Question

factorise x^2+1/x^2+3-2x-2/x​

Answer 1

Step-by-step explanation:

x^{2}+\frac{1}{x^{2}}+2-2x-\frac{2}{x}\\=\big(x+\frac{1}{x}\big)\big(x+\frac{1}{x}-2\big)

Step-by-step explanation:

Given ,\\x^{2}+\frac{1}{x^{2}}+2-2x-\frac{2}{x}

=x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}-2x-\frac{2}{x}

=\big(x+\frac{1}{x}\big)^{2}-2\big(x+\frac{1}{x}\big)

=\big(x+\frac{1}{x}\big)\big(x+\frac{1}{x}-2\big)

Therefore,

x^{2}+\frac{1}{x^{2}}+2-2x-\frac{2}{x}\\=\big(x+\frac{1}{x}\big)\big(x+\frac{1}{x}-2\big)

•••♪

Answer 2

Answer:

Step-by-step explanation:

x^2+1/x^2+3-2x-2/x=0

⇒x^3/x^5-2x-2/x=0

⇒x^3/x^5*x^3-2x*x^3-2/x*x^3=0

⇒x^9-5-2x^4-2x^2=0

⇒x^4-2x^4-2x^2=0

⇒-x^4-2x^2=0

⇒ -x^2(x^2+2)=0

X= -√2 OR X= -√1