How do you divide #(-3x^3-23x^2-4x+1)/(x-4) #?

Question

How do you divide #(-3x^3-23x^2-4x+1)/(x-4)  #?

Savannah Anderson · Accepted Answer

#color(green)(=>"Quotient " -3x^2 - 35x - 144, " Remainder " -575/ (x-4) # #(3x^3 - 23x^2 - 4x + 1 ) / (x - 4)# #color(white)(aaa) - - - - - - - - - - - - - # #4 color(white)(aa) | color(white)(aa) -3 color(white)(aaa) -23 color(white)(aaaa) -4 color(white)(aaaa) + 1# #color(white)(aaa) | color(white)(aaaa) darr color(white)(aa) -12 color(white)(aa)-140 color(white)(aa) -576# #color(white)(aaa)|- - - - - - - - - - - - - # #color(white)(aaa)| color(white)(aa) -3 color(white)(aaa) -35 color(white)(aa) -144 color(white)(aa) -575# #color(green)(=>"Quotient " -3x^2 - 35x - 144, " Remainder " -575/ (x-4) #

Isaiah Anthony · Answer

undefined To divide (-3x^3-23x^2-4x+1) by (x-4), 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-4) by -3x^2, which gives -3x^3 + 12x^2.
3. Subtract this result from the numerator: (-3x^3-23x^2-4x+1) - (-3x^3 + 12x^2) = -35x^2 - 4x + 1.
4. Bring down the next term from the numerator, which is -35x^2. Now you have -35x^2 - 4x + 1.
5. Divide the first term of this new numerator (-35x^2) by the first term of the denominator (x). This gives -35x.
6. Multiply the entire denominator (x-4) by -35x, which gives -35x^2 + 140x.
7. Subtract this result from the numerator: (-35x^2 - 4x + 1) - (-35x^2 + 140x) = -144x + 1.
8. Bring down the next term from the numerator, which is -144x. Now you have -144x + 1.
9. Divide the first term of this new numerator (-144x) by the first term of the denominator (x). This gives -144.
10. Multiply the entire denominator (x-4) by -144, which gives -144x + 576.
11. Subtract this result from the numerator: (-144x + 1) - (-144x + 576) = -575.
12. There are no more terms left in the numerator, so the division is complete.

The quotient is -3x^2 - 35x - 144, and the remainder is -575.