How do you divide #( -3x^3+ 6x^2-13x-5 )/(x + 1 )#?

Question

How do you divide #( -3x^3+ 6x^2-13x-5 )/(x + 1  )#?

Abigail Allen · Accepted Answer

#-3x^2+9x-22+17/(x+1)# #"one way is to use the divisor as a factor in the numerator"# #"consider the numerator"# #color(red)(-3x^2)(x+1)color(magenta)(+3x^2)+6x^2-13x-5# #=color(red)(-3x^2)(x+1)color(red)(+9x)(x+1)color(magenta)(-9x)-13x-5# #=color(red)(-3x^2)(x+1)color(red)(+9x)(x+1)color(red)(-22)(x+1)color(magenta)(+22)-5# #=color(red)(-3x^2)(x+1)color(red)(+9x)(x+1)color(red)(-22)(x+1)+17# #"quotient "=color(red)(-3x^2+9x-22)," remainder "=17# #rArr(-3x^3+6x^2-13x-5)/(x+1)# #=-3x^2+9x-22+17/(x+1)#

Nathan Axel · Answer

undefined To divide (-3x^3 + 6x^2 - 13x - 5) by (x + 1), you can use long division. Here are the steps:

1. Divide the first term of the dividend (-3x^3) by the first term of the divisor (x). This gives -3x^2.
2. Multiply the divisor (x + 1) by the result from step 1 (-3x^2), and write the product (-3x^3 - 3x^2) below the dividend.
3. Subtract the product from the dividend: (-3x^3 + 6x^2 - 13x - 5) - (-3x^3 - 3x^2) = 9x^2 - 13x - 5.
4. Bring down the next term from the dividend, which is -13x.
5. Divide the first term of the new dividend (9x^2) by the first term of the divisor (x). This gives 9x.
6. Multiply the divisor (x + 1) by the result from step 5 (9x), and write the product (9x^2 + 9x) below the new dividend.
7. Subtract the product from the new dividend: (9x^2 - 13x - 5) - (9x^2 + 9x) = -22x - 5.
8. Bring down the next term from the dividend, which is -5.
9. Divide the first term of the new dividend (-22x) by the first term of the divisor (x). This gives -22.
10. Multiply the divisor (x + 1) by the result from step 9 (-22), and write the product (-22x - 22) below the new dividend.
11. Subtract the product from the new dividend: (-22x - 5) - (-22x - 22) = 17.
12. There are no more terms in the dividend, so the division is complete.

The quotient is -3x^2 + 9x - 22, and the remainder is 17.