Resolve 2x3 - 3x2 - 8x - 26-2x2 - 5x-12 into partial fractions-

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Integration by Partial Fractions

Resolve 2x3...

Question

Resolve $\frac{2 x^{3} - 3 x^{2} - 8 x - 26}{2 x^{2} - 5 x - 12}$ into partial fractions.

x + 1 + \frac{3}{2 x + 3} + \frac{2}{x - 4}

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x - 1 + \frac{5}{2 x + 3} + \frac{2}{x - 4}

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x + \frac{5}{2 x + 3} + \frac{2}{x - 4}

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x + 1 + \frac{5}{2 x + 3} + \frac{2}{x - 4}

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Solution

The correct option is C $x + 1 + \frac{5}{2 x + 3} + \frac{2}{x - 4}$
Here given fraction is an improper fraction then by actual division i.e.,
$2 x^{2} - 5 x - 12 ⇃ 2 x^{3} - 3 x - 8 x - 26 ⇂ x + 1$
$2 x^{3} - 5 x^{2} - 12 x$
$- + +$

$2 x^{2} + 4 x - 26$
$2 x^{2} - 5 x - 12$
$- + +$

$9 x - 14$
$∴ \frac{2 x^{3} - 3 x^{2} - 8 x - 26}{2 x^{2} - 5 x - 12} = x + 1 + \frac{9 x - 14}{2 x^{2} - 5 x - 12}$
$= x + 1 + \frac{(9 x - 14)}{(2 x + 3) (x - 4)}$
$∴ \frac{2 x^{3} - 3 x^{2} - 8 x - 26}{2 x^{2} - 5 x - 12} = x + 1 + \frac{9 x - 14}{(2 x + 3) (x - 4)}$ ....(i)
Let $\frac{9 x - 14}{(2 x + 3) (x - 4)} = \frac{A}{2 x + 3} + \frac{B}{x - 4}$ ....(ii)
$⟹ 9 x - 14 = A (x - 4) + B (2 x + 3)$ ....(iii)
Putting $x - 4 = 0$ or $x = 4$ in (iii), we get
$36 - 14 = 0 + B (8 + 3)$
$∴ B = 2$
Putting $2 x + 3 = 0$ or $x = - 3 / 2$ in (iii), we get
$9 \times - \frac{3}{2} - 14 = A (- \frac{3}{2} - 4) + 0$
$⟹ - \frac{55}{2} = - \frac{11}{2} A$
$∴ A = 5$
Substituting the value A and B in (ii), then
$\frac{9 x - 14}{(2 x + 3) (x - 4)} = \frac{5}{2 x + 3} + \frac{2}{x - 4}$ ....(iv)
From (i) & (iv) we get the required partial fractions
$\frac{2 x^{3} - 3 x^{2} - 8 x - 26}{2 x^{2} - 5 x - 12} = x + 1 + \frac{5}{2 x + 3} + \frac{2}{x - 4}$
Hence, option 'D' is correct.

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Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

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Resolve 2x3−3x2−8x−262x2−5x−12 into partial fractions.

Resolve $\frac{2 x^{3} - 3 x^{2} - 8 x - 26}{2 x^{2} - 5 x - 12}$ into partial fractions.