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

Question

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

Eva Adamson · Accepted Answer

I prefer long division.

#(-3x^3-3x^2-4x+1)/(3x-4)# Long division form: #color(white)( (3x-4)/color(black)(3x-4)) color(white)( (-3x^3-3x^2-4x+1))/(") "-3x^3-3x^2-4x+1)# Write #-x^2# in the quotient: #color(white)( (3x-4)/color(black)(3x-4)) (color(white)("....")-x^2color(white)(-3x^2-4x+1))/(") "-3x^3-3x^2-4x+1)# Multiply #-x^2(3x-4) = -3x^2+ 4x^2# and subtract from the dividend: #color(white)( (3x-4)/color(black)(3x-4)) (color(white)("....")-x^2color(white)(-3x^2-4x+1))/(") "-3x^3-3x^2-4x+1)# #color(white)(".....................")ul(3x^2- 4x^2)# #color(white)(".................... .")-7x^2-4x# Write #-7/3x# in the quotient: #color(white)( (3x-4)/color(black)(3x-4)) (color(white)("....")-x^2-7/3xcolor(white)(-4x+1))/(") "-3x^3-3x^2-4x+1)# #color(white)(".....................")ul(3x^2- 4x^2)# #color(white)(".................... .")-7x^2-4x# Multiply #-7/3x(3x-4) = -7x^2+ 28/3x# and subtract from the dividend: #color(white)( (3x-4)/color(black)(3x-4)) (color(white)("....")-x^2-7/3xcolor(white)(-4x+1))/(") "-3x^3-3x^2-4x+1)# #color(white)(".....................")ul(3x^2- 4x^2)# #color(white)(".................... .")-7x^2-4x# #color(white)("....................... .")ul(7x^2-28/3x)# #color(white)("............................... .")-40/3x+ 1# Write #-40/9# in the quotient: #color(white)( (3x-4)/color(black)(3x-4)) (color(white)("...")-x^2-7/3x-40/9color(white)(1))/(") "-3x^3-3x^2-4x+1)# #color(white)(".....................")ul(3x^2- 4x^2)# #color(white)(".................... .")-7x^2-4x# #color(white)("....................... .")ul(7x^2-28/3x)# #color(white)("............................... .")-40/3x+ 1# Multiply #-40/9(3x-4) = -40/9x+ 160/9# and subtract from the dividend: #color(white)( (3x-4)/color(black)(3x-4)) (color(white)("...")-x^2-7/3x-40/9color(white)(1))/(") "-3x^3-3x^2-4x+1)# #color(white)(".....................")ul(3x^2- 4x^2)# #color(white)(".................... .")-7x^2-4x# #color(white)("....................... .")ul(7x^2-28/3x)# #color(white)("............................... .")-40/3x+ 1# #color(white)("............................... .")ul(40/9x - 160/9)# #color(white)("........................................... .")-151/9# The quotient is #-x^2-7/3x-40/9# with a remainder of #-151/9#

Eva Abramson · Answer

undefined To divide (-3x^3-3x^2-4x+1) by (3x-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 (3x). The result is -x^2.

2. Multiply the entire denominator (3x-4) by -x^2, giving -x^2(3x-4) = -3x^3 + 4x^2.

3. Subtract this result (-3x^3 + 4x^2) from the original numerator (-3x^3-3x^2-4x+1). This gives (-3x^3-3x^2-4x+1) - (-3x^3 + 4x^2) = -7x^2 - 4x + 1.

4. Bring down the next term from the numerator, which is -7x^2. Now you have (-7x^2 - 4x + 1).

5. Divide the first term of this new numerator (-7x^2) by the first term of the denominator (3x). The result is -7/3x.

6. Multiply the entire denominator (3x-4) by -7/3x, giving -7/3x(3x-4) = -7x^2 + 28/3x.

7. Subtract this result (-7x^2 + 28/3x) from the current numerator (-7x^2 - 4x + 1). This gives (-7x^2 - 4x + 1) - (-7x^2 + 28/3x) = (-4x + 1 - 28/3x).

8. Simplify the expression (-4x + 1 - 28/3x) if needed.

The final result of the division is -x^2 - 7/3x + (-4x + 1 - 28/3x), or simplified as -x^2 - 25/3x + 1.