Need help with weird algebra problem!?

Question

Can anyone help me solve this for the value of x plz?


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

Answer 1

First we expand all the terms:
(x + 1)^2 * (x - 2) + (x + 1) * (x - 2)^2 = 0
(x^2 + 2*x + 1) * (x - 2) + (x + 1) * (x^2 - 4*x + 4) = 0
(x^2 + 2*x + 1) * (x - 2) + (x + 1) * (x^2 - 4*x + 4) = 0
(x^3 - 3*x - 2) + (x^3 - 3*x^2 + 4) = 0

Now combine like terms:
x^3 - 3*x - 2 + x^3 - 3*x^2 + 4 = 0
2*x^3 - 3*x^2 - 3*x + 2 = 0

Now factor the polynomial, leaving:
(-2 + x)*(1 + x)*(-1 + 2*x) = 0

Any of these three terms could be zero by itself, so either:
(-2 + x) = 0
(1 + x) = 0
(-1 + 2*x) = 0

Leading the three solutions:
x = 2
x = -1
x = 1/2

OK, this was the long way around...clearly sometime can be saved if you cancel some factors at the beginning (for this problem). However, this method works for all polynomials that have only real solutions.

Answer 2

Calculate all of it out first.
(x + 1)^2 (x – 2) + (x + 1) (x – 2)^2 = 0
(x+1)(x+1)(x-2) + (x+1)(x-2)(x-2) = 0
You can factor out x+1 from both parts of this on the left side to make it easier to solve.
(x+1) [(x+1)(x-2) + (x-2)(x-2)] = 0

So the first part:
(x+1) = 0
x = -1

The other part of the equation is
[(x+1)(x-2) + (x-2)(x-2)] = 0
You can factor out (x-2) from this now because it's in both parts of the equation on the left.
(x-2) [(x+1) + (x-2)] = 0

So this part is:
(x-2) = 0
x = 2

Lastly, the rest of the equation is:
[(x+1) + (x-2)] = 0
Combine like terms:
2x -1 = 0
x = 1/2

So you have 3 answers
x= -1
x= 2
x= 1/2
Check them in the original equation.
Keep only the answers that work in the original equation.

Answer 3

3(d-8)-5 = 9(d+2) 3-D-24-5 = 9d+18 3-D-29 = 9d+18 3-D-40 seven = 9d -40 seven = 6d d = -40 seven/6 that's a adverse fraction So the two i'm incorrect or you copied the project incorrect Or your instructor and the different student are incorrect Or no longer an definitely project, as in it is a trick Like D is a distinctive variable from d yet i'm undecided in case you probably did certainly replica it incorrect, it could have been: 3(d-8)-5=9(d+2)+a million Then d=-8 which might artwork

Answer 4

take (x+1)(x-2) as common
(x+1)(x-2)[( x + 1) + (x - 2)] = 0
(x+1)(x-2)[ x + 1 + x - 2] = 0
(x+1)(x-2)(2x - 1) = 0

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

or

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

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

so x = -1,2,1/2

Answer 5

take (x+1)(x-2) common from both the terms
(x+1)(x-2){(x+1)+(x-2)}=0
(x+1)(x-2)(2x-1)=0
x+1=0--> x=-1
x-2=0--> x=2
2x-1=0--> x=1/2

Answer 6

they have a website online where you can type in your algebra problem and they will automatically answer it for you and show exactly how to get the answer. just google it. its free, sorry i dont have the link. good luck.