factorise the following in detail:?

Question

1)x^2+1/x^2+2-5x-5/x

Answer 1

This is a 4th order polynomial and coefficient of x^2 and 1/x^2 are same also the coefficeint of x and 1/x are same .so combine the 2 and both same sign
x^2+1/x^2 = (x+1/x)^2 -2
-5x-5/x = -5(x+1/x)

so we get given expression

(x+1/x)^2 -2 +2 -5(x+1/x)
= (x+1/x)^2 -5(x+1/x)
= (x+1/x)(x+1/x-5)

Answer 2

OK it seems like noone knows how to use brackets etc on here so everyone gets different versions of the question.

from what you've written this is neatest factorisation i can see

1/x^2(x^4+1+2x^2-5x^3-5x)
= 1/ x^2 (x^4-5x^3+2x^2-5x+1)

Answer 3

Combine like factors x^2+2+x^-2 - 5 x -5x^-1 note patten
x+x^-1. What is (x+x^-1)^2. Why is xx^-1 ~ 1. a^2 +2 ab + b^2

Y^2-5Y= Y(Y-5) i will leave it to you to substitute

Answer 4

x^2+1/x^2+2
=(x)^2+2(x)(1/x)+(1/x)^2
=(x+1/x)^2
-5x-5/x=-5(x+1/x)
now the expression
(x+1/x)^2-5(x+1/x)
taking out (x+1/x) as a common factor
(x+1/x)[(x+1/x)-5] answer

Answer 5

nicely the query seems to be in 2 factors. 9x^2 + 4 + 12x = x^2 + 4/9 + (4/3)x (divide via 9) = ( x + 2/3 )^2 (factorise) a^2 + 2ab - b^2 = ab (a + 2 - b) So the functionality finally ends up as (x + 2/3)^2 - ab(a + 2 - b)

Answer 6

Sorry, don't do home work? I graduated a long time ago and did my own.