
Resolve the fraction $\dfrac{{2{x^2} + 3x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}}$ into partial fractions.
Answer
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Hint-In partial fraction problem first we see denominator part in fraction and according to denominator we factor the fraction.
Complete Step-by-Step solution:
Here, In denominator $\left( {{x^2} + 2} \right)$ cannot be factored so, numerator of $\left( {{x^2} + 2} \right)$ will be $\left( {Bx + C} \right)$ here $B$and$C$ are constant and $\left( {x - 1} \right)$ is in factored form so numerator will be constant$A$.
$\dfrac{{2{x^2} + 3x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}} = \dfrac{A}{{x - 1}} + \dfrac{{Bx + C}}{{{x^2} + 2}}$ …… (1)
Now cross multiply,
$2{x^2} + 3x + 4 = A\left( {{x^2} + 2} \right) + \left( {Bx + C} \right)\left( {x - 1} \right)$
In cross multiply both side denominator parts will be cancel,
Now, we simply solve right hand side and make quadratic form like left hand side
$2{x^2} + 3x + 4 = A{x^2} + 2A + B{x^2} - Bx + Cx - C$
$2{x^2} + 3x + 4 = \left( {A + B} \right){x^2} + \left( {C - B} \right)x + 2A - C$
Comparing the coefficients,
$A + B = 2$ ……. (2)
$C - B = 3$ …….. (3)
$2A - C = 4$……. (4)
Adding the three equations 2,3&4 we get
$3A = 9$
Value of $A = 3$
Value of $A$ put in equation 4 to get value of $C$
$2 \times 3 - C = 4$
Value of $C = 2$
Value of $C$ put in equation 3 to get value of $B$
$2 - B = 3$
Value of $B = - 1$
Now, Value of$A$,$B$,$C$ put in equation 1
$\dfrac{{2{x^2} + 3x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}} = \dfrac{3}{{x - 1}} + \dfrac{{ - x + 2}}{{{x^2} + 2}}$
Now final partial fraction is $\dfrac{{2{x^2} + 3x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}} = \dfrac{3}{{x - 1}} - \dfrac{{x - 2}}{{{x^2} + 2}}$
Note- In partial fraction problem,
Step 1- Factor the denominator
Step 2- Write one partial fraction for each of those factors according to denominator factor
Step 3- Cross multiply
Step 4- Compare coefficient in both side and evaluate constant value
Step 5- Put the value of constants in partial fraction which get in step 2
Complete Step-by-Step solution:
Here, In denominator $\left( {{x^2} + 2} \right)$ cannot be factored so, numerator of $\left( {{x^2} + 2} \right)$ will be $\left( {Bx + C} \right)$ here $B$and$C$ are constant and $\left( {x - 1} \right)$ is in factored form so numerator will be constant$A$.
$\dfrac{{2{x^2} + 3x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}} = \dfrac{A}{{x - 1}} + \dfrac{{Bx + C}}{{{x^2} + 2}}$ …… (1)
Now cross multiply,
$2{x^2} + 3x + 4 = A\left( {{x^2} + 2} \right) + \left( {Bx + C} \right)\left( {x - 1} \right)$
In cross multiply both side denominator parts will be cancel,
Now, we simply solve right hand side and make quadratic form like left hand side
$2{x^2} + 3x + 4 = A{x^2} + 2A + B{x^2} - Bx + Cx - C$
$2{x^2} + 3x + 4 = \left( {A + B} \right){x^2} + \left( {C - B} \right)x + 2A - C$
Comparing the coefficients,
$A + B = 2$ ……. (2)
$C - B = 3$ …….. (3)
$2A - C = 4$……. (4)
Adding the three equations 2,3&4 we get
$3A = 9$
Value of $A = 3$
Value of $A$ put in equation 4 to get value of $C$
$2 \times 3 - C = 4$
Value of $C = 2$
Value of $C$ put in equation 3 to get value of $B$
$2 - B = 3$
Value of $B = - 1$
Now, Value of$A$,$B$,$C$ put in equation 1
$\dfrac{{2{x^2} + 3x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}} = \dfrac{3}{{x - 1}} + \dfrac{{ - x + 2}}{{{x^2} + 2}}$
Now final partial fraction is $\dfrac{{2{x^2} + 3x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}} = \dfrac{3}{{x - 1}} - \dfrac{{x - 2}}{{{x^2} + 2}}$
Note- In partial fraction problem,
Step 1- Factor the denominator
Step 2- Write one partial fraction for each of those factors according to denominator factor
Step 3- Cross multiply
Step 4- Compare coefficient in both side and evaluate constant value
Step 5- Put the value of constants in partial fraction which get in step 2
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