Question

multiply: $(2 + y)^4$
multiply: $(x + 2y)^7$
multiply: $(x - 3)^7$

Explanation:

We use the Binomial Theorem: For any positive integer $n$,
$$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}$$
where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

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Problem 1: Expand $(2 + y)^4$

Step1: Apply Binomial Theorem

$$(2+y)^4 = \binom{4}{0}2^4y^0 + \binom{4}{1}2^3y^1 + \binom{4}{2}2^2y^2 + \binom{4}{3}2^1y^3 + \binom{4}{4}2^0y^4$$

Step2: Calculate coefficients/terms

$\binom{4}{0}2^4=16$, $\binom{4}{1}2^3=32$, $\binom{4}{2}2^2=24$, $\binom{4}{3}2=8$, $\binom{4}{4}1=1$
$$=16 + 32y + 24y^2 + 8y^3 + y^4$$

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Problem 2: Expand $(x + 2y)^7$

Step1: Apply Binomial Theorem

$$(x+2y)^7 = \sum_{k=0}^{7} \binom{7}{k}x^{7-k}(2y)^k$$

Step2: Compute each term

$\binom{7}{0}x^7(2y)^0 = x^7$
$\binom{7}{1}x^6(2y)^1 = 14x^6y$
$\binom{7}{2}x^5(2y)^2 = 84x^5y^2$
$\binom{7}{3}x^4(2y)^3 = 280x^4y^3$
$\binom{7}{4}x^3(2y)^4 = 560x^3y^4$
$\binom{7}{5}x^2(2y)^5 = 672x^2y^5$
$\binom{7}{6}x^1(2y)^6 = 448xy^6$
$\binom{7}{7}x^0(2y)^7 = 128y^7$

Step3: Combine terms

$$=x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7$$

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Problem 3: Expand $(x - 3)^7$

Step1: Apply Binomial Theorem

$$(x-3)^7 = \sum_{k=0}^{7} \binom{7}{k}x^{7-k}(-3)^k$$

Step2: Compute each term

$\binom{7}{0}x^7(-3)^0 = x^7$
$\binom{7}{1}x^6(-3)^1 = -21x^6$
$\binom{7}{2}x^5(-3)^2 = 189x^5$
$\binom{7}{3}x^4(-3)^3 = -945x^4$
$\binom{7}{4}x^3(-3)^4 = 2835x^3$
$\binom{7}{5}x^2(-3)^5 = -5103x^2$
$\binom{7}{6}x^1(-3)^6 = 5103x$
$\binom{7}{7}x^0(-3)^7 = -2187$

Step3: Combine terms

$$=x^7 - 21x^6 + 189x^5 - 945x^4 + 2835x^3 - 5103x^2 + 5103x - 2187$$

Answer:

1. $(2 + y)^4 = y^4 + 8y^3 + 24y^2 + 32y + 16$
2. $(x + 2y)^7 = x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7$
3. $(x - 3)^7 = x^7 - 21x^6 + 189x^5 - 945x^4 + 2835x^3 - 5103x^2 + 5103x - 2187$