Question

factorise a^2+1/a^2+2-5a-5/a​

Answer 1

hope it's help you

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Answer 2

Answer:

$(a+\frac{1}{a})(a+\frac{1}{a}-5)$

Step-by-step explanation:

$Given\\\\a^2+\frac{1}{a^2}+2-5a-\frac{5}{a}\\\\=(a^2+\frac{1}{a^2}+2)-(5a+\frac{5}{a})\\\\=(a^2+\frac{1}{a^2}+2\times a^2\times\frac{1}{a^2})-5(a+\frac{1}{a})\\\\=(a+\frac{1}{a})^2-5(a+\frac{1}{a})\\\\=(a+\frac{1}{a})(a+\frac{1}{a}-5)\\\\\large{\bold{=(a+\frac{1}{a})(a+\frac{1}{a}-5)}}\\\\\\\\\large{\bold{FORMULAE}}\\\\\bold{\boxed{(x+y)^2=x^2+y^2+2xy}}$

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