Question

Factorise $x^2+\frac{1}{x^2}+2-5x-\frac{5}{x}$

• A. $(x+\frac{1}{x})(x+\frac{1}{x}-5)$
• B. $(x+\frac{1}{x})(x+\frac{1}{x}+5)$
• C. $(x-\frac{1}{x})(x-\frac{1}{x}-5)$
• D. $(x+\frac{1}{x})(x-5)$

Answer 1

The correct option is A $(x+\frac{1}{x})(x+\frac{1}{x}-5)$

$x^2+\frac{1}{x^2}+2-5x-\frac{5}{x}$

$a^2+2ab+b^2=(a+b{)}^2$ and taking 5 common ,we have
= $(x^2+\frac{1}{x^2}{)}^2+-5(x+\frac{1}{x}$)
Taking $x+\frac{1}{x}$ common
= $(x+\frac{1}{x})(x+\frac{1}{x}-5)$