Binomial Theorem For Quadratic Equation

[Air]

Question:
Find the coefficient of $x^5$ in $(1+x+x^2)^4$.


Problem:
I have not come across expanding brackets which have $x^2$. I know how to apply the binomial theorem for $(a+b)^n$ or $(1+a)^n$ but have not come across $(1 + ax + ax^2)^n$. They are not explained in my textbooks so I was wondering if you could provide hints or redirect me to a useful link. Thanks.

 

[rock.freak667]

Just use $b=x+x^2$ and if you need to expand out $b^2,b^3,b^4$ just use the terms that will give you $x^5$

 

[Architeuthis Dux]

The binomial theorem can be applied to any polynomial expression, including (1+ax+ax^2)^n. The key is to think of the expression as a binomial with two terms: (1+ax)^n and (ax^2)^n. Then, you can use the binomial theorem to expand each of these terms separately and then combine the results. For example, (1+ax)^n can be expanded using the binomial theorem as (1+ax)^n = ∑ (n choose k) * 1^(n-k) * (ax)^k = ∑ (n choose k) * a^k * x^k. Similarly, (ax^2)^n can be expanded as (ax^2)^n = ∑ (n choose k) * a^k * (x^2)^k = ∑ (n choose k) * a^k * x^(2k). Then, to find the coefficient of x^5 in (1+x+x^2)^4, you would need to find the terms in the expanded form that have x^5 as a factor and add them together. I hope this helps! You can also refer to this link for more information and examples on applying the binomial theorem to expressions with higher powers: https://www.mathsisfun.com/algebra/binomial-theorem.html