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## Q. 10.5.3

Using the Binomial Theorem

Expand:    $(2 x-y)^5$.

## Verified Solution

Because the Binomial Theorem involves the addition of two terms raised to a power, we rewrite $(2 x-y)^5$ as $[2 x+(-y)]^5$. We use the Binomial Theorem

$(a+b)^n=\left(\begin{array}{l}n \\0\end{array}\right) a^n+\left(\begin{array}{l}n \\1\end{array}\right) a^{n-1} b+\left(\begin{array}{l}n \\2\end{array}\right) a^{n-2} b^2+\left(\begin{array}{l}n \\3\end{array}\right) a^{n-3} b^3+\cdots+\left(\begin{array}{l}n \\n\end{array}\right) b^n$

to expand $[2 x+(-y)]^5$. In $[2 x+(-y)]^5$, a = 2x, b = -y, and n = 5. In the expansion, powers of 2x are in descending order, starting with $(2 x)^5$. Powers of -y are in ascending order, starting with $(-y)^0$. [Because $(-y)^0$ = 1, a – y is not shown in the first term.] The sum of the exponents on 2x and -y in each term is equal to 5 , the exponent in the expression $(2 x-y)^5$.

$(2 x-y)^5=[2 x+(-y)]^5$   $=1\left(32 x^5\right)+5\left(16 x^4\right)(-y)+10\left(8 x^3\right) y^2+10\left(4 x^2\right)\left(-y^3\right)+5(2 x) y^4+1\left(-y^5\right)$

Multiplying factors in each of the six terms gives us the desired expansion:

$(2 x-y)^5=32 x^5-80 x^4 y+80 x^3 y^2-40 x^2 y^3+10 x y^4-y^5$.