# Multinomial Theorem application

Here is my question:

Find the coefficient of $x^3$ in the expansion of $(1 + 2x - 3x^2)^5.$

I know that I am suppose to use the multinomial theorem, but I don't know how to. The proof that's in my textbook approaches it using partial derivatives, as opposed to combinations and I was expecting an example of how to find the coefficient of some expression, but there wasn't any. I've been reading a couple other threads about how to find coefficients of terms and I came across this link and this link as well. After reading through everything, here is my approach:

$$(1)^j(2x)^k(-3x^2)^p,$$ where $(j + k + p )= 5$. Since I want to look for the coefficient of the $x^3$ term, do I have to find combinations such that $(j + k + p) = 5?$ So for example, let $j = 1, k = 1$ and $p = 1$. But this doesn't add up to $5$... and this is where I got stuck Perhaps there is a different approach. By the way, wolfram gave me a coefficient of $-40$.

I don't know why this problem in particular is hard for me. I can do a similar problem of the same type.

EDIT So as the answer below suggests, this problem can be solved using the binomial theorem, but how would it look like if we used the multinomial theorem?

• Well, first, you think about HOW you can form a x^3. Feb 25 '17 at 23:07
• You can achieve this with an x and an x^2, or 3 x's. Feb 25 '17 at 23:08
• Rewrite your expression as (p+q+r)^5 and compute two cases, then multiply by appropriate poewrs (the coefficients of x and x^2 aren't 1 Feb 25 '17 at 23:08
• What two cases? Feb 25 '17 at 23:17
• Since I want to look for the coefficient of the x^3 term, do I have to find combinations such that (j+k+p)=5? You are looking for the $3^{rd}$ power term. Why =5?
– dxiv
Feb 26 '17 at 0:29

Ok but you could just use the Binomial Theorem instead

$$(1+x(2-3x))^5=1+5x(2-3x)+10x^2(2-3x)^2+10x^3(2-3x)^3+...$$

So you don't have to look very hard to find the coefficient of $x^3$ which is $$-120+80$$

Problem: Find the coefficient of $$x^3$$ in $$(1+2x-3x^2)$$.

The first step is to set up a term as $${(1)^a}{(2x)^b}{(-3x^2)^c}$$. $$a+b+c=5$$. Then you create 2 cases where the terms when multiplied, have the factor $$x^3$$.

Case 1: $$a=3, b=1,$$ and $$c=1$$.

This case states $${(1)^3}{(2x)^1}{(-3x^2)}^1=cx^3$$, where c is a real number. First, you need to find the number of ways you can order these cases. That is like asking how many permutations there are of "$$aaabc$$". That is equal to $$\frac{5!}{3!1!1!}=20$$. Then you have to find the products of the coefficients of the monomials(raised to their respective powers) in $${(1)^3}{(2x)^1}{(-3x^2)}^1$$. That is equal to $$(1^3)(2^1)(-3)^1=(2)(-3)=-6$$. Then you multiply $$-6$$ by $$20$$ to get $$-120$$, the coefficient of $$x^3$$ in Case 1.

Case 2: $$a=2, b=3,$$ and $$c=0$$

Using the method from Case 1, You get $$\frac{5!}{2!3!0!}=10$$ and $$(1^2)(2^3)((-3)^0)=8$$. Multiply those 2 together to get $$80$$, the coefficent of $$x^3$$ in Case 2.

To get the final answer, add the results from Case 1 and Case 2, which gives you $$-120+80=$$

$$-40$$