Question

solve for x where x is a real number (x+1)^2(x-2) + (x+1)(x-2)^2 =0

Answer 1

Solve for x over the real numbers:
(-2+x)^2 (1+x)+(-2+x) (1+x)^2 = 0
Write the cubic polynomial on the left hand side in standard form.
Expand out terms of the left hand side:
2-3 x-3 x^2+2 x^3 = 0
Factor the left hand side.
The left hand side factors into a product with three terms:
(-2+x) (1+x) (-1+2 x) = 0
Solve each term in the product separately.
Split into three equations:
-2+x = 0 or 1+x = 0 or -1+2 x = 0
Look at the first equation: Solve for x.
Add 2 to both sides:
x = 2 or 1+x = 0 or -1+2 x = 0
Look at the second equation: Solve for x.
Subtract 1 from both sides:
x = 2 or x = -1 or -1+2 x = 0
Look at the third equation: Isolate terms with x to the left hand side.
Add 1 to both sides:
x = 2 or x = -1 or 2 x = 1
Solve for x.
Divide both sides by 2:
Answer: |
| x = 2 or x = -1 or x = 1/2

Answer 2

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

Answer 3

=)

Answer 4

thank you!