AHL 1.11—Partial fractions

The numerator $4x^2-3x+7$ has degree 2. The denominator $(x-2)(3x+1)=3x^2-5x-2$ also has degree 2.

For partial fraction decomposition, the degree of the numerator must be strictly less than the degree of the denominator. Since deg(numerator) = deg(denominator) = 2, this condition is not satisfied.

When the degrees are equal (or the numerator degree is greater), polynomial long division must be performed first. This yields a polynomial quotient plus a proper remainder, and only the remainder is then decomposed.

The correct first step is to perform polynomial long division, giving $f(x)=\frac{4}{3}+\frac{R(x)}{(x-2)(3x+1)}$ where $R(x)$ has degree less than 2.

The question asks about the prerequisite check before decomposition — specifically whether the degree condition is met.

"Write $f(x)=\frac{A}{x-2}+\frac{B}{3x+1}$ directly" is incorrect because the degree condition (numerator degree < denominator degree) must be checked first; here degrees are equal so this form cannot be applied immediately.

"Factor the numerator" is not a required step for partial fractions — it is the denominator that must be factored, and it is already factored. Factoring the numerator does not help here.

"Multiply numerator and denominator by 3" is not a standard procedure for partial fractions. It would change the structure without resolving the degree issue.

The only valid first step is polynomial long division, since deg(numerator) = deg(denominator) = 2, violating the strict inequality requirement.

Numerator degree: 3. Denominator: $(x+2)(x^2-1)=(x+2)(x-1)(x+1)$, which is degree 3. Since degrees are equal, long division is needed.

Dividing $3x^3-2x^2+x-4$ by the degree-3 denominator $(x+2)(x-1)(x+1)=x^3+2x^2-x-2$ gives quotient 3 and a remainder of lower degree.

$x^2-1=(x-1)(x+1)$, so the denominator has three distinct linear factors: $(x+2)$, $(x-1)$, $(x+1)$. Each gives one partial fraction term.

After long division: $f(x)=3+\frac{R(x)}{(x+2)(x-1)(x+1)}$, then decompose the remainder as $\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{x+1}$. The full form is $f(x)=3+\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{x+1}$.

We need to check: (1) whether long division is required, and (2) whether $x^2-1$ factors into linear or irreducible quadratic parts.

"$\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{x+1}$" omits the polynomial quotient from long division. Since the numerator and denominator have equal degrees, a constant quotient (3) must appear in the result.

"$\frac{A}{x+2}+\frac{Bx+C}{x^2-1}$" treats $x^2-1$ as irreducible, but $x^2-1=(x-1)(x+1)$ is reducible. The numerator $Bx+C$ form is reserved for genuinely irreducible quadratics.

"$f(x)=3x+\dots$" would only arise if the quotient from long division were $3x$, which happens when the numerator degree exceeds the denominator degree by 1. Here both degrees are 3, so the quotient is a constant, not $3x$.

Option B correctly includes the constant quotient 3 from long division, and the three separate partial fraction terms for the three distinct linear factors.