(1+x+2 X^3)(3/2x^2-1/3x)^9

4 min read Jun 16, 2024

Expanding the Polynomial (1 + x + 2x^3)(3/2x^2 - 1/3x)^9

This article explores the expansion of the polynomial (1 + x + 2x^3)(3/2x^2 - 1/3x)^9. We will break down the process using the Binomial Theorem and discuss the key concepts involved.

Understanding the Binomial Theorem

The Binomial Theorem is a powerful tool for expanding expressions of the form (a + b)^n. It states that:

(a + b)^n = Σ (n choose k) a^(n-k) b^k

where:

n is a non-negative integer (the power).
k is an integer ranging from 0 to n.
(n choose k) represents the binomial coefficient, which is calculated as n! / (k! * (n-k)!).

Applying the Binomial Theorem

Let's apply the Binomial Theorem to the second part of our expression, (3/2x^2 - 1/3x)^9:

(3/2x^2 - 1/3x)^9 = Σ (9 choose k) (3/2x^2)^(9-k) (-1/3x)^k

Expanding this, we get:

(9 choose 0) (3/2x^2)^9 (-1/3x)^0 + (9 choose 1) (3/2x^2)^8 (-1/3x)^1 + (9 choose 2) (3/2x^2)^7 (-1/3x)^2 + ... + (9 choose 9) (3/2x^2)^0 (-1/3x)^9

Simplifying the Expansion

Now we can simplify each term by:

Calculating the binomial coefficients. For example, (9 choose 2) = 9! / (2! * 7!) = 36.
Simplifying the powers of x. For example, (3/2x^2)^8 = (3/2)^8 * x^16.
Combining the coefficients and powers of x.

This will result in a long polynomial expression with terms ranging from x^18 to x^0.

Multiplying by (1 + x + 2x^3)

Finally, we need to multiply this expanded polynomial by the first term (1 + x + 2x^3). This involves multiplying each term in the expanded polynomial by each term in (1 + x + 2x^3) and combining like terms.

This process will lead to a much larger polynomial with terms ranging from x^21 to x^0.

Conclusion

The expansion of (1 + x + 2x^3)(3/2x^2 - 1/3x)^9 is a complex process, but can be systematically carried out using the Binomial Theorem. The key steps involve expanding the binomial term, simplifying the resulting expression, and then multiplying by the remaining factor. The final result will be a lengthy polynomial with terms ranging from x^21 to x^0.