Question

Resolve into partial fractions: \[\dfrac{{26{x^2} + 208x}}{{\left( {{x^2} + 1} \right)\left( {x + 5} \right)}}\]
A) \[\dfrac{{41x + 3}}{{{x^2} + 1}} - \dfrac{{15}}{{x + 5}}\]
B) \[\dfrac{{4x + 3}}{{{x^2} - 1}} - \dfrac{5}{{x + 5}}\]
C) \[\dfrac{{ - 41x + 3}}{{{x^2} + 1}} + \dfrac{{15}}{{x + 5}}\]
D) \[\dfrac{{x - 3}}{{{x^2} - 1}} - \dfrac{5}{{x - 5}}\]

Answer 1

Hint: Here, we will Partial fraction decomposition method step by step as shown below:
For example, we have the expression \[\dfrac{{5x - 4}}{{{x^2} - x - 2}}\] which needs to resolve into partial fractions:
1) Decomposition of the denominator part:
\[ \Rightarrow \dfrac{{5x - 4}}{{\left( {x - 2} \right)\left( {x + 1} \right)}}\]
2) Writing one partial fraction for each of those factors:
\[ \Rightarrow \dfrac{{5x - 4}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} = \dfrac{{{A_1}}}{{x - 2}} + \dfrac{{{A_2}}}{{x + 1}}\]
3) Multiply through by the denominator part so that we have no longer fractions:
\[ \Rightarrow 5x - 4 = {A_1}\left( {x + 1} \right) + {A_2}\left( {x - 2} \right)\]
4) Now we will find the constants \[{A_1}\] and \[{A_2}\]by substituting the value of roots of \[\left( {x - 2} \right)\] and \[\left( {x + 1} \right)\]:
Roots of \[x\] from \[\left( {x + 1} \right)\] equal to \[x = - 1\], so we get:
\[ \Rightarrow 5\left( { - 1} \right) - 4 = {A_1}\left( {\left( { - 1} \right) + 1} \right) + {A_2}\left( {\left( { - 1} \right) - 2} \right)\]
By simplifying the brackets, we get:
\[ \Rightarrow - 9 = 0 + {A_2}\left( { - 3} \right)\]
Now, the value \[{A_2}\] will be:
\[ \Rightarrow {A_2} = 3\]
5) Similarly, for finding the value of constant \[{A_1}\] , we will substitute the root value of \[x\] from \[\left( {x - 2} \right)\] which is \[2\] and we get :
\[ \Rightarrow {A_1} = 2\]

Complete step by step answer:
Step 1: For decomposing the expression, we will do factorization of the denominator and by writing it in the form of partial fraction decomposition, we get:
\[\dfrac{{26{x^2} + 208x}}{{\left( {{x^2} + 1} \right)\left( {x + 5} \right)}} = \dfrac{{{\text{A}}x + {\text{B}}}}{{{x^2} + 1}} + \dfrac{{\text{C}}}{{x + 5}}\]
Step 2: Now, we will do multiplication through by the denominator part so that we have no longer fractions:
\[ \Rightarrow 26{x^2} + 208x = \left( {{\text{A}}x + {\text{B}}} \right)\left( {x + 5} \right) + {\text{C}}\left( {{x^2} + 1} \right)\] ……….. (1)
In the RHS side of the above equation (1), by multiplying inside the brackets, we get:
\[ \Rightarrow 26{x^2} + 208x = A{x^2} + 5{\text{A}}x + {\text{B}}x + 5{\text{B}} + {\text{C}}{x^2} + {\text{C}}\]
By adding the coefficients of constants, we get:
\[ \Rightarrow 26{x^2} + 208x = \left( {{\text{A}} + {\text{C}}} \right){x^2} + x\left( {5{\text{A}} + {\text{B}}} \right) + \left( {5{\text{B}} + {\text{C}}} \right)\] ……… (2)
Step 3: Now, for finding the value of constants, we will put some random values of
\[x\] , i.e. \[0\], \[1\] and \[ - 1\]:

(i) When \[x = 0\]substituting this value in the above equation (2):
\[ \Rightarrow 26{\left( 0 \right)^2} + 208\left( 0 \right) = \left( {{\text{A}} + {\text{C}}} \right){\left( 0 \right)^2} + 0\left( {5{\text{A}} + {\text{B}}} \right) + \left( {5{\text{B}} + {\text{C}}} \right)\] …. …………… (3)
By doing multiplication inside the brackets in the above equation (3) we get:
\[ \Rightarrow 5{\text{B}} + {\text{C = 0}}\] …………………….. (4)

