Q.Resolve $\dfrac{2x^2 + 3x + 4}{(x-1)(x^2+2)}$ into partial fraction.

Since $1-x+x^2=x^2-x+1$ is an irreducible quadratic (negative discriminant), use a linear numerator over it and a constant over the linear factor $(x+2)$.

Write $1-x+x^2 = x^2-x+1$; its discriminant is $(-1)^2-4(1)(1)=-3<0$, so it doesn't factor over the reals — it needs a linear numerator $Ax+B$.

Assume

$\frac{3x-1}{(x^2-x+1)(x+2)} = \frac{Ax+B}{x^2-x+1} + \frac{C}{x+2}$

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

$3x-1 = (Ax+B)(x+2) + C(x^2-x+1) \quad (\star)$

Find $C$: put $x=-2$ (which kills the first term):

$3(-2)-1 = C\big((-2)^2-(-2)+1\big) \Rightarrow -7 = C(4+2+1)=7C \Rightarrow C=-1$

Find $A,B$: expand $(\star)$:

$(Ax+B)(x+2) = Ax^2+(2A+B)x+2B$

$C(x^2-x+1) = -x^2+x-1 \quad(\text{since } C=-1)$

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Since $x^2+2$ has no real linear factors, write the partial fraction as $\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+2}$ and match coefficients.

Let $\dfrac{2x^2+3x+4}{(x-1)(x^2+2)} = \dfrac{A}{x-1} + \dfrac{Bx+C}{x^2+2}$.

Multiplying both sides by $(x-1)(x^2+2)$:

$2x^2+3x+4 = A(x^2+2) + (Bx+C)(x-1)$.

Substitute $x=1$: $2+3+4 = A(1+2) \Rightarrow 9 = 3A \Rightarrow A=3$.

Expand the right side: $A x^2 + 2A + Bx^2 - Bx + Cx - C = (A+B)x^2 + (C-B)x + (2A-C)$.

Matching coefficients:

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