How do you find the coefficient of #x^3y^2# in the expansion of #(x-3y)^5#?

Question

Lily Anderson · Accepted Answer

The coefficient of #x^3y^2# in #(x-3y)^5# is #90#.

Expansion of #(a+b)^n# gives us #(n+1)# terms which are given by binomial expansion #color(white)x ^nC_ra^((n-r))b^r#, where #r# ranges from #n# to #0#. Note that powers of #a# and #b# add up to #n# and in the given problem this #n=5#. In #(x-3y)^5#, we need coefficient of #x^3y^2#, we have #3^(rd)# power of #x# and as such #r=5-3=2# and as such the desired coefficient of #x^3y^2# is given by #color(white)x ^5C_2x^((5-2))(-3y)^2=(5xx4)/(1xx2)x^3(-3y)^2# = #10x^3xx9y^2=90x^3y^2# Hence, the coefficient of #x^3y^2# in #(x-3y)^5# is #90#.

Elijah Allen · Answer

undefined To find the coefficient of $x^3y^2$ in the expansion of $(x-3y)^5$, you can use the binomial theorem or Pascal's triangle. The coefficient can be calculated using the formula:

$$\binom{n}{k} \cdot a^{n-k} \cdot b^k$$

where $n$ is the power to which the binomial is raised, $k$ is the term number, $a$ and $b$ are the terms in the binomial, and $\binom{n}{k}$ is the binomial coefficient.

In this case, $n = 5$, $a = x$, $b = -3y$, and $k$ is determined by the power of $x$ and $y$ in the term, which is $3$ for $x^3$ and $2$ for $y^2$.

Substituting these values into the formula:

$$\binom{5}{3} \cdot (x)^{5-3} \cdot (-3y)^{3}$$

$$= \binom{5}{3} \cdot x^2 \cdot (-3)^3y^3$$

$$= 10 \cdot x^2 \cdot (-27)y^3$$

$$= -270x^2y^3$$

So, the coefficient of $x^3y^2$ in the expansion of $(x-3y)^5$ is $-270$.