P2 §1.4 Repeated factors

Pure mathematics Year 2

Table of contents

1. Partial fractions revisited
2. Partial fractions with repeated factors
3. Examples
4. Edexcel P2 Ch1 Exercise 1E

 

The main task of this section is to understand partial fractions with repeated factors. In order to do so, we need to consider and understand some general form of fractions.

1. Partial fractions revisited

In the previous section, P2 §1.3 Partial fractions, we studies partial fractions when there are two or three linear factors in the denominator.

1.1. Two linear factors

For example:

$$\begin{align} \frac{ A }{ x + 2 } + \frac{ B }{ x - 1 } &= \frac{ A (x - 1) + B (x + 2) }{ (x + 2) (x - 1) } \\ &= \frac{ (A+B)x + (-A+2B) }{ (x + 2) (x - 1) } \\ &= \frac{ \textrm{Linear} }{ \textrm{Quadratic} } \end{align}$$

1.2. Three linear factors

For example:

$$\begin{align} \frac{ A }{ x + 2 } + \frac{ B }{ x - 1 } + \frac{ C }{ x - 3 } &= \frac{ A (x - 1)(x - 3) + B (x + 2)(x - 3) + C (x + 2)(x - 1) }{ (x + 2) (x - 1) (x - 3) } \\ &= \frac{ A \left( x^2 - 4x + 3 \right) + B \left( x^2 - x - 6 \right) + C \left( x^2 + x - 2 \right) }{ (x + 2) (x - 1) (x - 3) } \\ &= \frac{ (A+B+C) x^2 + (-4A -B + C) x + (3A -6B -2C) }{ (x + 2) (x - 1) (x - 3) } \\ &= \frac{ \textrm{Quadratic} }{ \textrm{Cubic} } \end{align}$$

1.3. One quadratic factor and one linear factor

We may generalise this to the case where we have one quadratic factor and one linear factor.

 

A first attempt: $$\begin{align} \frac{ A }{ x^2 + x + 1 } + \frac{ B }{ x - 1 } &= \frac{ A (x - 1) + B \left( x^2 + x + 1 \right) }{ \left( x^2 + x + 1 \right) (x - 1) } \\ &= \frac{ B x^2 + (A + B) x + (B - A) }{ \left( x^2 + x + 1 \right) (x - 1) } \end{align}$$ However, there are three coefficients in the numerator expressed in terms of two variables $A$ and $B$. This could be problematic as we may not be able to find $A$ and $B$ that satisfy three conditions. (In the language of simultaneous equations, we need the same number of variables as the number of equations/conditions.)

 

Our second (and correct) attempt: $$\begin{align} \frac{ Ax + B }{ x^2 + x + 1 } + \frac{ C }{ x - 1 } &= \frac{ (Ax + B) (x - 1) + C \left( x^2 + x + 1 \right) }{ \left( x^2 + x + 1 \right) (x - 1) } \\ &= \frac{ ( A + B + C) x^2 + (-A + B + C) x + (- B + C) }{ \left( x^2 + x + 1 \right) (x - 1) } \end{align}$$ where the number of equations and that of variables are equal. The final result is of the form $$\frac{ Ax + B }{ \textrm{Quadratic} } + \frac{ C }{ \textrm{Linear} } = \frac{ \textrm{Quadratic} }{ \textrm{Quadratic} \times \textrm{Linear} } = \frac{ \textrm{Quadratic} }{ \textrm{Cubic} }$$

 

1.4. A general case

We may generalise this further to: $$\begin{align} \frac{ \textrm{Polynomial of order $n$} }{ \textrm{Polynomial of order $n+1$} } = \frac{ \textrm{Polynomial of order $p-1$} }{ \textrm{Polynomial of order $p$} } + \frac{ \textrm{Polynomial of order $q-1$} }{ \textrm{Polynomial of order $q$} } \end{align}$$ where $p+q=n+1$.

• The polynomial of order $n$ has $n+1$ coefficients.
• The polynomial of order $p-1$ has $p$ coefficients.
• The polynomial of order $q-1$ has $q$ coefficients.
• Thus, the number of coefficients on the LHS, $n+1$, matches the number of variables on the RHS, $p+q$.

