Question
Factorise $: x^2+\frac{1}{(4 x)^2}+1-7 x-\frac{7}{2 x}$
✓
Answer
$ x^2+\frac{1}{(4 x)^2}+1-7 x-\frac{7}{2 x}$
$ =(x)^2+\frac{1}{(2 x)^2}+2 \times x \times \frac{1}{2 x}-7\left(x+\frac{1}{2 x}\right)$
$ =\left(x+\frac{1}{2 x}\right)^2-7\left(x+\frac{1}{2 x}\right)$
$ =\left(x+\frac{1}{2 x}\right)\left(x+\frac{1}{2 x}-7\right)$
$ =\left(x+\frac{1}{2 x}\right)\left(x-7+\frac{1}{2 x}\right)$
$ =(x)^2+\frac{1}{(2 x)^2}+2 \times x \times \frac{1}{2 x}-7\left(x+\frac{1}{2 x}\right)$
$ =\left(x+\frac{1}{2 x}\right)^2-7\left(x+\frac{1}{2 x}\right)$
$ =\left(x+\frac{1}{2 x}\right)\left(x+\frac{1}{2 x}-7\right)$
$ =\left(x+\frac{1}{2 x}\right)\left(x-7+\frac{1}{2 x}\right)$
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