Question

rewrite the following expression in expanded form and simplify completely. $(2x + 4y)^5$

Explanation:

Step1: Apply Binomial Theorem

The binomial theorem states $$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}$$
For $(2x+4y)^5$, $a=2x$, $b=4y$, $n=5$:
$$(2x+4y)^5 = \binom{5}{0}(2x)^5(4y)^0 + \binom{5}{1}(2x)^4(4y)^1 + \binom{5}{2}(2x)^3(4y)^2 + \binom{5}{3}(2x)^2(4y)^3 + \binom{5}{4}(2x)^1(4y)^4 + \binom{5}{5}(2x)^0(4y)^5$$

Step2: Calculate binomial coefficients

$\binom{5}{0}=1$, $\binom{5}{1}=5$, $\binom{5}{2}=10$, $\binom{5}{3}=10$, $\binom{5}{4}=5$, $\binom{5}{5}=1$:
$$=1(2x)^5(4y)^0 + 5(2x)^4(4y)^1 + 10(2x)^3(4y)^2 + 10(2x)^2(4y)^3 + 5(2x)^1(4y)^4 + 1(2x)^0(4y)^5$$

Step3: Simplify each term

• Term1: $1\cdot 32x^5\cdot 1 = 32x^5$
• Term2: $5\cdot 16x^4\cdot 4y = 5\cdot 64x^4y = 320x^4y$
• Term3: $10\cdot 8x^3\cdot 16y^2 = 10\cdot 128x^3y^2 = 1280x^3y^2$
• Term4: $10\cdot 4x^2\cdot 64y^3 = 10\cdot 256x^2y^3 = 2560x^2y^3$
• Term5: $5\cdot 2x\cdot 256y^4 = 5\cdot 512xy^4 = 2560xy^4$
• Term6: $1\cdot 1\cdot 1024y^5 = 1024y^5$

Step4: Combine all terms

Add the simplified terms together.

Answer:

$32x^5 + 320x^4y + 1280x^3y^2 + 2560x^2y^3 + 2560xy^4 + 1024y^5$