Question

Expand each of the expressions in 1 to 2.
$(1-2x{)}^5$

${(\frac{2}{x}-\frac{x}{2})}^5$

Answer 1

We have,

Part (1):-

${(1-2x)}^5=1-5\left(2x\right)+\frac{5(5-1)}{2!}{\left(2x\right)}^2-\frac{5(5-1)(5-2)}{3!}{\left(2x\right)}^3+\frac{5(5-1)(5-2)(5-3)}{4!}{\left(2x\right)}^4-\frac{5(5-1)(5-2)(5-3)(5-4)}{5!}{\left(2x\right)}^5$

$=1-10x+40x^2-80x^3+80x^4-32x^5$

Part (2):-

${(\frac{2}{x}-\frac{x}{2})}^5={\left(\frac{2}{x}\right)}^5-5{\left(\frac{2}{x}\right)}^4\left(\frac{x}{2}\right)+\frac{5(5-1)}{2!}{\left(\frac{2}{x}\right)}^3{\left(\frac{x}{2}\right)}^2-\frac{5(5-1)(5-2)}{3!}{\left(\frac{2}{x}\right)}^2{\left(\frac{x}{2}\right)}^3+\frac{5(5-1)(5-2)(5-3)}{4!}\left(\frac{2}{x}\right){\left(\frac{x}{2}\right)}^4-\frac{5(5-1)(5-2)(5-3)(5-2)}{5!}{\left(\frac{2}{x}\right)}^0{\left(\frac{x}{2}\right)}^4$

$=\frac{32}{x^5}-\frac{5\times 8}{x^3}+\frac{20\times 2}{2x}-\frac{60x}{6\times 2}+\frac{120x^3}{24\times 8}-\frac{x^4}{16}$

$=\frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5x+\frac{5x^3}{8}-\frac{x^4}{16}$

Hence, this is the answer.