9. Partial Fractions

a2. General Decompositions

a. Long Division of Polynomials - Review

Recall from high school Algebra that the division of numbers can be written as: \[\begin{aligned} &\quad\;\text{quotient}\;+\;\dfrac{\text{remainder}}{\text{divisor}}\\ \text{divisor}\;&\overline{\left)\quad\text{dividend}\qquad\qquad\right.} \end{aligned}\] or \[ \dfrac{\text{dividend}}{\text{divisor}}=\text{quotient} +\dfrac{\text{remainder}}{\text{divisor}} \]

For example, the division \[\begin{aligned} &\quad31+\dfrac{7}{15} \\ 15&\overline{\left)\,472\right.\quad} \end{aligned}\] can also be written as \[ \dfrac{472}{15}=31+\dfrac{7}{15} \]

We check this by computing \[ \text{divisor}\cdot\text{quotient}+\text{remainder}=\text{dividend} \] For example, \[15\cdot31+7=472 \]

The above formulas work for polynomials as well as for numbers. Here's how:

Perform the division \(\dfrac{3x^3-x^2+12}{x^2+x+2}\). In other words, find the quotient and remainder when \(3x^3-x^2+12\) is divided by \(x^2+x+2\).

1. Arrange both polynomials in descending powers of the variable. Leave space in the dividend for missing powers of the variable. \[ x^2+x+2\overline{\left)3x^3-x^2+0\,x+12\right.} \]
2. Divide the leading term in the dividend by the leading term in the divisor; i.e. divide \(3x^3\) by \(x^2\). Write the quotient above the \(3x^3\) term. \[\begin{aligned} &\;\;3x \\ x^2+x+2&\overline{\left)3x^3-x^2+0\,x+12\right.} \end{aligned}\]
3. Multiply the divisor by \(3x\), lining up like terms below those in the dividend. \[\begin{aligned} &\;\;3x \\ x^2+x+2&\overline{\left)3x^3-x^2+0\,x+12\right.} \\ &\;\;3x^3+3x^2+6x \end{aligned}\]
4. Subtract and bring down all remaining terms in the dividend. \[\begin{aligned} &\;\;3x \\ x^2+x+2&\overline{\left)3x^3-x^2+0\,x+12\right.} \\ &\;\underline{\;3x^3+3x^2+6x\qquad} \\ &\qquad-4x^2-6x+12 \end{aligned}\]
5. Now divide the leading term on the bottom line \(-4x^2\) by the leading term in the divisor \(x^2\), writing the quotient \(-4\) above the next term in the dividend. \[\begin{aligned} &\;\;3x\;-\;4 \\ x^2+x+2&\overline{\left)3x^3-x^2+0\,x+12\right.} \\ &\;\underline{\;3x^3+3x^2+6x\qquad} \\ &\qquad-4x^2-6x+12 \end{aligned}\]
6. Multiply the divisor by \(-4\) and subtract \[\begin{aligned} &\;\;3x\;-\;4 \\ x^2+x+2&\overline{\left)3x^3-x^2+0\,x+12\right.} \\ &\;\underline{\;3x^3+3x^2+6x\qquad} \\ &\qquad-4x^2-6x+12 \\ &\qquad\;\,\underline{-4x^2-4x-8\quad} \\ &\qquad\qquad\;\;-2x+20 \end{aligned}\]
7. Repeat steps 5 and 6 until the result upon subtraction is a polynomial of degree less than that of the divisor (as has just occurred in this example.)
8. The quotient \(3x-4\) is at the top, and the remainder \(-2x+20\) is at the bottom. To display your result, add the remainder divided by the divisor to the quotient. \[\begin{aligned} &\;\;3x\;-\;4\;+\;\dfrac{-2x+20}{x^2+x+2} \\ x^2+x+2&\overline{\left)3x^3-x^2+0\,x+12\right.} \\ &\;\underline{\;3x^3+3x^2+6x\qquad} \\ &\qquad-4x^2-6x+12 \\ &\qquad\;\,\underline{-4x^2-4x-8\quad} \\ &\qquad\qquad\;\;-2x+20 \end{aligned}\]
9. Finally we can write this as \[ \dfrac{3x^3-x^2+12}{x^2+x+2}=3x-4+\dfrac{-2x+20}{x^2+x+2} \]
10. We check this by computing \[ (x^2+x+2)(3x-4)+(-2x+20)=3x^3-x^2+12 \]

Now you try one.