- Factorization of Polynomials to Linear/Quadratic Factors
- Method of Undetermined Coefficients
- Substitution Method
- Multiplication Method
- Heaviside Cover-up Method
- Partial Fraction Decomposition through Mathematical Software
- Problems

Before we start, we emphasize not to forget the following steps before decomposing the fraction:

- Ensure the numerator and denominator are both polynomials.
- Ensure the numerator of the fraction has an order less than the denominator.
- Ensure the denominator is completely reduced to irreducible factors.

Irreducible factor in denominator | Term in decomposition |
---|---|

$ax+b$ | $\frac{C_1}{ax+b}$ |

$(ax+b)^n$ | $\frac{C_1}{ax+b}+\frac{C_2}{(ax+b)^2}+...+\frac{C_n}{(ax+b)^n}$ |

$ax^2+bx+c$ | $\frac{C_1x+D_1}{ax^2+bx+c}$ |

$(ax^2+bx+c)^n$ | $\frac{C_1x+D_1}{ax^2+bx+c}+\frac{C_2x+D_2}{(ax^2+bx+c)^2}+...+\frac{C_nx+D_n}{(ax^2+bx+c)^n}$ |

## Factorization of Polynomials to Linear/Quadratic Factors

Example: Reduce the polynomial into irreducible linear and quadratic factors with real coefficients. $$f(x)=2x^7-9x^6+31x^4-21x^3+24x^2-9x+54$$ At first, this looks extremely intimidating, but initial factorization is simple. Graph the polynomial and it is immediately obvious that two roots exist: $x=-2$ with a multiplicity of 1 and $x=3$ with a multiplicity of 2. Therefore, the linear factors, which are comprised of the real roots, are $x+2$ and $(x-3)^2$ (multiplicity is incorporated as a power). Long division of $f(x)$ with the product of the linear factors produces $g(x)=2x^4-x^3+2x^2+3$. With an order of 4, this function is therefore a product of two quadratics. We denote these two quadratics as: \begin{align*} ax^2+bx+c\\ ux^2+vx+w\end{align*} Multiplying the two polynomials and setting it equal to $g(x)$ results in: $$(au)x^4+(av+bu)x^3+(aw+bv+cu)x^2+(bw+cv)x+cw=2x^4-x^3+2x^2+3$$ Comparison of coefficients yields: \begin{align*} au&=2 \\ av+bu&=-1 \\ aw+bv+cu&=2 \\ bw+cv&=0 \\ cw&=3\end{align*} There are five equations and six unknowns yielding one degree of freedom. We arbitratily set $a=1$ making $u=2$ through the first equation. With some algebra, it can be shown that there are two sets of polynomials that satisfy the equations above. One solution produces the two polynomials $x^2+x+1$ and $2x^2-3x+3$. So, the final answer is: $$f(x)=(x-3)^2(x+2)(x^2+x+1)(2x^2-3x+3)$$ It is significantly easier to do this factorization through a mathematical software. MATLAB provides a function,

*factor*, that can be used for this purpose.

syms x F = factor(2*x^7-9*x^6+31*x^4-21*x^3+24*x^2-9*x+54)

```
F = [ x + 2, x^2 + x + 1, 2*x^2 - 3*x + 3, x - 3, x - 3]
```

## Method of Undetermined Coefficients

## Substitution Method

## Multiplication Method

Now, let's consider another more complicated partial fraction. Consider the function and its partial fraction expansion: $$\frac{6x^2+9x+3}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2} \label{eq:4} \tag{4}$$ $A$ can be found through this method. The steps for showing $A=2$ is omitted. However, note we cannot solve for $B$ using this method. If we proceed to solve for $B$ using this method, we first need to multiply both sides by $x+2$. $$\frac{6x^2+9x+3}{(x-1)(x+2)}=\frac{A(x+2)}{x-1}+B+\frac{C}{x+2}$$ If we substitute $x=-2$, we get a division by zero on the left and right-hand side. However, if we instead solve for $C$, multiplying both sides by $(x+2)^2$ yields: $$\frac{6x^2+9x+3}{x-1}=\frac{A(x+2)^2}{x-1}+B(x+2)+C$$ Substituting $x=-2$ now works: $$\frac{6(-2)^2+9(-2)+3}{-2-1}=\frac{A(-2+2)^2}{-2-1}+B(-2+2)+C$$ $$\therefore C=\frac{9}{-3}=-3$$ To solve for $B$, one may use the Method of Undetermined Coefficients, however it is easier in this case just to use the Substitution method and substitute one number into equation \ref{eq:4}. Any number that does not result in a division by zero can be substituted, so we substitute $x=-1$, which is the root of the numerator in the original function. Note this value is in the domain of the original function. There is no division by zero as a result and the left-hand side of equation \ref{eq:4} goes to zero: $$0=\frac{A}{-1-1}+\frac{B}{-1+2}+\frac{C}{(-1+2)^2}$$ Values of $A$ and $C$ have already been solved using the Multiplication method. $$0=\frac{2}{-1-1}+\frac{B}{-1+2}-\frac{3}{(-1+2)^2}$$ $$\therefore B=1+3=4$$ $$\frac{6x^2+9x+3}{(x-1)(x+2)^2}=\frac{2}{x-1}+\frac{4}{x+2}-\frac{3}{(x+2)^2}$$

