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Chapter 11: Problem 20

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (y-4)^{4} $$

Short Answer

Expert verified
The expansion of \( (y-4)^4 \) given by Binomial Theorem is \( y^4 - 16y^3 + 96y^2 - 256y + 256 \)

Step by step solution

01

Understand Binomial Theorem

The binomial theorem is given by \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \), where \( \binom{n}{k} \) is the binomial coefficient, equal to \( \frac{n!}{k!(n-k)!} \), n! is the factorial of n, k is the number of items to choose, and \( a^{n-k} b^k \) is the binomial expression.
02

Substituting the values into the Binomial Theorem

For \( (y-4)^4 \), the terms are: (i) 'a' is 'y', (ii) 'b' is '-4' and (iii) 'n' is '4'. Substituting in the formula, the expanded form becomes \( \sum_{k=0}^{4} \binom{4}{k} y^{4-k} (-4)^k \).\n\nThis will give us 5 terms:\nTerm 1: \( \binom{4}{0} y^4 (-4)^0 \)\nTerm 2: \( \binom{4}{1} y^3 (-4)^1 \)\nTerm 3: \( \binom{4}{2} y^2 (-4)^2 \)\nTerm 4: \( \binom{4}{3} y^1 (-4)^3 \)\nTerm 5: \( \binom{4}{4} y^0 (-4)^4 \)
03

Expand and Simplify

Now calculating each term and combining, we get:\nTerm 1: \( \binom{4}{0} y^4 (-4)^0 = 1*y^4*1 = y^4 \)\nTerm 2: \( \binom{4}{1} y^3 (-4)^1 = 4*y^3*(-4) = -16y^3 \)\nTerm 3: \( \binom{4}{2} y^2 (-4)^2 = 6*y^2*16 = 96y^2 \)\nTerm 4: \( \binom{4}{3} y^1 (-4)^3 = 4*y*(-64) = -256y \)\nTerm 5: \( \binom{4}{4} y^0 (-4)^4 = 1*1*256 = 256 \)\n\nSo, \( (y-4)^4 = y^4 - 16y^3 + 96y^2 - 256y + 256 \)

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Most popular questions from this chapter

A single die is rolled twice. Find the probability of rolling a 5 the first time and a 1 the second time.

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

Will help you prepare for the material covered in the next section. Use the formula \(a_{n}=a_{1} 3^{n-1}\) to find the seventh term of the sequence \(11,33,99,297, \ldots\)

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Exercises \(116-118\) will help you prepare for the material covered in the next section. In Exercises \(116-117\) show that $$1+2+3+\cdots+n=\frac{n(n+1)}{2}$$ is true for the given value of \(n\) $$ n=3: \text { Show that } 1+2+3=\frac{3(3+1)}{2} $$
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