damon and bob I am not so bright. When I originally posted my question I mistakenly put an = sign where there should have been a + sign. the equation should read like this:
(x^2 - 2x)(x + 3) = -2x(x + 1)
What does this do to the solution now???
x^3 + 3 x^2 - 2 x^2 - 6 x + 2 x^2 + 2 x = 0
No change. Reiny guessed what you meant way back
x^3 + 3 x^2 - 4 x = 0
x(x^2 + 3 x - 4 ) = 0
x (x+4)(x-1) = 0
x = 0
x = -4
x = 1
Damon thanks for the help. I had the sign wrong. I am having trouble with the whole sign thing. I just submitted my test and I missed 4 of the multiple choice questions and each one was because I had the signs wrong. I had the correct numbers just the wrong signs.
Go very slowly then
x^3 + 3 x^2 - 2 x^2 - 6 x = - 2 x^2 - 2 x
then add the quantity + 2 x^2 + 2 x to both sides to make the right side zero.
No problem! Let's work through the equation correctly and see what happens.
First, let's expand both sides of the equation:
(x^2 - 2x)(x + 3) = -2x(x + 1)
Expanding the left-hand side:
(x^2 - 2x)(x + 3) = x(x + 3) - 2x(x + 3)
= x^2 + 3x - 2x^2 - 6x
= -x^2 - 3x
Expanding the right-hand side:
-2x(x + 1) = -2x^2 - 2x
Now, we can simplify the equation:
-x^2 - 3x = -2x^2 - 2x
To further simplify, let's move all the terms to one side of the equation:
-x^2 - 3x + 2x^2 + 2x = 0
Combining like terms:
x^2 - x = 0
Next, let's factor out the common term 'x':
x(x - 1) = 0
Now, observe that there are two factors on the left-hand side: x and (x - 1). In order for the equation to be satisfied, at least one of the factors must be equal to zero. Therefore, there are two potential solutions:
1) x = 0
2) x - 1 = 0, which simplifies to x = 1
So, the solutions to the equation (x^2 - 2x)(x + 3) = -2x(x + 1), after correcting the equation, are x = 0 and x = 1.