%5E5.jpeg "(1+x+x^2+x^3)^5")
Expansion of (1+x+x^2+x^3)^5
In algebra, the expansion of expressions involving polynomials can be a challenging task. One such expression is (1+x+x^2+x^3)^5, which involves raising a polynomial to a power of 5. In this article, we will explore the expansion of this expression and discuss the resulting polynomial.
Binomial Theorem
To expand (1+x+x^2+x^3)^5, we can use the Binomial Theorem, which states that for any positive integer n and real numbers a and b:
$(a+b)^n = \sum_{k=0}^n {n \choose k} a^{n-k} b^k$
where {n \choose k} is the binomial coefficient.
In our case, we have a = 1+x+x^2+x^3 and n = 5. Applying the Binomial Theorem, we get:
$(1+x+x^2+x^3)^5 = \sum_{k=0}^5 {5 \choose k} (1+x+x^2+x^3)^{5-k} (x^3)^k$
Expansion
Now, let's expand the expression using the Binomial Theorem:
$(1+x+x^2+x^3)^5 = {5 \choose 0} (1+x+x^2+x^3)^5 + {5 \choose 1} (1+x+x^2+x^3)^4 x^3 + {5 \choose 2} (1+x+x^2+x^3)^3 x^6 + {5 \choose 3} (1+x+x^2+x^3)^2 x^9 + {5 \choose 4} (1+x+x^2+x^3) x^{12} + {5 \choose 5} x^{15}$
Simplifying the expression, we get:
$(1+x+x^2+x^3)^5 = 1 + 5x + 15x^2 + 35x^3 + 70x^4 + 126x^5 + 210x^6 + 330x^7 + 495x^8 + 792x^9 + 1260x^{10} + 1985x^{11} + 3003x^{12} + 4389x^{13} + 6279x^{14} + 8439x^{15}$
Resulting Polynomial
The resulting polynomial is a polynomial of degree 15 with 16 terms. The coefficients of the polynomial can be calculated using the Binomial Theorem.
In conclusion, the expansion of (1+x+x^2+x^3)^5 is a polynomial of degree 15 with 16 terms. The Binomial Theorem provides a powerful tool for expanding expressions involving polynomials.
