Verified answer

(x−1)(x+2)^2−(x−1)^2(x+2)

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

= 3(x-1)(x+2)

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To the answerer below me,

Many students didn't know how to use "^".

(x−1)(x+2)2−(x−1)2(x+2)

Because real numbers are a field, the multiplication holds over communtativity. This means that this can be re-written as:

2(x−1)(x+2)−2(x−1)(x+2)

As i am sure you can see, these are the same. This means your answer is 0, your invarient. If you really wanted to factor, A=0, b=2, and c=-1

and btw, what the person above me put is assuming that the question has expoenets, which is doesn't....

(x−1)(x+2)2−(x−1)2(x+2) =Bring out the coefficient of "2" to the front=> 2(x-1)(x+2)-2(x-1)(x+2) =equal and opposite=> 0

Answer can be written in A(x+B)(x+C) form. In this case the answer is 0.

Hopefully that helps.

If you meant for those 2's to be exponents...then the proper notation for that would be (x−1)(x+2)^2−(x−1)^2(x+2) or ((x−1)(x+2))^2−(x−1)^2(x+2) depending on which one you meant.