Problem 28
Question
In Exercises 9-30, use the Binomial Theorem to expand each binomial and express the result in simplified form. $$(x-3 y)^{5}$$
Step-by-Step Solution
Verified Answer
The expanded form of \((x-3y)^5\) in simplified form using the Binomial Theorem is \(x^5 - 15x^4y + 90x^3y^2 - 270x^2y^3 + 405xy^4 - 243y^5\).
1Step 1: Apply the Binomial Theorem
The Binomial Theorem states that \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In our case, \(a = x\), \(b = -3y\), and \(n = 5\). Applying the theorem, we get: \((x-3y)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (-3y)^{k}\).
2Step 2: Expand with Pascal's Triangle
Now expand each term in the equation by applying the coefficients from Pascal's triangle (specifically, the row starting with 1, 5). The expanded form is: \( \binom{5}{0}x^{5}(-3y)^0 + \binom{5}{1}x^{4}(-3y)^1 + \binom{5}{2}x^{3}(-3y)^2 + \binom{5}{3}x^{2}(-3y)^3 + \binom{5}{4}x(-3y)^4 + \binom{5}{5}(-3y)^5\).
3Step 3: Simplify the Expression
Simplify each term, taking \(\binom{5}{k}\) as the coefficient, \(x\) raised to a power, and \((-3y)\) raised to a power as well. The simplified expression is \(x^5 - 15x^4y + 90x^3y^2 - 270x^2y^3 + 405xy^4 - 243y^5\).
Key Concepts
Pascal's TriangleBinomial ExpansionAlgebraic Expressions
Pascal's Triangle
Pascal's Triangle is a triangular array of numbers that provides coefficients for expanding binomials. Each number is the sum of the two directly above it, creating a beautiful pattern.
To construct it, start with a "1" at the top. The second row has "1"s on both ends, and each subsequent row begins and ends with "1"s too.
The numbers between the "1"s are determined by adding the two numbers above them. So, the third row is "1, 2, 1", the fourth is "1, 3, 3, 1", and so on.
For expanding a binomial expression like \((x-3y)^5\), we use the 5th row: "1, 5, 10, 10, 5, 1." These are the coefficients in our expansion.
To construct it, start with a "1" at the top. The second row has "1"s on both ends, and each subsequent row begins and ends with "1"s too.
The numbers between the "1"s are determined by adding the two numbers above them. So, the third row is "1, 2, 1", the fourth is "1, 3, 3, 1", and so on.
For expanding a binomial expression like \((x-3y)^5\), we use the 5th row: "1, 5, 10, 10, 5, 1." These are the coefficients in our expansion.
- 1st term has coefficient 1
- 2nd term has coefficient 5
- 3rd term has coefficient 10
- 4th term also has coefficient 10
- 5th term has coefficient 5
- 6th term returns to 1
Binomial Expansion
The binomial expansion uses the Binomial Theorem to expand expressions of the form \((a + b)^n\). This theorem helps us transform binomials into a sum of terms involving powers of \(a\) and \(b\).
Here's the formula: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). It uses the binomial coefficient \(\binom{n}{k}\), which is found using Pascal's Triangle or the combination formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).
For \((x-3y)^5\), with \(n=5\), \(a=x\), and \(b=-3y\), the expansion becomes: \(\sum_{k=0}^{5} \binom{5}{k} x^{5-k} (-3y)^{k}\). Each term is constructed as a combination of powers of \(x\) and \(-3y\), multiplied by their respective coefficients from Pascal's Triangle or using the combination formula.
Here's the formula: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). It uses the binomial coefficient \(\binom{n}{k}\), which is found using Pascal's Triangle or the combination formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).
For \((x-3y)^5\), with \(n=5\), \(a=x\), and \(b=-3y\), the expansion becomes: \(\sum_{k=0}^{5} \binom{5}{k} x^{5-k} (-3y)^{k}\). Each term is constructed as a combination of powers of \(x\) and \(-3y\), multiplied by their respective coefficients from Pascal's Triangle or using the combination formula.
- 1st term: \(x^5 \times 1\)
- 2nd term: \(x^4 \times (-3y) \times 5\)
- and so on...
Algebraic Expressions
Algebraic expressions like \((x-3y)^5\) can be rewritten into multiple terms using expansion methods. These expressions are combinations of numbers, variables, and mathematical operations like addition or multiplication.
Simplification refers to the process of reducing the expression into a standard form. Once expanded using the Binomial Theorem or Pascal's Triangle, each term is simplified to make calculations or further algebraic manipulations easier.
In this context:
Simplification refers to the process of reducing the expression into a standard form. Once expanded using the Binomial Theorem or Pascal's Triangle, each term is simplified to make calculations or further algebraic manipulations easier.
In this context:
- Combine the coefficients from the Binomial Theorem with your variables, i.e. \(x\) and \(-3y\).
- Restructure powers as per their expansion result.
- Combine any like terms if necessary, to get the final expression.
Thus, \((x-3y)^5\) simplifies to: \(x^5 - 15x^4y + 90x^3y^2 - 270x^2y^3 + 405xy^4 - 243y^5\). Each term clearly indicates a degree or power of \(x\) and \(y\) present in the original expression.
Other exercises in this chapter
Problem 28
Evaluate each expression. $$ \frac{_{5} C_{1} \cdot_{7} C_{2}}{_{12} C_{3}} $$
View solution Problem 28
In Exercises \(23-34,\) write a formula for the general term (the nth term) of each arithmetic sequence. Do not use a recursion formula. Then use the formula fo
View solution Problem 28
Use the formula for the sum of the first n terms of a geometric sequence to solve. Find the sum of the first 11 terms of the geometric sequence: $$4,-12,36,-108
View solution Problem 29
Involve a deck of 52 cards. A poker hand consists of five cards. a. Find the total number of possible five-card poker hands. b. A diamond flush is a five-card h
View solution