If (x³ − 6x² + 10x − 2)/(x² − 5x + 6) = f(x) + A/(x−2) + B/(x−3), Then f(x) Is

A standard partial fractions problem, not tied to a specific exam paper.

Problem

If $\dfrac{x^3-6x^2+10x-2}{x^2-5x+6} = f(x)+\dfrac{A}{x-2}+\dfrac{B}{x-3}$, then $f(x)$ is?

Key insight. Since the numerator’s degree (3) exceeds the denominator’s degree (2), the fraction as given isn’t “proper” — the standard partial-fraction terms $A/(x-2)$ and $B/(x-3)$ can only ever produce a proper fraction on their own. That means $f(x)$ must be exactly the polynomial quotient obtained by dividing numerator by denominator, with the leftover proper fraction matching the $A/(x-2)+B/(x-3)$ part.

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Approach

Perform polynomial long division of $x^3-6x^2+10x-2$ by $x^2-5x+6$. The quotient from this division is $f(x)$, and the remainder (divided by the original denominator) is the proper fraction that decomposes into $A/(x-2)+B/(x-3)$.

Solution

Step 1 — Set up the long division

Divide $x^3-6x^2+10x-2$ by $x^2-5x+6$.

Step 2 — First term of the quotient

$$\frac{x^3}{x^2} = x$$

Multiply: $x(x^2-5x+6) = x^3-5x^2+6x$

Subtract: $(x^3-6x^2+10x-2)-(x^3-5x^2+6x) = -x^2+4x-2$

Step 3 — Second term of the quotient

$$\frac{-x^2}{x^2} = -1$$

Multiply: $-1(x^2-5x+6) = -x^2+5x-6$

Subtract: $(-x^2+4x-2)-(-x^2+5x-6) = -x+4$

Step 4 — Stop the division

The remainder $-x+4$ has degree $1$, less than the divisor’s degree $2$ — the division is complete.

Step 5 — Write the division result

$$x^3-6x^2+10x-2 = (x^2-5x+6)(x-1) + (-x+4)$$

$$\frac{x^3-6x^2+10x-2}{x^2-5x+6} = (x-1) + \frac{4-x}{(x-2)(x-3)}$$

Step 6 — Identify f(x)

Comparing with the given form $f(x)+\dfrac{A}{x-2}+\dfrac{B}{x-3}$, the polynomial part is:

$$f(x) = x-1$$

Answer

Common mistakes

• Trying to set up the partial fraction decomposition before performing the division. Standard partial-fraction techniques (comparing coefficients, or substituting specific $x$ values) only work directly on a proper fraction — attempting them on the original improper fraction leads to inconsistent equations, since $A/(x-2)+B/(x-3)$ alone can never account for the numerator’s higher degree.
• Stopping the division one step early. After the first division step, the remainder $-x^2+4x-2$ still has the same degree as the divisor ($2$) — the division isn’t finished until the remainder’s degree drops strictly below the divisor’s, which requires the second step.

Practise next

• If $\dfrac{x^3-4x^2+5x-1}{x^2-3x+2} = g(x)+\dfrac{C}{x-1}+\dfrac{D}{x-2}$, find $g(x)$, using the same long-division approach.