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Chapter 14: Problem 17

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$\left(x^{2}+2 y\right)^{4}$$

Short Answer

Expert verified
The expanded form of \((x^{2}+2 y)^{4}\) using the Binomial Theorem is \(x^8 + 32x^6y + 96x^4y^2 + 128x^2y^3 + 16y^4\)

Step by step solution

01

Identify the components of binomial

Identify the components of your binomial. In our case, a = \(x^2\) and b= \(2y\), and n = 4
02

Apply the formula

Apply the Binomial theorem formula and expand the binomial as given: \((a + b)^n = \sum_{k=0}^n {n\choose k} a^{n-k} b^k\)So that makes it:\((x^2 + 2y)^4=\sum_{k=0}^4 {4\choose k} (x^2)^{4-k} (2y)^k\)
03

Expand the binomial

Plug in the values and simplify to get:= {4\choose 0} \(x^{2*4}\) * \(2y)^0 + {4\choose 1} \(x^{2*3}\) * \(2y)^1 +{4\choose 2} \(x^{2*2}\) * \(2y)^2 + {4\choose 3} \(x^{2*1}\) * \(2y)^3 +{4\choose 4} \(x^{2*0}\) * \(2y)^4 = \(x^8\) + \(4*8x^6y\) + \(6*16x^4y^2\) + \(4*32x^2y^3\) + \(16y^4\)= \(x^8\) + \(32x^6y\) + \(96x^4y^2\) + \(128x^2y^3\) + \(16y^4\)

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Most popular questions from this chapter

Expand and write the answer as a single logarithm with a coefficient of 1. $$\sum_{i=1}^{4} \log (2 i)$$

Each exercise involves observing a pattern in the expanded form of the binomial expression \((a+b)^{n}\).$$\begin{array}{l}(a+b)^{1}=a+b \\\\(a+b)^{2}=a^{2}+2 a b+b^{2} \\\\(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\\\(a+b)^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \\\\(a+b)^{5}=a^{5}+5 a^{4} b+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a b^{4}+b^{5}\end{array}$$ Describe the pattern for the exponents on \(b\).

Factor: \(27 x^{3}-8\) (Section 6.4, Example 8)

Explain how to find \(n !\) if \(n\) is a positive integer.

What is a geometric sequence? Give an example with your explanation.
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