**Therefore, for the partial fractions that result from repeated linear factors, the Multiplication Method should be used on the associated partial fraction with the denominator raised to the highest power. This method cannot be solely used to solve for the other factors raised to a lower power.**

For fractions where the denominator has quadratics or quadratic terms raised to a power,

*all*unknown parameters can be calculated with this method since a division by zero is mitigated. However, the use of complex numbers is required. For example, observe the function and its decomposition below: $$\frac{-x^2 + 3 x - 11}{(x - 2) (x^2 - 2 x + 3)}=\frac{A}{x-2}+\frac{Bx+C}{x^2-2x+3}$$ $A$ is solved easily to yield $A=-3$. To solve for $B$ and $C$, multiply both sides by $x^2-2x+3$. $$\frac{-x^2 + 3 x - 11}{x - 2}=\frac{A(x^2-2x+3)}{x-2}+Bx+C$$ The roots to $x^2-2x+3$ are $1+i\sqrt{2}$ and $1-i\sqrt{2}$. For the first root: $$\frac{-(1+i\sqrt{2})^2+3(1+i\sqrt{2})-11}{1+i\sqrt{2}-2}=\frac{A(0)}{x-2}+B(1+i\sqrt{2})+C$$ $$\frac{-7+i\sqrt{2}}{-1+i\sqrt{2}}=B(1+i\sqrt{2})+C$$ $$3 + 2 i \sqrt{2}=B(1+i\sqrt{2})+C \label{eq:5} \tag{5}$$ At this point, we know $B$ and $C$ must be real numbers, so that automatically forces $B$ to equal 2. Then it follows that $C=1$. A review of complex number operations is located here.

Complex number algebra may be a bit tedious so in this case, the Method of Undetermined Coefficients to solve for $B$ and $C$ may have produced a quicker answer. However, if a fraction has many linear factors in its denominator and only one or two quadratic terms, this method for partial fraction decomposition may be faster compared to the Method of Undetermined Coefficients since this method does not require the tedious step of combining all partial fractions into one fraction.

With the same analysis, it can be shown that

**for partial fractions that result from repeated quadratic factors, the Multiplication Method should be used on the associated partial fraction with the denominator raised to the highest power. This method cannot be solely used to solve for the other factors raised to a lower power.**This is the same conclusion as the case for repeated linear factors.

## Heaviside Cover-up Method

Irreducible factor in denominator | Heaviside method restriction |
---|---|

$ax+b$ | None. |

$(ax+b)^n$ | May only be used on associated partial fraction with denominator raised to highest power. |

$ax^2+bx+c$ | None, complex numbers required. May be quicker to use Method of Undetermined Coefficients. |

$(ax^2+bx+c)^n$ | May only be used on associated partial fraction with denominator raised to highest power. Complex numbers required. May be quicker to use Method of Undetermined Coefficients. |

## Partial Fraction Decomposition through Mathematical Software

*partfrac*, that can be used. The function accepts the function, and optionally the independent variable. For the same function given first in the Methods of Undetermined Coefficient method $\frac{4x^2+6x+6}{x^3-7x-6}$, the partial fraction is calculated through the code:

syms x f=partfrac((4*x^2+6*x+6)/(x^3-7*x-6))

`f = 2/(x + 2) - 1/(x + 1) + 3/(x - 3)`

Use the optional second parameter to specify the independent variable. In the case of $\frac{a}{(x+a)(x+2)}$, $x$ is the independent variable and $a$ is a constant:

syms x a f=partfrac(a/((x+a)*(x+2)),x)

`f = a/((a - 2)*(x + 2)) - a/((a + x)*(a - 2))`

## Problems

- Decompose $f(x)=\frac{4x^2+6x+8}{x^2(x+4)}$ into partial fractions using all the methods shown.
- Decompose $f(x)=\frac{a}{(x-b)(x-c)}$ into partial fractions, where $a,b,c$ are real numbers and $b \neq c$.