Problem 35 - Algebra 2

Question

Expand each binomial. $$ (2 x+3 y)^{4} $$

Step-by-Step Solution

Answer

The expansion of the binomial \( (2x+3y)^{4} \) is: \( 16x^{4} + 96x^{3}y + 216x^{2}y^{2} + 216xy^{3} + 81y^{4} \)

Step 1: Write out binomial theorem formula

For any binomial \( (a+b)^{n} \), the expansion can be written using binomial theorem as: \[(a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^{k}\]where \(\binom{n}{k}\) is the binomial coefficient, which is the coefficient of each term in the expansion and can be calculated as: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]Note: n! means the factorial of n.

Step 2: Insert the given binomial into formula

Given binomial is \( (2x+3y)^{4} \). Here \( a=2x, b=3y, n=4 \). Replacing these into the Binomial theorem formula, we get: \[(2x+3y)^{4} = \sum_{k=0}^{4} \binom{4}{k} (2x)^{4-k}(3y)^{k} \]

Step 3: Expand the binomial

We calculate each term of the expansion as follows:when k = 0,\[\binom{4}{0} (2x)^{4-0}(3y)^{0} = 1*(16x^{4})*1 = 16x^{4}\]when k = 1,\[\binom{4}{1} (2x)^{4-1}(3y)^{1} = 4*(8x^{3})*3y = 96x^{3}y\]when k = 2,\[\binom{4}{2} (2x)^{4-2}(3y)^{2} = 6*(4x^{2})*9y^{2} = 216x^{2}y^{2}\]when k = 3,\[\binom{4}{3} (2x)^{4-3}(3y)^{3} = 4*(2x)*27y^{3} = 216xy^{3}\]when k = 4,\[\binom{4}{4} (2x)^{4-4}(3y)^{4} = 1*1*81y^{4} = 81y^{4}\] Combining all the above terms, the expansion of \( (2x+3y)^{4} \) will be: \[(2x+3y)^{4} = 16x^{4} + 96x^{3}y + 216x^{2}y^{2} + 216xy^{3} + 81y^{4}\]

Key Concepts

Binomial Expansion

The concept of binomial expansion is crucial in algebra, especially when dealing with expressions raised to a power. It involves expanding an expression of the form \((a+b)^n\), where \(a\) and \(b\) are any numbers or variables, and \(n\) is a positive integer.
The binomial expansion process results in a polynomial expressed as the sum of several terms derived from \((a+b)\).

• Each term in the expansion is made up of the powers of \(a\) and \(b\) from 0 up to \(n\).
• The number of terms you get from this expansion is \((n+1)\).
• These powers decrease for \(a\) and increase for \(b\) across each term.

Understanding binomial expansion allows us to express lengthy polynomial multiplication succinctly using the binomial theorem. This theorem provides a foolproof way to expand not just small, but even very large powers of binomials quickly.

Binomial Coefficients

Binomial coefficients appear as the leading numbers in the terms of a binomial expansion. They determine how each term contributes to the final expanded polynomial. These coefficients are represented using the symbol \(\binom{n}{k}\), often read as "n choose k".
The value of these coefficients depends on both the total number of terms \(n\), and the specific term you're calculating, represented by \(k\). You can calculate a binomial coefficient using the formula:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(!\) denotes factorial, which means multiplying all whole numbers from \(1\) up to that number.

This structured approach helps manage the complexity of different powers in each term, thus playing an essential role in the expansion process. Using binomial coefficients is advantageous because it can simplify otherwise daunting calculations.

Polynomial Expansion

Polynomial expansion comes into play when we express a binomial raised to a power as a sum involving several terms, each of which is a product of coefficients, variables raised to an appropriate power, and factorial calculations. Applying binomial expansion to \((a+b)^n\) translates a compact expression into a more detailed polynomial.
This concept emphasizes breaking down polynomials into manageable parts:

• You start with smaller expressions (binomials) and use expansion to get individual terms.
• Each term ends up being a product of binomial coefficients and variable powers.
• As seen in expanding \((2x+3y)^4\), this results in a polynomial with variable terms like \(16x^4\), \(96x^3y\), and so on.

With polynomial expansion, analyzing and solving algebraic problems becomes easier, as it converts potentially unruly expressions into a sequence of straightforward calculations. Such detailed expansions make these expressions simpler to compute and understand.