Question

How do you evaluate `(- 2x ^ { 3} + 5x ^ { 2} - 3x + 19) \div ( - x ^ { 2} - 4)`?

Answer 1

`-2x^3 + 5x^2 - 3x + 19` = (`-x^2-4`) (`2x-5`) + `5x - 1`

Explanation:

You have to use long division.

NOTE:
`-2x^3 + 5x^2 - 3x + 19` = Dividend
`-x^2-4` = Divisor
`2x-5` = Quotient
`5x - 1` = Remainder

Step 1:

• You take the first value, `-2x^3` and divide it by `-x^2` which gives you `2x`
• Then to get the `-2x^2` in the first line, you have to multiply the quotient, `2x` by the divisor which is `-x^2-4`

That will give you `-2x^3 - 8x`, the reason there's an empty space under `5x^2` is because there's no value in the divisor you can multiply the quotient by to get a x^2, so you leave the space BLANK, and bring it down to the next line

Step 2:

• You have to imagine there's a subtract sign on the side, so you SUBTRACT the top by the bottom, `-2x^2` - `-2x^2` = 0
• Then you would bring down 19, giving you `5x^2 + 5x + 19`
• Now, you take ONLY the first value, `5x^2` and divide that by the divisor again, and you take that value and bring it up to the top.

Step 3:

• Then multiply ONLY the value you got from dividing, 5, by the divisor once again. You would do, `-5 *` `-x^2`, put it under `5x^2`.

• Then do `-5 *`-4 which gives you 20, and put it under 19.

WHY PUT 20 UNDER 19?
Because you put values that have similar variables together, or in this case, constants with constants.

Stetp 4:
- Subtract like usual, and bring down the `5x` since it was basically, `5x` - 0x which equals, `5x`

Answer:
`(−2x3+5x2−3x+19)÷(−x2−4)` = `2x - 5`
and since it has a remainder you would include it as well giving you,
`-2x^3 + 5x^2 - 3x + 19` = (`-x^2-4`) (`2x-5`) + `5x - 1`