Question

factorise x4/4+4/x 4+1

Answer 1

you can factories it by taking 4 as common number
4(1÷1+1÷x4)

Answer 2

The factorisation of $\dfrac{x^4}{4}+\dfrac{4}{x^4}+1$ is $(\dfrac{x^2}{2}+\dfrac{2}{x^2}+1)(\dfrac{x^2}{2}+\dfrac{2}{x^2}-1).$

Step-by-step explanation:

We have,

$\dfrac{x^4}{4}+\dfrac{4}{x^4}+1$

To find, the factorisation of $\dfrac{x^4}{4}+\dfrac{4}{x^4}+1=?$

∴ $\dfrac{x^4}{4}+\dfrac{4}{x^4}+1$

$=(\dfrac{x^2}{2})^2+(\dfrac{2}{x^2})^2+1$

$=(\dfrac{x^2}{2}+\dfrac{2}{x^2})^2-2.\dfrac{x^2}{2}.\dfrac{2}{x^2}+1$

Using algebraic identity,

$(a+b)^{2}=a^{2}+b^{2}+2ab$

⇒ $(a+b)^{2}-2ab=a^{2}+b^{2}$

$=(\dfrac{x^2}{2}+\dfrac{2}{x^2})^2-2+1$

$=(\dfrac{x^2}{2}+\dfrac{2}{x^2})^2-1^2$

Using algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

$=(\dfrac{x^2}{2}+\dfrac{2}{x^2}+1)(\dfrac{x^2}{2}+\dfrac{2}{x^2}-1)$

Hence, the factorisation of $\dfrac{x^4}{4}+\dfrac{4}{x^4}+1$ is $(\dfrac{x^2}{2}+\dfrac{2}{x^2}+1)(\dfrac{x^2}{2}+\dfrac{2}{x^2}-1).$