How do you divide #(-x^4-4x^3-3x^2-6x-2)/(x^2+4) #?

Question

How do you divide #(-x^4-4x^3-3x^2-6x-2)/(x^2+4)  #?

Julia Adams · Accepted Answer

If you go to Topics (https://tutor.hix.ai) and click on the subgroup as listed above (currently within group no 12) you will find lots of examples. There are some different approaches. Have a look at some of mine

Jameson Allen · Answer

undefined To divide (-x^4-4x^3-3x^2-6x-2) by (x^2+4), you can use long division. Here are the steps:

1. Divide the first term of the numerator (-x^4) by the first term of the denominator (x^2). The result is -x^2.

2. Multiply the entire denominator (x^2+4) by -x^2, and subtract the result from the numerator. This gives you a new numerator: (-4x^3-3x^2-6x-2) - (-x^2)(x^2+4).

3. Repeat the process by dividing the first term of the new numerator (-4x^3) by the first term of the denominator (x^2). The result is -4x.

4. Multiply the entire denominator (x^2+4) by -4x, and subtract the result from the new numerator. This gives you a new numerator: (-3x^2-6x-2) - (-4x)(x^2+4).

5. Continue this process until you have divided all terms of the numerator. The final quotient will be (-x^2-4x) and the remainder will be (-2).

Therefore, the division of (-x^4-4x^3-3x^2-6x-2) by (x^2+4) is equal to (-x^2-4x) with a remainder of -2.