Resolve into partial fractions. $\dfrac{2x-1}{x^2-5x+6}$

Resolve into partial fractions. (2x-1)/(x²-5x+6)

Question 48 of 48

\dfrac{2x-1}{x^2-5x+6}

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2025Subjective· 3mImportance★★★★★

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Factor $x^2-5x+6=(x-2)(x-3)$ , set $\dfrac{2x-1}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}$ ; $A=-3$ , $B=5$ .

Step 1 — factorise the denominator.

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

Step 2 — set up the partial fractions.

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

Multiplying through:

$2x-1=A(x-3)+B(x-2).$

Step 3 — find $A$ and $B$ by substitution.

Put $x=2$ : $\;2(2)-1=A(2-3)\Rightarrow 3=-A\Rightarrow A=-3.$ …