An example: $$\begin{align} \frac{ 11x^4 - 7 x^3 + 5 x^2 + x - 3 }{ \left( x^3 + x^2 + x + 1 \right) \left( x^2 + x + 1 \right) } &= \frac{ A x^2 + B x + C }{ x^3 + x^2 + x + 1 } + \frac{ D x + E }{ x^2 + x + 1 } \end{align}$$

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2. Partial fractions with repeated factors

We will study partial fractions with repeated factors with an example.

 

Step 1. For partial fractions with repeated factors, it is essentially the case where we have a quadratic factor and a linear factor in the denominator. For example, $$\begin{align} \frac{ 2x + 9 }{ (x - 5)(x + 3)^2 } &= \frac{ 2x + 9 }{ (x - 5) \left( x^2 + 6x + 9 \right) } \\ &= \frac{A}{x-5} + \frac{Bx + C}{x^2 + 6x + 9} \\ &= \frac{ A (x + 3)^2 + (Bx + C) (x - 5) }{ (x - 5) \left( x^2 + 6x + 9 \right) } \end{align}$$

By the method of substitution, we find: $$\begin{align} \textrm{(i)}&& x &= -3 &&\Rightarrow& & -8(-3B + C) = 2(-3) + 9 = 3 \\ \\ \textrm{(ii)}&& x &= 5 &&\Rightarrow& & 64A = 2(5) + 9 = 19 \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& & 9A - 5C = 9 \end{align}$$ We see that:

• Unlike linear factors, even with the method of substitution, the coefficients appear together and we need to solve them as simultaneous equations.
• The numerator on the RHS suggests only two convenient choices for $x$. For the third one, we may choose any convenient number, usually $x=0,\pm 1$ and so on.

This gives

$$\begin{align} A &= \frac{19}{64} \\ \\ C &= \frac{9(A-1)}{5} = \frac95 \left( - \frac{45}{64} \right) = - \frac{81}{64} \\ \\ B &= - \frac13 \left( -\frac38 - C \right) = - \frac13 \left( -\frac{24}{64} + \frac{81}{64} \right) = - \frac{19}{64} \end{align}$$ and thus, for partial fractions, we obtain : $$\begin{align} \Rightarrow \qquad \frac{ 2x + 9 }{ (x - 5)(x + 3)^2 } &= \frac{19}{64(x-5)} - \frac{19x + 81}{ 64( x + 3 )^2 } \end{align}$$

 

Step 2. We can simplify the second term with the repeated factor as follows. $$\begin{align} \frac{19x + 81}{ 64( x + 3 )^2 } &= \frac{19(x+3-3)+ 81}{ 64( x + 3 )^2 } \\ &= \frac{19(x+3) + 24}{ 64( x + 3 )^2 } \\ &= \frac{19}{64(x+3)} + \frac{3}{8(x+3)^2} \end{align}$$ and thus the partial fractions now read: $$\begin{align} \frac{ 2x + 9 }{ (x - 5)(x + 3)^2 } &= \frac{19}{64(x-5)} - \frac{19}{64(x+3)} - \frac{3}{8(x+3)^2} \end{align}$$

 

Step 3. In general, for partial fractions with repeated factors, we find: $$\begin{align} \frac{ ax^2 + bx + c }{ (x - 5) (x + 3)^2 } = \frac{ A }{ x - 5 } + \frac{ B }{ (x + 3) } + \frac{ C }{ (x + 3)^2 } \end{align}$$

A generalisation

If more factors are repeated, we can go through the same steps and find a similar result. For example, $$\begin{align} \frac{ 2x^4 + 9x - 7 }{ (x - 5)^2 (x + 3)^3 } &= \frac{ Ax + b }{ (x - 5)^2 } + \frac{ Cx^2 + dx + e }{ (x + 3)^3 } \\ &= \frac{ A(x-5) + 5a + b }{ (x - 5)^2 } + \frac{ C(x+3)^2 + D(x+3) + E }{ (x + 3)^3 } \\ &= \frac{ A }{ x - 5 } + \frac{ B }{ (x - 5)^2 } + \frac{ C }{ x + 3 } + \frac{ D }{ x + 3)^2 } + \frac{ E }{ (x + 3)^3 } \end{align}$$

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3. Examples

Example. Show that $$\frac{ 11x^2 + 14x + 5 }{ (x+1)^2 (2x+1) }$$ can be written in the form $$\frac{ A }{ x + 1 } + \frac{ B }{ (x+1)^2 } + \frac{ C }{ 2x + 1 },$$ where $A,B$ and $C$ are constants to be found.

