Choose and solve partial fractions

Distinguish proper and improper rational functions

Compare the numerator and denominator degrees before decomposing. Equal degrees require division too.

The source defines a polynomial as $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0$, with real coefficients, $a_n\ne0$ and nonnegative integer degree n. A rational function is P(x)/Q(x), where P,Q are polynomials and the denominator is nonzero at each allowed input.

It is proper when $\deg P<\deg Q$, as in the source example

$$\frac{4x^2+5x-1}{3x^3+2x-4}.$$

It is improper when $\deg P\ge\deg Q$. Both printed examples are retained:

$$\frac{3x^2+2x-1}{5x-4},\qquad \frac{3x^4+2x-1}{(x+1)(x+3)(x-1)^2}.$$

In the latter denominator the degrees add to 4, so equal degrees still mean improper. Divide first to write an improper expression as a polynomial plus a proper rational remainder; the polynomial part must not be lost.

Choose the right partial-fraction form

Build the complete template before calculating coefficients. A missing repeated-factor term cannot be repaired by choosing different coefficient values.

The source’s four-step procedure is: divide first if improper; factorise the denominator as far as possible; assign the correct terms for its factor types; then solve their unknown coefficients.

Case 1: distinct linear factors.

$$\frac{px+q}{(ax+b)(cx+d)}=\frac A{ax+b}+\frac B{cx+d}.$$

The displayed example form is

$$\frac{x+1}{(x-2)(x+3)}=\frac A{x-2}+\frac B{x+3}.$$

Case 2: a repeated linear factor. Include every power up to its multiplicity:

$$\frac{px^2+qx+r}{(ax+b)(cx+d)^2} =\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2}.$$

Both printed example forms are

$$\begin{aligned} \frac{x+1}{(x-2)(x+3)^2}&=\frac A{x-2}+\frac B{x+3}+\frac C{(x+3)^2},\\ \frac{x+1}{(x-2)x^2}&=\frac A{x-2}+\frac Bx+\frac C{x^2}. \end{aligned}$$

Case 3: a non-repeated irreducible quadratic. Use a linear numerator over the quadratic:

$$\frac{px^2+qx+r}{(ax+b)(x^2+c^2)} =\frac A{ax+b}+\frac{Bx+C}{x^2+c^2}.$$

Here c must be nonzero for $x^2+c^2$ to be irreducible over the reals. The two printed example forms are

$$\begin{aligned} \frac{x+1}{(x-2)(x^2+3)}&=\frac A{x-2}+\frac{Bx+C}{x^2+3},\\ \frac{x+1}{(x-2)(x^2+3x+4)}&=\frac A{x-2}+\frac{Bx+C}{x^2+3x+4}. \end{aligned}$$

The latter quadratic has discriminant −7; it also has no real linear factors. The three template cases are labelled “in MF27” in the source; that is a retained source reference. Linear-factor templates assume nonzero leading coefficients and distinct roots between different factors, while inputs must avoid every denominator zero. No coefficient values are printed for these illustrative templates; Examples 12–15 calculate coefficients in full.

Solve coefficients without requiring cover-up knowledge

Clear denominators first, then use convenient inputs or compare like powers. Check the final decomposition by recombining its fractions.

The source permits substitution, comparing coefficients, or the optional direct cover-up rule if previously taught. Substitution and coefficient comparison suffice. First clear denominators to obtain a polynomial identity. It is legitimate to substitute a denominator zero into that identity; the original rational function is still undefined there.

Example 12 isolates each of three distinct linear-factor coefficients. Example 13 includes both powers of a repeated factor and uses a third convenient input for the remaining coefficient. Example 14 finds its simple linear-factor coefficient by cover-up, then compares x² and constant coefficients for the quadratic numerator. Its source callout reassures students that substitution is an equally valid method.

Direct cover-up isolates simple linear-factor coefficients and the highest power of a repeated linear factor; it does not directly give the other repeated-factor coefficients or all coefficients over a quadratic. These restrictions preserve the source remark without making cover-up compulsory.

Example 15 retains the complete division with a zero x² placeholder, then decomposes a positive remainder numerator before restoring the minus sign around the whole remainder. Continue with the existing, distinct Tutorial Question 12, which asks for three different decompositions.