Question

x-2 whole square=(x+1) (x-1)​

Answer 1

Answer:

(x-2)^2 = (×+1)(×-1)

Step-by-step explanation:

x-2 )^2

=x^2-4x+4

=x^2-2x-2x+4

=x (x-2)-2(x-2)

=x-2 x-2

therefore x=2

(x+1)(x-1)

=x-1)^2

=x^2-2x+1

=x^2-x-x+1

=x (x-1)-1(x-1)

=x-1 x-1

therefore x=1

x=1

x=2

now apply both the values one at a time whether lhs=rhs or not

the one which gives this condition corrects that is the zero of the polynomial

if both are correct then both are the zeroes or else only one of the above

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