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Question: Resolve into partial fractions: \[\dfrac{{26{x^2} + 208x}}{{\left( {{x^2} + 1} \right)\left( {x + 5}...

Resolve into partial fractions: 26x2+208x(x2+1)(x+5)\dfrac{{26{x^2} + 208x}}{{\left( {{x^2} + 1} \right)\left( {x + 5} \right)}}
A) 41x+3x2+1−15x+5\dfrac{{41x + 3}}{{{x^2} + 1}} - \dfrac{{15}}{{x + 5}}
B) 4x+3x2−1−5x+5\dfrac{{4x + 3}}{{{x^2} - 1}} - \dfrac{5}{{x + 5}}
C) −41x+3x2+1+15x+5\dfrac{{ - 41x + 3}}{{{x^2} + 1}} + \dfrac{{15}}{{x + 5}}
D) x−3x2−1−5x−5\dfrac{{x - 3}}{{{x^2} - 1}} - \dfrac{5}{{x - 5}}

Explanation

Solution

Here, we will Partial fraction decomposition method step by step as shown below:
For example, we have the expression 5x−4x2−x−2\dfrac{{5x - 4}}{{{x^2} - x - 2}} which needs to resolve into partial fractions:

  1. Decomposition of the denominator part:
    ⇒5x−4(x−2)(x+1)\Rightarrow \dfrac{{5x - 4}}{{\left( {x - 2} \right)\left( {x + 1} \right)}}
  2. Writing one partial fraction for each of those factors:
    ⇒5x−4(x−2)(x+1)=A1x−2+A2x+1\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:
    ⇒5x−4=A1(x+1)+A2(x−2)\Rightarrow 5x - 4 = {A_1}\left( {x + 1} \right) + {A_2}\left( {x - 2} \right)
  4. Now we will find the constants A1{A_1} and A2{A_2}by substituting the value of roots of (x−2)\left( {x - 2} \right) and (x+1)\left( {x + 1} \right):
    Roots of xx from (x+1)\left( {x + 1} \right) equal to x=−1x = - 1, so we get:
    ⇒5(−1)−4=A1((−1)+1)+A2((−1)−2)\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:
    ⇒−9=0+A2(−3)\Rightarrow - 9 = 0 + {A_2}\left( { - 3} \right)
    Now, the value A2{A_2} will be:
    ⇒A2=3\Rightarrow {A_2} = 3
  5. Similarly, for finding the value of constant A1{A_1} , we will substitute the root value of xx from (x−2)\left( {x - 2} \right) which is 22 and we get :
    ⇒A1=2\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:
26x2+208x(x2+1)(x+5)=Ax+Bx2+1+Cx+5\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:
⇒26x2+208x=(Ax+B)(x+5)+C(x2+1)\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:
⇒26x2+208x=Ax2+5Ax+Bx+5B+Cx2+C\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:
⇒26x2+208x=(A+C)x2+x(5A+B)+(5B+C)\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
xx , i.e. 00, 11 and −1 - 1:

(i) When x=0x = 0substituting this value in the above equation (2):
⇒26(0)2+208(0)=(A+C)(0)2+0(5A+B)+(5B+C)\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:
⇒5B+C = 0\Rightarrow 5{\text{B}} + {\text{C = 0}} …………………….. (4)

(ii) When x=1x = 1substituting this value in the above equation (2):
⇒26(1)2+208(1)=(A+C)(1)2+1×(5A+B)+(5B+C)\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:
⇒234=6A + 6B + 2C\Rightarrow 234 = 6{\text{A + 6B + 2C}} …………………………… (5)

(iii) When x=−1x = - 1substituting this value in the above equation (2):
⇒26(−1)2+208(−1)=(A+C)(−1)2+(−1)×(5A+B)+(5B+C)\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:
⇒−182=2C - 4A + 4B\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 C{\text{C}}:
⇒C = - 5B\Rightarrow {\text{C = - 5B}}
Putting the value C = - 5B{\text{C = - 5B}} in the equation (5) and (6), we get:
⇒234=6A + 6B + 2(−5B)\Rightarrow 234 = 6{\text{A + 6B + 2}}\left( { - 5{\text{B}}} \right) and ⇒−182=2(−5B) - 4A + 4B \Rightarrow - 182 = 2\left( { - 5{\text{B}}} \right){\text{ - 4A + 4B}}
By simplifying the above equations, we get:
⇒6A - 4B = 234\Rightarrow 6{\text{A - 4B = 234}} and ⇒4A + 6B = 182 \Rightarrow 4{\text{A + 6B = 182}}
By comparing the equations 6A - 4B = 2346{\text{A - 4B = 234}} and 4A + 6B = 1824{\text{A + 6B = 182}}, we will find the value of constants A{\text{A}} and B{\text{B}}:
First of all, we will multiply the equation 6A - 4B = 2346{\text{A - 4B = 234}} with 44 and equation 4A + 6B = 1824{\text{A + 6B = 182}} with 66 for making the coefficients of constant A{\text{A}} and subtracting them:

6A - 4B = 234 4A + 6B = 182 2‾4A−16B=936 24A+36B=1092 0−52B=−156‾   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 5252 into the RHS side, and dividing it with 156156 we get:
⇒B=15652\Rightarrow {\text{B}} = \dfrac{{156}}{{52}}
⇒B=3\Rightarrow {\text{B}} = 3
Now, by putting the value of B=3{\text{B}} = 3 in the equation ⇒5B+C = 0 \Rightarrow 5{\text{B}} + {\text{C = 0}}, we get the value of C{\text{C}}:
⇒5(3)+C = 0\Rightarrow 5\left( 3 \right) + {\text{C = 0}}
By simplifying the brackets, we get:
⇒15+C = 0\Rightarrow 15 + {\text{C = 0}}
Bringing 1515 into the RHS side, we get:
⇒C=−15\Rightarrow {\text{C}} = - 15
Similarly, for finding the value of A{\text{A}}, we will put the values C=−15{\text{C}} = - 15 and B=3{\text{B}} = 3 in the equation 234=6A + 6B + 2C234 = 6{\text{A + 6B + 2C}} we get:
⇒234=6A + 6(3) + 2(−15)\Rightarrow 234 = 6{\text{A + 6}}\left( 3 \right){\text{ + 2}}\left( { - 15} \right)
By simplifying the brackets, we get:
⇒234=6A - 12\Rightarrow 234 = 6{\text{A - 12}}
By taking 1212 into the LHS side and dividing the with the coefficient of A{\text{A}}, we get:
⇒A=41\Rightarrow {\text{A}} = 41
Step 4: So, by substituting the values of constants in the equation 26x2+208x(x2+1)(x+5)=Ax+Bx2+1+Cx+5\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:
⇒26x2+208x(x2+1)(x+5)=41x+3x2+1−15x+5\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.