How do you divide #( -x^4 + 3x^3 + 9x^2 +4x)/(2x^2-3x)#?

Question

Avery Alexander · Accepted Answer

Trigonometry long division gives quotient of #(-4x^2 +6x +9)/8# and remainder of #59/8# This can be simplified a bit by first taking out common factors. After that you need to do the trigonometry version of long division. #(x(-x^3 +3x^2 +9x +4))/(x(2x - 3)) = (-x^3 +3x^2 +9x +4)/(2x -3)# Divide the first element by the first element of the divisor: #-x^3/(2x) = **-x^2/2** # Multiply this result by the divisor: #-x^2/2 *(2x - 3) = -x^3+3x^2/2# Then subtract this from the dividend: #(-x^3 +3x^2 +9x +4) - (-x^3+3x^2/2) = 3x^2/2 +9x +4# Repeat this process with the remaining dividend: #(3x^2/2) / (2x) = **3x/4** # #(3x/4)*(2x-3) = (6x^2 -9x)/4# #((3x^2)/2 +9x +4) - ((6x^2 - 9x)/4) = (9x)/4 +4# And repeat again #((9x)/4)/(2x) = **9/8** # #(9/8)*(2x-3) = (9x)/4 -27/8# #((9x)/4 +4) - ((9x)/4 -27/8) = **59/8** # The quotient is then the sum of the factors in bold and the remainder is #59/8# #-x^2/2 + 3x/4 + 9/8 = (-4x^2 +6x +9)/8#

Ethan Adkins · Answer

undefined To divide (-x^4 + 3x^3 + 9x^2 + 4x) by (2x^2 - 3x), we can use polynomial long division.

First, we divide the highest degree term of the numerator by the highest degree term of the denominator. In this case, (-x^4) ÷ (2x^2) gives us -0.5x^2.

Next, we multiply the entire denominator (2x^2 - 3x) by the result we obtained (-0.5x^2), giving us (-x^4 + 1.5x^3).

We subtract this result from the numerator (-x^4 + 3x^3 + 9x^2 + 4x) to get (4.5x^3 + 9x^2 + 4x).

We repeat the process by dividing the highest degree term of the new numerator (4.5x^3) by the highest degree term of the denominator (2x^2), which gives us 2.25x.

We multiply the entire denominator (2x^2 - 3x) by the result we obtained (2.25x), giving us (4.5x^3 - 6.75x^2).

We subtract this result from the new numerator (4.5x^3 + 9x^2 + 4x) to get (15.75x^2 + 4x).

We continue this process until we have no more terms to divide. In this case, we divide (15.75x^2 + 4x) by (2x^2 - 3x).

The final result of the division is (-0.5x^2 + 2.25x + 7.875) with a remainder of (4x + 7.875).

Eva Adamson · Answer

undefined To divide $ \frac{-x^4 + 3x^3 + 9x^2 + 4x}{2x^2 - 3x} $, we perform polynomial long division or synthetic division. Let's use polynomial long division in this case.

Step 1: Divide the leading term of the numerator by the leading term of the denominator. This gives us the first term of the quotient. In this case, $ \frac{-x^4}{2x^2} = -\frac{1}{2}x^2 $.

Step 2: Multiply the entire denominator by the first term of the quotient and subtract the result from the numerator. This gives us the remainder after dividing by the leading term.

Step 3: Repeat the process with the new polynomial obtained in the previous step.

Performing these steps iteratively, we can find the quotient and the remainder. Once we have the quotient, we can write the division as:

$$ \frac{-x^4 + 3x^3 + 9x^2 + 4x}{2x^2 - 3x} = -\frac{1}{2}x^2 + \frac{3}{4}x + \frac{33}{8} + \frac{4}{8x - 6} $$

This is the quotient, and $ \frac{4}{8x - 6} $ is the remainder.