If one doesn't know of the polynomial can be factored, one could still do long division with polynomials or synthetic division. Both for me are sort of difficult to explain. So, I put in the attachments representations of how to do them. I can say, an important part for each is, make sure you put the zeros in as I have them. You would have to have this expression written as 2x^3 - 3x^2 + 0x - 4 for long division or synthetic division to work.

For the synthetic division, you can construct the lines most anyway you would like to.

For the long division, in short, it's exactly like doing long division with regular numbers. But, each term would be like one digit in the number.

The expression 2x^3 -3x^2-4 has to be divided by x-2. First factor 2x^3 - 3x^2 - 4.

2x^3 - 3x^2 - 4

= 2x^3 + x^2 + 2x - 4x^2 - 2x - 4

= x(2x^2 + x + 2) - 2(2x^2 + x + 2)

= (x - 2)(2x^2 + x + 2)

`(2x^3 -3x^2-4)/(x - 2)`

= `((x - 2)(2x^2 + x + 2))/(x - 2)`

= `2x^2 + x + 2`

In order to solve for the following equ: `(2x^(3)-3x^2-4)/(x-2)`

You will need find a common factor that you can cancel out from the numerator AND the denominator. And, from this equation, it seems like the numerator and denominator has (x-2) in common!

- `((x-2)(2x^(2)+x+2))/(x-2)`

And what happens, if you divide `(x-2)/(x-2)?`

- It will equal 1!
- THINK OF IT AS: What will happen if you divide `(4)/(4)= 1`

Therefore, you will be left with `2x^(2)+x+2`

as your final answer!

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Set the problem up so that it your quadratic is under the long division box, with and additional +0x.

1) Determine what you would have to multiply by your x in your x-2 term to produce 2x^3, which would be 2x^2, this # will go in the answer section above the division box

2) multiply that # by the x-2 and put the answer below the division box, just below your quadratic so that you can subtract them

3) Once everything has been subtracted out, repeat this process until there are no more terms, or you can no longer multiply to cancel out. When this happens you have a remainder.

In this case you end up with 2x^2 +x +2, no remainder

There are different ways to divide a polynomial by a binomial. But I think it is always safe to use the long division method. I attached a photo of my solution. The final answer is `2x^2 + x + 2`

To get this you must divide the monomial with the highest degree with your x and multiply this to the number in your divisor. It sounds very confusing, but it works the same way as dividing non-algebraic numbers.

Notice also, that I added 0x to complete the polynomial. If there is a missing part of the polynomial, you will end up with a different answer.

I always use long division for these types of problems. When doing this you just have to think logically even though it seems hard at first. Make sure that you put a +0x where an x term would normally be or else the answer will be different.

First you must think "what can I multiply by x to get 2x^3?" 2x^2 multiplied by x will give you what you need so that is the first part of your answer. Don't forget to also multiply your 2x^2 by the -2 as well. After you multiply you will get 2x^3-4x^2 which must be subtracted from your original equation. This makes a new equation and the steps repeat.

I know this sounds confusing when written so I will attach a picture. There are several other ways you can do this, but I find this easier for me. Anyway, the final answer ends up being **2x^2+x+2**