How do i do this using long division? This is part of a Partial Fractions Question. Please help, thanks!
(x^3) / (x^3 - 2x^2 + 3x - 18)
I tried using long division, but I could only get a quotient of 1, and a remainder of 2x^2 - 3x + 18
Did i do something wrong?
One does not use polynomial long division to apply the method of partial fractions. Try factoring x^3 - 2x^2 + 3x - 18 instead. The factors are (x-3)and (x^2 +x -6).
The next step is, I believe to express your ratio in terms of
C/(x-3) + (Ax + B)/(x^2 +x -6).
where A, B and C are constants. There should be three values that work.
Change (x^2 +x -6) to (x^2 +x +6)
To divide the numerator (x^3) by the denominator (x^3 - 2x^2 + 3x - 18) using long division, follow these steps:
1. Write the numerator (x^3) and the denominator (x^3 - 2x^2 + 3x - 18) in descending order of powers of x.
2. Start by dividing the highest power term in the numerator (x^3) by the highest power term in the denominator (x^3). The quotient for this division is the first term in the quotient polynomial.
x^3 / x^3 = 1
3. Multiply the quotient term (1) by the denominator (x^3 - 2x^2 + 3x - 18) and subtract the result from the numerator (x^3). This step is similar to subtracting polynomials.
(x^3) - (x^3 - 2x^2 + 3x - 18) = 2x^2 - 3x + 18 (remainder)
4. Now, bring down the next term from the numerator (none in this case), and repeat step 3.
Since the remainder obtained after one division is not zero, it means that the denominator (x^3 - 2x^2 + 3x - 18) is not a factor of the numerator (x^3).
So, you haven't done anything wrong. The quotient you obtained is correct. The remainder (2x^2 - 3x + 18) cannot be further reduced using long division, and it remains as the remainder.