How do you divide #(-x^3-x^2+9x-3)/(x-3) #?

Question

How do you divide #(-x^3-x^2+9x-3)/(x-3)   #?

Nathan Axel · Accepted Answer

#-x^2 + 2x +1# You write the answer using everything you know. The division of a degree 3 by a degree one polynomial will be degree two. So you write #-x^3 - x^2 + 9x -3 = (x-3)(-x^2 +1 + ax)# This gives you #a - 3= -1# #a=2#

Nathan Axel · Answer

undefined To divide (-x^3-x^2+9x-3) by (x-3), you can use long division. Here are the steps:

1. Divide the first term of the numerator (-x^3) by the first term of the denominator (x). The result is -x^2.
2. Multiply the entire denominator (x-3) by -x^2, giving you (-x^3+3x^2).
3. Subtract (-x^3+3x^2) from the numerator (-x^3-x^2+9x-3). This gives you (-4x^2+9x-3).
4. Bring down the next term from the numerator, which is 0x^2. Now you have (-4x^2+9x-3).
5. Divide the first term of the new numerator (-4x^2) by the first term of the denominator (x). The result is -4x.
6. Multiply the entire denominator (x-3) by -4x, giving you (-4x^2+12x).
7. Subtract (-4x^2+12x) from the new numerator (-4x^2+9x-3). This gives you (-3x-3).
8. Bring down the next term from the numerator, which is -3. Now you have (-3x-3).
9. Divide the first term of the new numerator (-3x) by the first term of the denominator (x). The result is -3.
10. Multiply the entire denominator (x-3) by -3, giving you (-3x+9).
11. Subtract (-3x+9) from the new numerator (-3x-3). This gives you (-12).
12. There are no more terms left in the numerator, so the division is complete.

The quotient is -x^2 - 4x - 3, and the remainder is -12.