(ii) When \[x = 1\]substituting this value in the above equation (2):
\[ \Rightarrow 26{\left( 1 \right)^2} + 208\left( 1 \right) = \left( {{\text{A}} + {\text{C}}} \right){\left( 1 \right)^2} + 1 \times \left( {5{\text{A}} + {\text{B}}} \right) + \left( {5{\text{B}} + {\text{C}}} \right)\]
By multiplying inside the brackets above, we get:
\[ \Rightarrow 234 = 6{\text{A + 6B + 2C}}\] …………………………… (5)

(iii) When \[x = - 1\]substituting this value in the above equation (2):
\[ \Rightarrow 26{\left( { - 1} \right)^2} + 208\left( { - 1} \right) = \left( {{\text{A}} + {\text{C}}} \right){\left( { - 1} \right)^2} + \left( { - 1} \right) \times \left( {5{\text{A}} + {\text{B}}} \right) + \left( {5{\text{B}} + {\text{C}}} \right)\]
By multiplying inside the brackets above, we get:
\[ \Rightarrow - 182 = 2{\text{C - 4A + 4B}}\] …………………………… (6)

Step 4: By comparing the equations (4), (5), and (6), we get:
First of all, from equation (4), we will find the value of constant \[{\text{C}}\]:
\[ \Rightarrow {\text{C = - 5B}}\]
Putting the value \[{\text{C = - 5B}}\] in the equation (5) and (6), we get:
\[ \Rightarrow 234 = 6{\text{A + 6B + 2}}\left( { - 5{\text{B}}} \right)\] and \[ \Rightarrow - 182 = 2\left( { - 5{\text{B}}} \right){\text{ - 4A + 4B}}\]
By simplifying the above equations, we get:
\[ \Rightarrow 6{\text{A - 4B = 234}}\] and \[ \Rightarrow 4{\text{A + 6B = 182}}\]
By comparing the equations \[6{\text{A - 4B = 234}}\] and \[4{\text{A + 6B = 182}}\], we will find the value of constants \[{\text{A}}\] and \[{\text{B}}\]:
First of all, we will multiply the equation \[6{\text{A - 4B = 234}}\] with \[4\] and equation \[4{\text{A + 6B = 182}}\] with \[6\] for making the coefficients of constant \[{\text{A}}\] and subtracting them:
\[
  6{\text{A - 4B = 234}} \\
  4{\text{A + 6B = 182}} \\
  \overline
  24{\text{A}} - 16{\text{B}} = 936 \\
  24{\text{A}} + 36{\text{B}} = 1092 \\
  \overline {0 - 52{\text{B}} = - 156} \\
    \\
 \]
Now, by bringing \[52\] into the RHS side, and dividing it with \[156\] we get:
 \[ \Rightarrow {\text{B}} = \dfrac{{156}}{{52}}\]
\[ \Rightarrow {\text{B}} = 3\]
Now, by putting the value of \[{\text{B}} = 3\] in the equation \[ \Rightarrow 5{\text{B}} + {\text{C = 0}}\], we get the value of \[{\text{C}}\]:
\[ \Rightarrow 5\left( 3 \right) + {\text{C = 0}}\]
By simplifying the brackets, we get:
\[ \Rightarrow 15 + {\text{C = 0}}\]
Bringing \[15\] into the RHS side, we get:
\[ \Rightarrow {\text{C}} = - 15\]
Similarly, for finding the value of \[{\text{A}}\], we will put the values \[{\text{C}} = - 15\] and \[{\text{B}} = 3\] in the equation \[234 = 6{\text{A + 6B + 2C}}\] we get:
\[ \Rightarrow 234 = 6{\text{A + 6}}\left( 3 \right){\text{ + 2}}\left( { - 15} \right)\]
By simplifying the brackets, we get:
\[ \Rightarrow 234 = 6{\text{A - 12}}\]
By taking \[12\] into the LHS side and dividing the with the coefficient of \[{\text{A}}\], we get:
\[ \Rightarrow {\text{A}} = 41\]
Step 4: So, by substituting the values of constants in the equation \[\dfrac{{26{x^2} + 208x}}{{\left( {{x^2} + 1} \right)\left( {x + 5} \right)}} = \dfrac{{{\text{A}}x + {\text{B}}}}{{{x^2} + 1}} + \dfrac{{\text{C}}}{{x + 5}}\] , we get:
\[ \Rightarrow \dfrac{{26{x^2} + 208x}}{{\left( {{x^2} + 1} \right)\left( {x + 5} \right)}} = \dfrac{{41x + 3}}{{{x^2} + 1}} - \dfrac{{15}}{{x + 5}}\]

Hence, option (A) is correct.

Note:
In these types of questions students need to remember that if the denominator part is not in factorized form then first change it into factor form.
Also, there are four different simplest denominator types:
- Linear factors
- Irreducible factors of degree two.
- Repeated linear factors
- Repeated irreducible factors of degree two.