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Finding the Coefficient of (x^4) in the Expansion of (1 x x^2 x^8)

Finding the Coefficient of (x^4) in the Expansion of (1 x x^2 x^8)

January 08, 2025 Education

How to Find the Coefficient of (x^4) in the Expansion of (1 x x^2 x^8)

To find the coefficient of (x^4) in the expansion of (1 x x^2 x^8), we can use the multinomial expansion theorem. This theorem allows us to expand expressions that are the sum of multiple terms, similar to the binomial theorem but for more than two terms.

Identify the Structure

Our expression is (1 x x^2 x^8). We will use the multinomial theorem, which states:

n  a_1   a_2   a_3^n  sum_{k_1   k_2   k_3  n} frac{n!}{k_1! k_2! k_3!} a_1^{k_1} a_2^{k_2} a_3^{k_3}

In this problem, our terms are (a_1 1), (a_2 x), (a_3 x^2), and (n 8). We need to find the combinations of (k_1), (k_2), and (k_3) such that:

(k_2 2k_3 4) (k_1 k_2 k_3 8)

Set Up the Equations

Expressing (k_1) in terms of (k_2) and (k_3), we get:

(k_1 8 - k_2 - k_3)

Now we have two equations:

(k_2 2k_3 4) (8 - k_2 - k_3 k_1)

Solve for Possible Values of (k_2) and (k_3)

Rearranging the first equation gives:

k_2  4 - 2k_3

Substitute this into the second equation:

8 - (4 - 2k_3) - k_3  8 - 4   k_3  4   k_3  k_1

Thus, we get:

(k_1 4 - k_3) (k_2 4 - 2k_3)

The non-negative integer constraints are:

(k_3 leq 2) since (k_3) must be a non-negative integer.

Possible values for (k_3) are 0, 1, and 2.

Calculate Coefficients for Each Case

Case 1: (k_3 0) (k_2 4), (k_1 4) (Coefficient frac{8!}{4!4!0!} binom{8}{4} 70) Case 2: (k_3 1) (k_2 2), (k_1 5) (Coefficient frac{8!}{5!2!1!} frac{40320}{120 times 2 times 1} 168) Case 3: (k_3 2) (k_2 0), (k_1 6) (Coefficient frac{8!}{6!0!2!} frac{40320}{720 times 2} 28)

Sum the Coefficients

(70 168 28 266)

Thus, the coefficient of (x^4) in the expansion of (1 x x^2 x^8) is 266.

Another viewpoint simplifies the process by focusing on the different ways to form (x) from the polynomial factors. We can form (x) using:

(1^6, x^0, x^2^2) (1^4, x^2, x^2, x^4) (1^4, x^4, x^2^0)

The number of ways to choose two copies of (x^2) from eight factors is (_8C_2 28). There are (8 times _7C_2 168) ways to choose one (x^2) and two (x)'s from the remaining seven factors. Finally, there are (_8C_4 70) ways to choose four (x)'s from the eight factors. Summing these gives 28 168 70 266, confirming the coefficient of (x^4) is 266.

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