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Resolve into partial fractions.
\[\dfrac{46+13x}{12{{x}^{2}}-11x-15}\]
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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.
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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