(x-2)(x+1)^2 + (x+1)(x-2)^2 = 0?

(x-2)(x+1)^2 + (x+1)(x-2)^2 = 0

(x-2)(x+1)[(x+1)+(x-2)]=0

(x-2)(x+1)(2x-1)=0

Therefore, three possible answers for x:

(2x-1) = 0, x = 1/2

(x+1) = 0, x = -1

(x-2) = 0, x = 2

X = 2, -1 or 1/2

Proof: Substitute these 3 numbers to the equation, will prove the equation correct

(x - 2)(x + 1)^2 + (x + 1)(x - 2)^2 = 0
(x - 2)[(x + 1)^2 + (x + 1)(x - 2)] = 0
(x - 2)(x + 1)[(x + 1) + (x - 2)] = 0
(x - 2)(x + 1)(x + 1 + x - 2) = 0
(x - 2)(x + 1)(x + x + 1 - 2) = 0
(2x - 1)(x - 2)(x + 1) = 0

2x - 1 = 0
x = 1/2 (0.5)
x = 0

x - 2 = 0
x = 2

x + 1 = 0
x = -1

∴ x = -1, 1/2 (0.5), 2

This IS correct so far.

Now set each of the three factors equal to zero:
2x-1 = 0 and x-2 = 0 and x+1 = 0

Solve each equation for x:
x = 1/2 and x = 2 and x = -1.

(2x-1)(x-2)(x+1)=o

=>2x-1=o /OR/ x-2=0 /OR/ x+1=0

x=1/2
OR
x=2
OR
x=-1

2x-1 = 0
x= 1/2

x-2 = 0
x=2

x+1 = 0
x= (-1)

(x-2)(x+1)[(x+1)+(x-2)]=0
x=2,x=-1 &
2x-1=0,x=1/2

(2x-1)[(x-2)(x+1)] = 0

so 2x-1 = 0
or x-2 = 0
or x+ 1= 0

ie x =-1, 1/2 or 2 are solutions
check
(x-2)(x+1)[ x+1 + x-2] = 0
(x-2)(x+1)[2x -1]=0

what you have is not quite correct

the first thing to see is the two terms are the same.

the second thing to do is to expand it inside:

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

then since there is two of the same term, (1) x 2

0 = 2x^2 - 2x - 4

quadratic equation with b = -2, c = -4, a = 2:

x = 0.5 +/- 0.25*sqrt(4-(4)(-4)(2))
x = 0.5 +/- 0.25*sqrt(36)
x = 0.5 +/- 0.25*6
x1 = 1/2 + 3/2 = 2
x2 = 1/2 - 3/2 = -1

so
2x-1=0 , x= 1/2

x-2=0, x=2

x+1=0, x=-1
so x= 1/2,2 and -1

(x-2)(x+1)^2 + (x+1)(x-2)^2 = 0

First we have to rewrite it like this since the items in parentheses are squared, we must write them twice:
(x-2)(x+1)(x+1) + (x+1)(x-2)(x-2) = 0

Now we have to FOIL these out:
(x² - x - 2)(x+1) + (x² - x - 2)(x-2) = 0

Go and multiply these out:
(x³ - x² + x² - x - 2x - 2) + (x³ - 2x² - x² - 2x - 2x + 4) = 0

Now combine like terms:
(x³ - 3x - 2) + (x³ - 3x² - 4x + 4) = 0

Get rid of the parentheses and combine like terms again:
2x³ - 3x² - 7x + 2 = 0

Now you can plug it into a calculator and graph it to find the zeros:
x = -1.4319 , .26151 , and 2.6704