/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 400 Expand \((x+2 y)^{5}\)... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 400

Expand \((x+2 y)^{5}\)

Short Answer

Expert verified
The expanded form of \((x+2y)^5\) is \(x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5\).

Step by step solution

01

Identify the variables

In this problem, we have the expression \((x + 2y)^5\). We will let: - \(a = x\) - \(b = 2y\) - \(n = 5\)
02

Apply the binomial theorem

Using the binomial theorem formula, we can expand the expression like this: $$(x + 2y)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (2y)^k$$
03

Expand each term in the sum

Now we can expand each term in the sum using the combination formula and substitute the values of \(a\), \(b\), and \(n\): \begin{align*} (x + 2y)^5 &= \binom{5}{0} x^{5-0} (2y)^0 + \binom{5}{1} x^{5-1} (2y)^1 + \binom{5}{2} x^{5-2} (2y)^2 \\ &\phantom{{}=}+ \binom{5}{3} x^{5-3} (2y)^3 + \binom{5}{4} x^{5-4} (2y)^4 + \binom{5}{5} x^{5-5} (2y)^5 \end{align*}
04

Calculate the binomial coefficients and simplify

Now we compute the binomial coefficients and simplify each term: \begin{align*} (x + 2y)^5 &= \frac{5!}{0!5!} x^5 (1) + \frac{5!}{1!4!} x^4 (2y) + \frac{5!}{2!3!} x^3 (4y^2) \\ &\phantom{{}=}+ \frac{5!}{3!2!} x^2 (8y^3) + \frac{5!}{4!1!} x (16y^4) + \frac{5!}{5!0!} (32y^5) \\ & = 1 x^5 + 5 x^4(2y) + 10 x^3 (4y^2) \\ &\phantom{{}=}+ 10 x^2 (8y^3) + 5 x (16y^4) + 1 (32y^5) \\ & = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5 \end{align*} Thus, the expanded form of \((x+2y)^5\) is \(x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5\).

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