Question

Factorise: ${(x+\frac{1}{x})}^2+3(x+\frac{1}{x})+2$

• A. $(x+\frac{1}{x}-1)(x+\frac{1}{x}+2)$
• B. $(x+\frac{1}{x}+2)(x+\frac{1}{x}+2)$
• C. $(x+\frac{1}{x}+1)(x+\frac{1}{x}+2)$
• D. $(x+\frac{1}{x}+1)(x-\frac{1}{x}+2)$

Answer 1

The correct option is C $(x+\frac{1}{x}+1)(x+\frac{1}{x}+2)$

On substituting $x+\frac{1}{x}$ as $t$,
We get $t^2+3t+2$
By middle term splitting we can write,
$t^2+2t+t+2$
$t(t+2)+(t+2)=(t+1)(t+2)$; As $x+\frac{1}{x}=t$,
Thus the final answer will be; $(x+\frac{1}{x}+1)(x+\frac{1}{x}+2)$ which is option (c)