How do you divide #( -3x^3+ x^2-3x-28 )/(x + 1 )#?

Question

How do you divide #( -3x^3+ x^2-3x-28 )/(x + 1  )#?

Ethan Adkins · Accepted Answer

The quotient is #=-3x^2+4x-7# and the remainder is #=-21#

Now let's divide that long way.

#color(white)(aaaa)##x+1##|##-3x^3+x^2-3x-28##color(white)(aaaa)##|##-3x^2+4x-7#

#color(white)(aaaaaaaaaa)##-3x^3-3x^2#

#color(white)(aaaaaaaaaaaaa)##0+4x^2-3x#

#color(white)(aaaaaaaaaaaaaaa)##+4x^2+4x#

#color(white)(aaaaaaaaaaaaaaaaa)##+0-7x-28#

#color(white)(aaaaaaaaaaaaaaaaaaaaa)##-7x-7#

#color(white)(aaaaaaaaaaaaaaaaaaaaaa)##-0-21#

The quotient is #=-3x^2+4x-7# and the remainder is #=-21#

Eli Assouline · Answer

undefined To divide (-3x^3 + x^2 - 3x - 28) by (x + 1), you can use long division. Here are the steps:

1. Divide the first term of the numerator (-3x^3) by the first term of the denominator (x). This gives -3x^2.
2. Multiply the entire denominator (x + 1) by -3x^2, which gives -3x^3 - 3x^2.
3. Subtract this result from the numerator: (-3x^3 + x^2 - 3x - 28) - (-3x^3 - 3x^2) = 4x^2 - 3x - 28.
4. Bring down the next term from the numerator, which is -3x.
5. Divide the first term of the new numerator (4x^2) by the first term of the denominator (x). This gives 4x.
6. Multiply the entire denominator (x + 1) by 4x, which gives 4x^2 + 4x.
7. Subtract this result from the new numerator: (4x^2 - 3x - 28) - (4x^2 + 4x) = -7x - 28.
8. Bring down the next term from the numerator, which is -28.
9. Divide the first term of the new numerator (-7x) by the first term of the denominator (x). This gives -7.
10. Multiply the entire denominator (x + 1) by -7, which gives -7x - 7.
11. Subtract this result from the new numerator: (-7x - 28) - (-7x - 7) = -21.
12. There are no more terms left in the numerator, so the division is complete.

The quotient is -3x^2 + 4x - 7, and the remainder is -21.