How do you divide #(-2x^3-23x^2-4x+11)/(x-5) #?

Question

How do you divide #(-2x^3-23x^2-4x+11)/(x-5)  #?

Genesis Allen · Accepted Answer

#-2x^2-33x-169-834/(x-5)# #"one way is to use the divisor as a factor in the numerator"# #"consider the numerator"# #color(red)(-2x^2)(x-5)color(magenta)(-10x^2)-23x^2-4x+11# #=color(red)(-2x^2)(x-5)color(red)(-33x)(x-5)color(magenta)(-165x)-4x+11# #=color(red)(-2x^2)(x-5)color(red)(-33x)(x-5)color(red)(-169)(x-5)color(magenta)(-845)+11# #=color(red)(-2x^2)(x-5)color(red)(-33x)(x-5)color(red)(-169)(x-5)-834# #"quotient "=color(red)(-2x^2-33x-169)," remainder "=-834# #rArr(-2x^3-23x^2-4x+11)/(x-5)# #=-2x^3-33x-169-834/(x-5)#

Savannah Anderson · Answer

undefined To divide (-2x^3-23x^2-4x+11) by (x-5), you can use long division. Here are the steps:

1. Divide the first term of the numerator (-2x^3) by the first term of the denominator (x). The result is -2x^2.
2. Multiply the entire denominator (x-5) by -2x^2, giving -2x^3 + 10x^2.
3. Subtract this result from the original numerator: (-2x^3-23x^2-4x+11) - (-2x^3 + 10x^2) = -33x^2 - 4x + 11.
4. Bring down the next term from the numerator, which is -33x^2. Now you have -33x^2 - 4x + 11.
5. Divide the first term of this new numerator (-33x^2) by the first term of the denominator (x). The result is -33x.
6. Multiply the entire denominator (x-5) by -33x, giving -33x^2 + 165x.
7. Subtract this result from the new numerator: (-33x^2 - 4x + 11) - (-33x^2 + 165x) = -169x + 11.
8. Bring down the next term from the numerator, which is -169x. Now you have -169x + 11.
9. Divide the first term of this new numerator (-169x) by the first term of the denominator (x). The result is -169.
10. Multiply the entire denominator (x-5) by -169, giving -169x + 845.
11. Subtract this result from the new numerator: (-169x + 11) - (-169x + 845) = -834.
12. There are no more terms left in the numerator, so the division is complete.

The quotient is -2x^2 - 33x - 169, and the remainder is -834.