Question

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

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.