Solve (-X^2+2X+1)^2 step by step?

Answer 1

There's nothing to SOLVE!

You SOLVE equations or inequalities, but you EXPAND, RE-EXPRESS, SIMPLIFY or EVALUATE stand-alone mathematical expressions.

Your desired result is: x^4 - 4 x^3 + 2 x^2 + 4 x + 1.

The simplest way to do this is to consider the GENERAL CASE:

(ay + bz + c)^2

= a^2 y^2 + b^2z^2 + c^2 + 2 bc z + 2 ca y + 2 ab yz.

[Clearly the individual squared terms will start this off, followed by twice all the cross terms --- just re-write the original expression as

(ay + bz + c)(ay + bz + c),

then multiply each term in the second parenthesis in turn by ay, then by bz, and finally by c. You'll naturally obtain the square of each original term this way, and cross terms will combine as (bz)*c + c*(bz) = 2 bc z, etc. Hence what is written above.]

In your case, we can just think of a = - 1, b = 2, and c = 1, with y = x^2, and z = x.

Then (a^2 y^2 + b^2z^2 + c^2) + (2 bc z + 2 ca y + 2 ab yz)

= (x^4 + 4 x^2 + 1) + (2*2 x - 2 x^2 - 2*2 x^3)

= x^4 - 4 x^3 + 2 x^2 + 4 x + 1.

Live long and prosper.

Answer 2

There is nothing to solve, but if you want to simplify this:
(-X^2+2X+1)^2 =

(x-(1-sqrt(2)))^2*
(x-(1+sqrt(2)))^2 =

(x-1+sqrt(2))^2*
(x-1-sqrt(2))^2.

To find the factors of -x^2+2x+1, use the general solution to the binomial where
x=(1/(2a))*(-b+/-sqrt(b^2-4ac))
and in this case a=-1, b=2, and c=1.

OR...
(-X^2+2X+1)^2 =

(-X^2+2X+1)(-X^2+2X+1) =

(-x^2)*(-x^2+2x+1)+
(2x)*((-X^2+2X+1)+
(-X^2+2X+1) =

(x^4-2x^3-x^2)+
(-2x^3+4x^2+2x)+
(-X^2+2X+1) =

x^4+ -2x^3-2x^3
-x^2+4x^2-x^2 +
2x+1 =

x^4-4x^3+2x^2+2x+1

Answer 3

okay, it would work without the negative sign before x^2

anyway here it is:

if problem is:
(x^2 + 2x + 1)^2
simplify be getting the sqare root of everything
so what's left will be:
x^2 + 2x + 1
simplify further:
(x+1) (x+1)

so x = -1

Answer 4

(-X^2+2X+1)^2
= -x^2(-X^2+2X+1) +2x(-X^2+2X+1)+1(-X^2+2X+1)
x^4 -2x^3 -x^2 -2x^3 +4x^2 +2x-x^2+2x+1
= x^4 - 4x^3 +2x^2 +4x +1