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

Resolve into partial fractions: (2x²+3x+4)/((x-1)(x²+2)).

Question 6 of 8

\dfrac{2x^2+3x+4}{(x-1)(x^2+2)}

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For an irreducible quadratic factor like $x^2+2$ , the numerator over it must be a general LINEAR expression $Bx+C$ (not a plain constant), since a constant alone can't match every possible remainder.

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

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

At $x=1$ : $9=A(3)\Rightarrow A=3$

Comparing $x^2$ coefficients: $2=A+B\Rightarrow B=-1$ …

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Q.Resolve into partial fractions: 2x2+3x+4(x−1)(x2+2)\dfrac{2x^2+3x+4}{(x-1)(x^2+2)}(x−1)(x2+2)2x2+3x+4​.

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

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