Resolve into partial fractions. \[\dfrac{46+13x}{12{{x}^{2}}-11x-15}\]
Answer
Verified
Hint: First of all factorise the denominator of the given expression that is \[12{{x}^{2}}-11x-15\] in its linear factors. Then write the numerator that is 46 + 13x in terms of these linear factors of the denominator.
Here, we have to resolve the following expression into partial fractions. Let us consider the given expression as \[D=\dfrac{46+13x}{12{{x}^{2}}-11x-15}\] Let us consider, \[A=46+13x\] And \[B=12{{x}^{2}}-11x-15\] Therefore, \[D=\dfrac{A}{B}....\left( i \right)\] First of all, we will factorize B. In B, we can write \[11x=20x-9x\] Therefore, we get \[B=12{{x}^{2}}-\left( 20x-9x \right)-15\] \[\Rightarrow B=12{{x}^{2}}-20x+9x-15\] Or, \[B=\left( 12{{x}^{2}}-20x \right)+\left( 9x-15 \right)\] By taking 4x common from \[\left( 12{{x}^{2}}-20x \right)\] and 3 common from \[\left( 9x-15 \right)\], we get \[\Rightarrow B=4x\left( 3x-5 \right)+3\left( 3x-5 \right)\] By taking \[\left( 3x-5 \right)\] common, we get, \[B=\left( 3x-5 \right).\left( 4x+3 \right)\] By putting the values of A and B in equation (i), we get \[D=\dfrac{A}{B}=\dfrac{46+13x}{\left( 3x-5 \right).\left( 4x+3 \right)}\] Now, to resolve D into partial fractions, let us consider, \[D=\dfrac{P}{\left( 3x-5 \right)}+\dfrac{Q}{\left( 4x+3 \right)}.....\left( ii \right)\] Or, \[D=\dfrac{46+13x}{\left( 3x-5 \right).\left( 4x+3 \right)}=\dfrac{P}{\left( 3x-5 \right)}+\dfrac{Q}{\left( 4x+3 \right)}\] Now, we will solve for P and Q to find the partial fractions. By simplifying RHS, we get \[\dfrac{46+13x}{\left( 3x-5 \right).\left( 4x+3 \right)}=\dfrac{P\left( 4x+3 \right)+Q\left( 3x-5 \right)}{\left( 3x-5 \right)\left( 4x+3 \right)}\] By cancelling the like terms from both sides, we get \[46+13x=P\left( 4x+3 \right)+Q\left( 3x-5 \right)\] By simplifying the above equation, we get \[46+13x=4Px+3P+3Qx-5Q\] We can write it as, \[46+13x=\left( 4P+3Q \right)x+\left( 3P-5Q \right)\] By equating the coefficient of x and constant terms from LHS and RHS, we get \[4P+3Q=13....\left( iii \right)\] And, \[3P-5Q=46....\left( iv \right)\] Now, we will multiply equation (iii) by 3 and equation (iv) by 4. We get, \[12P+9Q=39....\left( v \right)\] \[12P-20Q=184....\left( vi \right)\] By subtracting equation (v) from equation (vi), we get, \[\left( 12P-20Q \right)-\left( 12P+9Q \right)=184-39\] \[\Rightarrow -20Q-9Q=145\] \[\Rightarrow -29Q=145\] Or, \[Q=\dfrac{145}{-29}\] Therefore, we get \[Q=-5\] By putting the values of Q in equation (iii), We get, \[4P+3\left( -5 \right)=13\] \[\Rightarrow 4P-15=13\] \[\Rightarrow 4P=13+15\] \[\Rightarrow 4P=28\] \[\Rightarrow P=\dfrac{28}{4}\] Therefore, we get P = 7. By putting the values of P and Q in equation (ii). We get, \[D=\dfrac{7}{\left( 3x-5 \right)}+\dfrac{\left( -5 \right)}{\left( 4x+3 \right)}\] Therefore, \[\dfrac{46+13x}{12{{x}^{2}}-11x-15}=\dfrac{7}{\left( 3x-5 \right)}-\dfrac{5}{\left( 4x+3 \right)}\] has been resolved into partial fraction.
Note: Whenever the degree of denominator is greater than numerator, always first try to factorize the denominator in these types of questions. Then solve for A and B by solving \[\dfrac{mx+n}{\left( x-a \right)\left( x-b \right)}=\dfrac{A}{\left( x-a \right)}+\dfrac{B}{\left( x-b \right)}\] like in the above solution. Finally, students should always cross check if the partial fractions are giving the same expression as given in the question by solving it.
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