Calculate the 2001 power of -2 plus the 2002 power of -2

Calculate the 2001 power of -2 plus the 2002 power of -2

2002 Power of -2 equals 2002 power of 2
2001 Power of -2 equals 2000 power of 2
-2*-2*-2*-2*-2…… *-2+2*2*2*2*2*2…… *2
2001 Counts-2 2002 Counts 2
=2^2000*(-2)+2^2000*4
=2^2000*(-2+4)
=2^2000*2
=2^2001

Factorization (1) of the cubic power of x -3x squared +4(2) x squared +2001x squared +200x squared +2001 Factorization (1) cubic of x-3x square +4(2) x fourth power +2001x cubic +200x square +2001 Factorization of (1) x to the third power -3x squared +4(2) x to the fourth power +2001x to the third power +200x squared +2001

1) X3-3 x2+4
=X3-2x2-x2+4[ split -3x2 into -2x2-x2 factor x-2]
=(X3-2x2)-(x2-4)
= X2(x-2)-(x+2)(x-2)
=(X-2)[ x2-(x+2)]
=(X-2)(x2-x-2)
=(X-2)(x-2)(x-1)
=(X-2)2(x-1)
2) Is the title wrong? Should be x^4+2001x2+2000x+2001
[X^4 means "x to the fourth power "]
This problem can start from the general, because 2000,2001 is larger, set 2000= n,2001= n +1
Original expression
X^4+(n+1) x2+nx+(n+1)
=X^4+nx2+x2+nx+n+1
=(X^4+x2+1)+(nx2+nx+1)
=[(X^4+2x2+1)-x2]+n (x2+x+1)
=[(X2+1)2-x2]+n (x2+x+1)
=(X2-x+1)(x2+x+1)+n (x2+x+1)
=(X2+x+1)(x2-x+1+n)
Substitute n+1=2001
X^4+2001 x2+2000 x2+2001=(x2+x+1)(x2-x+2001)