Find partial fraction of (2x+3)/(x²+3x+2)

Question

Resolve $\dfrac{2x+3}{x^2+3x+2}$ into partial fractions.

Solution — Step by Step

We need to factor $x^2 + 3x + 2$ .

Looking for two numbers that multiply to 2 and add to 3: those are 1 and 2.

x^2 + 3x + 2 = (x+1)(x+2)

These are distinct linear factors, so we use the standard partial fraction form.

Since the denominator has two distinct linear factors, write:

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

where $A$ and $B$ are constants to be determined.

Multiply both sides by $(x+1)(x+2)$ :

2x + 3 = A(x+2) + B(x+1)

This equation must hold for all values of $x$ .

Let $x = -1$ : (makes the $B$ term vanish)

2(-1) + 3 = A(-1+2) + B(0)

1 = A(1) \Rightarrow \boxed{A = 1}

Let $x = -2$ : (makes the $A$ term vanish)

2(-2) + 3 = A(0) + B(-2+1)

-1 = B(-1) \Rightarrow \boxed{B = 1}

\frac{2x+3}{x^2+3x+2} = \frac{1}{x+1} + \frac{1}{x+2}

Verification: $\frac{1}{x+1} + \frac{1}{x+2} = \frac{(x+2) + (x+1)}{(x+1)(x+2)} = \frac{2x+3}{x^2+3x+2}$ ✓

Why This Works

Partial fractions works because every rational function with distinct linear factors in the denominator can be expressed as a sum of simpler fractions. This is a fundamental theorem in algebra.

The “cover-up” method works because at $x = -1$ , the factor $(x+1)$ in the denominator cancels with the $A(x+2)$ expression, leaving a direct equation for $A$ . It’s just substituting a root of each linear factor.

This technique is heavily used in integration: $\int \frac{2x+3}{x^2+3x+2}dx = \ln|x+1| + \ln|x+2| + C$ .

Alternative Method — Equating Coefficients

Instead of substituting specific values, expand and compare:

2x + 3 = A(x+2) + B(x+1) = (A+B)x + (2A+B)

Comparing coefficients:

Coefficient of $x$ : $A + B = 2$
Constant term: $2A + B = 3$

Subtracting: $A = 1$ . Then $B = 1$ .

Same answer. The cover-up method is faster; equating coefficients is more methodical for repeated roots.

Common Mistake

When the degree of the numerator equals or exceeds the degree of the denominator, you must perform polynomial long division FIRST. For example, $\frac{x^2 + 5x + 6}{x^2 + 3x + 2}$ cannot be directly decomposed — divide first to get $1 + \frac{2x+4}{x^2+3x+2}$ , then decompose the fractional part. In our problem, the numerator (degree 1) is less than the denominator (degree 2), so we can proceed directly.

Question

Solution — Step by Step

Why This Works

Alternative Method — Equating Coefficients

Common Mistake

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