(-2)^{2}=6\times1\times4 = 24\).

Answer:

24