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Chapter 11: Problem 28

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (x-3 y)^{5} $$

Short Answer

Expert verified
The expression \((x-3 y)^5\) expands and simplifies to \(x^{5}-15x^{4}y+90x^{3}y^{2}-270x^{2}y^{3}+405xy^{4}-243y^{5}\) according to the Binomial theorem.

Step by step solution

01

Identify 'a', 'b' and 'n'

From the binomial \((x-3 y)^5\), we can identify 'a' as 'x', 'b' as '-3y', and 'n' as 5.
02

Calculate binomial coefficients, and powers of 'a' and 'b'

We can calculate the binomial coefficients \({5\choose k}\) for 'k' from 0 to 5, along with the corresponding powers of 'a' and 'b': \((x)^{5-k}\) and \((-3y)^{k}\). Together, we get five terms: \({5\choose 0} x^{5-0} (-3y)^{0}, {5\choose 1} x^{5-1} (-3y)^{1}, {5\choose 2} x^{5-2} (-3y)^{2}, {5\choose 3} x^{5-3} (-3y)^{3}, {5\choose 4} x^{5-4} (-3y)^{4}, {5\choose 5} x^{5-5} (-3y)^{5}\)
03

Simplify obtained results

The expression simplifies to: \(x^{5}-15x^{4}y+90x^{3}y^{2}-270x^{2}y^{3}+405xy^{4}-243y^{5}\)
04

Application of Binomial Theorem

Now, we apply the Binomial Theorem, and the expanded formula for \((x-3 y)^5\) becomes \(x^{5}-15x^{4}y+90x^{3}y^{2}-270x^{2}y^{3}+405xy^{4}-243y^{5}\).

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