 

(Edexcel 2017 Specifications, P2 Ch1 Example 10.)

 

Solution.

Let $$\begin{align} \frac{ 11x^2 + 14x + 5 }{ (x+1)^2 (2x+1) } &\equiv \frac{ A }{ x + 1 } + \frac{ B }{ (x+1)^2 } + \frac{ C }{ 2x + 1 } \\ &\equiv \frac{ A (x+1)(2x+1) + B (2x+1) + C (x+1)^2 }{ (x+1)^2 (2x+1) } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& -B&=11(-1)^2 + 14(-1) + 5 = 11 - 14 + 5 = 2 \\ && & && & B&= -2 \\ \\ \textrm{(ii)}&& x &= -\frac12 &&\Rightarrow& \frac14 C&=11 \left( -\frac12 \right)^2 + 14 \left( -\frac12 \right) + 5 = \frac{11}{4} - 7 + 5 = \frac34 \\ && & && & C&= 3 \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& A+B+C&=5 \\ && & && & A&= 5-B-C=5+2-3=4 \end{align}$$ Hence, we find: $$A = 4, \qquad B = - 2, \qquad C = 3 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \frac{ 11x^2 + 14x + 5 }{ (x+1)^2 (2x+1) } \equiv \frac{ 4 }{ x + 1 } - \frac{ 2 }{ (x+1)^2 } + \frac{ 3 }{ 2x + 1 } \qquad \checkmark$$

 

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4. Edexcel P2 Ch1 Exercise 1E

Question 1. (E) $$\textrm f(x) = \frac{ 3x^2 + x + 1 }{ x^2 (x+1) }, \qquad x \ne 0, \quad x \ne -1.$$ Given that $\textrm f(x)$ can be expressed in the form $$\frac{ A }{ x } + \frac{ B }{ x^2 } + \frac{ C }{ x + 1 },$$ find the values of $A,B$ and $C$. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1E Q1.)

 

Solution.

Let $$\begin{align} \textrm f(x) = \frac{ 3x^2 + x + 1 }{ x^2 (x+1) } &\equiv \frac{ A }{ x } + \frac{ B }{ x^2 } + \frac{ C }{ x + 1 } \\ &\equiv \frac{ A x(x+1) + B (x+1) + C x^2 }{ x^2 (x+1) } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& B&=1 \\ \\ \textrm{(ii)}&& x &= -1 &&\Rightarrow& C&= 3 (-1)^2 + (-1) + 1 \\ && & && &&= 3 \\ \\ \textrm{(iii)}&& x &= 1 &&\Rightarrow& 2A+2B+C&=5 \\ && & && & 2A &= 5 - 2B - C \\ && & && &&= 5 - 2 - 3 \\ && & && &&= 0 \\ && & && & A &= 0 \end{align}$$ Hence, we find: $$A = 0, \qquad B = -1, \qquad C = - 3 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \textrm f(x) = \frac{ 3x^2 + x + 1 }{ x^2 (x+1) } \equiv \frac{ 1 }{ x^2 } + \frac{ 3 }{ x + 1 } \qquad \checkmark$$

 

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Question 2. (E) $$\textrm g(x) = \frac{ - x^2 - 10x - 5 }{ x^2 (x+1) }, \qquad x \ne -1, \quad x \ne 1.$$ Find the values of the constants $D,E$ and $F$ such that $$\textrm g(x) = \frac{ D }{ x+1 } + \frac{ E }{ (x+1)^2 } + \frac{ F }{ x - 1 }.$$ [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1E Q2.)

 

Solution.

