Resolve the fraction $\dfrac{{2{x^2} + 3x + 4}}{{\left( {x - 1} \right)\left( {{x^2} + 2} \right)}}$ into partial fractions.
Answer
639k+ views
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
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

In cricket, what is the term for a bowler taking five wickets in an innings?

Who Won 36 Oscar Awards? Record Holder Revealed

What is the name of Japan Parliament?

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

