Solving Equivalent Expressions Using Subtraction Method

[eleventhxhour]

How do you determine if the two expressions are equivalent using the subtraction method?\(\displaystyle \frac{2x^3+7x^2-5x-6}{x+1} - [(2x^2+5x-10) + \frac{4}{x+1}] \)

The textbook says that they're equivalent. I know that, using the subtraction method, they have to equal 0 in order to be equivalent. But, I'm not getting that. This is what I've done so far:

\(\displaystyle 2x^2+7x-5-\frac{6}{x+1} - [(2x^2+5x-10) + \frac{4}{x+1}] \)

Then I'm not sure what to do...

 

[Opalg]

How do you determine if the two expressions are equivalent using the subtraction method?\(\displaystyle \frac{2x^3+7x^2-5x-6}{x+1} - [(2x^2+5x-10) + \frac{4}{x+1}] \)

The textbook says that they're equivalent. I know that, using the subtraction method, they have to equal 0 in order to be equivalent. But, I'm not getting that. This is what I've done so far:

\(\displaystyle 2x^2+7x-5-\frac{6}{x+1} - [(2x^2+5x-10) + \frac{4}{x+1}] \)

Then I'm not sure what to do...

Start by putting everything over the denominator $x+1$. You have already slipped up by writing \(\displaystyle 2x^2+7x-5-\frac{6}{x+1}\) as though it is the same as \(\displaystyle \frac{2x^3+7x^2-5x-6}{x+1}\) (it isn't!). What you should find is that the expression is equal to $$\frac{2x^3+7x^2-5x-6 - \bigl[(2x^2+5x-10)(x+1) + 4\bigr]}{x+1}.$$

The next step (in fact, maybe this should have been the first step) is to get rid of the square brackets, noticing that there is a minus sign in front of them, which changes the sign of every term inside.

 

[eleventhxhour]

Start by putting everything over the denominator $x+1$. You have already slipped up by writing \(\displaystyle 2x^2+7x-5-\frac{6}{x+1}\) as though it is the same as \(\displaystyle \frac{2x^3+7x^2-5x-6}{x+1}\) (it isn't!). What you should find is that the expression is equal to $$\frac{2x^3+7x^2-5x-6 - \bigl[(2x^2+5x-10)(x+1) + 4\bigr]}{x+1}.$$

The next step (in fact, maybe this should have been the first step) is to get rid of the square brackets, noticing that there is a minus sign in front of them, which changes the sign of every term inside.

This is what I got, but it's not equivalent...
\(\displaystyle \frac{2x^3+7x^2-5x-6}{x+1}-\frac{2x^3+5x^2-10x+4}{x+1}\)

 

[Fermat1]

This is what I got, but it's not equivalent...
\(\displaystyle \frac{2x^3+7x^2-5x-6}{x+1}-\frac{2x^3+5x^2-10x+4}{x+1}\)

You're doing it wrong. Forget about the minus sign for a minute and multiply out the brackets and collect 'like' terms.

 

[MarkFL]

How is it not the subtraction method? Isn't Opalg's expression the difference between the two expressions, just written with a common denominator?

 

[Fermat1]

I don't think my last post was very clear. My advice to multiply out brackets is to find all terms with the highest power of x, then then the terms with the next highest power and so on. So $(2x^2+5x-10)(x+1)+4$=$2x^3+7x^2-5x-6$ which is what you want

 

[Deveno]

I think you are trying to "skip steps".

To get everything over a common denominator, we have to write:

$2x^2 + 5x - 10 = \dfrac{\text{something}}{x + 1}$.

The way to do this is to multiply by $1 = \dfrac{x+1}{x+1}$. That is:

$2x^2 + 5x - 10 = \dfrac{(2x^2 + 5x - 10)(x+1)}{x+1} = \dfrac{(2x^2 + 5x - 10)x + (2x^2 + 5x - 10)(1)}{x+1}$

$= \dfrac{2x^3 + 5x^2 - 10x + 2x^2 + 5x - 10}{x+1} = \dfrac{2x^3 + 7x^2 - 5x - 10}{x+1}$

ONLY NOW can we add the two terms:

$2x^2 + 5x - 10 + \dfrac{4}{x+1} = \dfrac{2x^3 + 7x^2 - 5x - 10}{x+1} + \dfrac{4}{x + 1} = \dfrac{2x^3 + 7x^2 - 5x - 6}{x+1}$.

Now what do you get when you subtract this from $\dfrac{2x^3 + 7x^2 - 5x - 6}{x+1}$?