Let $$\begin{align} \textrm g(x) = \frac{ - x^2 - 10x - 5 }{ x^2 (x+1) } &\equiv \frac{ D }{ x+1 } + \frac{ E }{ (x+1)^2 } + \frac{ F }{ x - 1 } \\ &\equiv \frac{ D (x+1)(x-1) + E (x-1) + F (x+1)^2 }{ x^2 (x+1) } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& -2E &= - (-1)^2 - 10(-1) - 5 \\ && & && &&= -1 + 10 - 5 \\ && & && &&= 4 \\ && & && & E &= -2 \\ \\ \textrm{(ii)}&& x &=1 &&\Rightarrow& 4F &= - 1 - 10 - 5 \\ && & && &&= -16 \\ && & && & F &= -4 \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& - D - E + F &= - 5 \\ && & && & - D &= - 5 + E - F \\ && & && &&= - 5 - 2 + 4 \\ && & && &&= -3 \\ && & && & D &= 3 \end{align}$$ Hence, we find: $$D = 3, \qquad E = -2, \qquad F = - 4 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \textrm g(x) = \frac{ - x^2 - 10x - 5 }{ x^2 (x+1) } \equiv \frac{ 3 }{ x+1 } - \frac{ 2 }{ (x+1)^2 } - \frac{ 4 }{ x - 1 } \qquad \checkmark$$

 

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Question 3. (E) Given that, for $x < 0$, $$\frac{ 2x^2 + 2x - 18 }{ x(x-3)^2 } \equiv \frac{P}{x} + \frac{Q}{x-3} + \frac{R}{(x-3)^2},$$ where $P,Q$ and $R$ are constants, find the values of $P,Q$ and $R$. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1E Q3.)

 

Solution.

We have $$\begin{align} \frac{ 2x^2 + 2x - 18 }{ x(x-3)^2 } &\equiv \frac{P}{x} + \frac{Q}{x-3} + \frac{R}{(x-3)^2} \\ &\equiv \frac{ P (x-3)^2 + Q x(x-3) + R x }{ x(x-3)^2 } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& 9P &= - 18 \\ && & && & P &= -2 \\ \\ \textrm{(ii)}&& x &= 3 &&\Rightarrow& 3R &= 2(3)^2 + 2(3) - 18 \\ && & && &&= 6 \\ && & && & R &= 2 \\ \\ \textrm{(iii)}&& x &= 1 &&\Rightarrow& 4P - 2Q + R &= 2 + 2 - 18 \\ && & && &&= -14 \\ && & && & -2Q &= -14 - 4P - R \\ && & && &&= -14 + 8 - 2 \\ && & && &&= -8 \\ && & && & Q &= 4 \end{align}$$ Hence, we find: $$P = -2, \qquad Q = 4, \qquad R = 2 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \frac{ 2x^2 + 2x - 18 }{ x(x-3)^2 } \equiv - \frac{2}{x} + \frac{4}{x-3} + \frac{2}{(x-3)^2} \qquad \checkmark$$

 

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Question 4. (E) Show that $$\frac{ 5x^2 - 2x - 1 }{ x^3 - x^2 }$$ can be written in the form $$\frac{C}{x} + \frac{D}{x^2} + \frac{E}{x-1},$$ where $C,D$ and $E$ are constants to be found. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1E Q4.)

 

Solution.

We factorise the denominator first. $$\begin{align} x^3 - x^2 = x^2 (x - 1) \end{align}$$ Let $$\begin{align} \frac{ 5x^2 - 2x - 1 }{ x^3 - x^2 } &\equiv \frac{ 5x^2 - 2x - 1 }{ x^2 (x-1) } \\ &\equiv \frac{ C }{ x } + \frac{ D }{ x^2 } + \frac{ E }{ x-1 } \\ &\equiv \frac{ C x(x-1) + D (x-1) + E x^2 }{ x^2 (x-1) } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& - D &= - 1 \\ && & && & D &= 1 \\ \\ \textrm{(ii)}&& x &= 1 &&\Rightarrow& E &= 5 - 2 - 1 \\ && & && &&= 2 \\ \\ \textrm{(iii)}&& x &= -1 &&\Rightarrow& 2C - 2D + E &= 5(-1)^2 - 2(-1) - 1 \\ && & && &&= 5 + 2 - 1 \\ && & && &&= 6 \\ && & && & 2C &= 6 + 2D - E = 6 + 2 - 2 = 6 \\ && & && & C &= 3 \end{align}$$ Hence, we find: $$C = 3, \qquad D = 1, \qquad E = 2 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \frac{ 5x^2 - 2x - 1 }{ x^3 - x^2 } \equiv \frac{ 5x^2 - 2x - 1 }{ x^2 (x-1) } \equiv \frac{ 3 }{ x } + \frac{ 1 }{ x^2 } + \frac{ 2 }{ x-1 } \qquad \checkmark$$

 

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Question 5. (E) $$\textrm p(x) = \frac{ 2x }{ (x+2)^2 }, \qquad x \ne -2.$$ Find the values of the constants $A$ and $B$ such that $$\textrm p(x) = \frac{A}{x+2} + \frac{B}{(x+2)^2}$$ [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1E Q5.)

 

Solution.

Let $$\begin{align} \textrm p(x) = \frac{ 2x }{ (x+2)^2 } &\equiv \frac{A}{x+2} + \frac{B}{(x+2)^2} \\ &\equiv \frac{ A(x+2) + B }{(x+2)^2} \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= -2 &&\Rightarrow& B &= - 4 \\ \\ \textrm{(ii)}&& x &= 0 &&\Rightarrow& 2A+B &= 0 \\ && & && & A &= - \frac{B}{2} = 2 \end{align}$$ [Note: Here, the method of equating coefficients is equally convenient as it gives $A=2$ and $2A+B=0$.]

 

Hence, we find: $$A = 2, \qquad B = -4 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \textrm p(x) = \frac{ 2x }{ (x+2)^2 } \equiv \frac{2}{x+2} - \frac{4}{(x+2)^2} \qquad \checkmark$$

 

Aside. An alternative way: $$\begin{align} \textrm p(x) &= \frac{ 2x }{ (x+2)^2 } \\ &= \frac{ 2(x+2-2) }{ (x+2)^2 } \\ &= \frac{ 2(x+2) - 4 }{ (x+2)^2 } \\ &= \frac{ 2 }{ x+2 } - \frac{ 4 }{ (x+2)^2 } \qquad \checkmark \end{align}$$

 

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Question 6. (E) $$\frac{ 10x^2 - 10x + 17 }{ (2x+1)(x-3)^2 } \equiv \frac{A}{2x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}, \qquad x > 3.$$ Find the values of the constants $A,B$ and $C$. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1E Q6.)

 

Solution.

We have $$\begin{align} \frac{ 10x^2 - 10x + 17 }{ (2x+1)(x-3)^2 } &\equiv \frac{A}{2x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2} \\ &\equiv \frac{ A (x-3)^2 + B (2x+1)(x-3) + C (2x+1) }{ (2x+1)(x-3)^2 } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= - \frac12 &&\Rightarrow& \left( - \frac12 - 3 \right)^2 A &= 10 \left( - \frac12 \right)^2 - 10 \left( - \frac12 \right) + 17 \\ && & && & \frac{49}{4} A &= \frac52 + 5 + 17 = \frac{49}{2} \\ && & && & A &= 2 \\ \\ \textrm{(ii)}&& x &= 3 &&\Rightarrow& 7 C &= 10(3)^2 - 10(3) + 17 \\ && & && &&= 90 - 30 + 17 \\ && & && &&= 77 \\ && & && & C &= 11 \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& 9A - 3B + C &= 17 \\ && & && & - 3B &= 17 - 9A - C \\ && & && &&= 17 - 18 - 11 \\ && & && &&= -12 \\ && & && & B &= 4 \end{align}$$ Hence, we find: $$A = 2, \qquad B = 4, \qquad C = 11 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \frac{ 10x^2 - 10x + 17 }{ (2x+1)(x-3)^2 } \equiv \frac{2}{2x+1} + \frac{4}{x-3} + \frac{11}{(x-3)^2} \qquad \checkmark$$

 

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Question 7. (E) Show that $$\frac{ 39x^2 + 2x + 59 }{ (x+5)(3x-1)^2 }$$ can be written in the form $$\frac{A}{x+5} + \frac{B}{3x-1} + \frac{C}{ (3x-1)^2 },$$ where $A,B$ and $C$ are constants to be found. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1E Q7.)

 

Solution.

Let $$\begin{align} \frac{ 39x^2 + 2x + 59 }{ (x+5)(3x-1)^2 } &\equiv \frac{A}{x+5} + \frac{B}{3x-1} + \frac{C}{ (3x-1)^2 } \\ &\equiv \frac{ A (3x-1)^2 + B (x+5)(3x-1) + C (x+5) }{ (x+5)(3x-1)^2 } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= - 5 &&\Rightarrow& 16^2 A &= 39(-5)^2 + 2(-5) + 59 \\ && & && &&= 39 \times 25 + 49 \\ && & && &&= 1024 \\ && & && & 2^8 A &= 2^{10} \\ && & && & A &= 2^2 = 4 \\ \\ \textrm{(ii)}&& x &= \frac13 &&\Rightarrow& \left( \frac13 + 5 \right) C &= 39 \left( \frac13 \right)^2 + 2 \left( \frac13 \right) + 59 \\ && & && & \frac{16}{3} C &= \frac{13}{3} + \frac{2}{3} + 59 \\ && & && &&= 64 \\ && & && & C &= 12 \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& A - 5B + 5C &= 59 \\ && & && & - 5B &= 59 - A - 5C \\ && & && &&= 59 - 4 - 60 \\ && & && &&= -5 \\ && & && & B &= 1 \end{align}$$ Hence, we find: $$A = 4, \qquad B = 1, \qquad C = 12 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \frac{ 39x^2 + 2x + 59 }{ (x+5)(3x-1)^2 } \equiv \frac{4}{x+5} + \frac{1}{3x-1} + \frac{12}{ (3x-1)^2 } \qquad \checkmark$$

 

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Question 8. (P) Express the following as partial fractions: $$\begin{align} &\textbf{(a)} \qquad \frac{ 4x + 1 }{ x^2 + 10x + 25 } \\ \\ &\textbf{(b)} \qquad \frac{ 6x^2 - x + 2 }{ 4x^3 - 4x^2 + x } \end{align}$$

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1E Q8.)

 

Solution.

(a) We factorise the denominator first. $$x^2 + 10x + 25 = (x + 5)^2$$ Let $$\begin{align} \frac{ 4x + 1 }{ x^2 + 10x + 25 } &= \frac{ 4x + 1 }{ (x+5)^2 } \\ &\equiv \frac{A}{x+5} + \frac{B}{(x+5)^2} \\ &\equiv \frac{ A(x+5) + B }{ (x+5)^2 } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= - 5 &&\Rightarrow& B &= -19 \\ \\ \textrm{(ii)}&& x &= 0 &&\Rightarrow& 5A + B &= 1 \\ && & && & 5A &= 1 - B = 20 \\ && & && & A &= 4 \end{align}$$ [Note: Here, the method of equating coefficients is equally convenient as it gives $A=4$ and $5A+B=1$.]

 

Hence, we find: $$A = 4, \qquad B = -19 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \frac{ 4x + 1 }{ x^2 + 10x + 25 } = \frac{ 4x + 1 }{ (x+5)^2 } \equiv \frac{4}{x+5} - \frac{19}{(x+5)^2} \qquad \checkmark$$

 

Aside. An alternative way: $$\begin{align} \frac{ 4x + 1 }{ x^2 + 10x + 25 } &= \frac{ 4x + 1 }{ (x+5)^2 } \\ &= \frac{ 4(x+5-5) + 1 }{ (x+5)^2 } \\ &= \frac{ 4(x+5) - 19 }{ (x+5)^2 } \\ &= \frac{ 4 }{ x+5 } - \frac{ 19 }{ (x+5)^2 } \qquad \checkmark \end{align}$$

 

(b) We factorise the denominator first. $$\begin{align} 4x^3 - 4x^2 + x &= x \left( 4x^2 - 4x + 1 \right) \\ &= x (2x - 1)^2 \end{align}$$ Let $$\begin{align} \frac{ 6x^2 - x + 2 }{ 4x^3 - 4x^2 + x } &= \frac{ 6x^2 - x + 2 }{ x (2x - 1)^2 } \\ &\equiv \frac{A}{x} + \frac{B}{ 2x-1 } + \frac{C}{ (2x-1)^2 } \\ &\equiv \frac{ A (2x-1)^2 + B x(2x-1) + C x }{ x (2x - 1)^2 } \end{align}$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& A &= 2 \\ \\ \textrm{(ii)}&& x &= \frac12 &&\Rightarrow& \frac{C}{2} &= 6 \left( \frac12 \right)^2 - \left( \frac12 \right) + 2 \\ && & && &&= \frac32 - \frac12 + 2 \\ && & && &&= 3 \\ && & && & C &= 6 \\ \\ \textrm{(ii)}&& x &= 1 &&\Rightarrow& A + B + C &= 6 - 1 + 2 = 7 \\ && & && & B &= 7 - A - C \\ && & && &&= 7 - 2 - 6 \\ && & && &&= -1 \end{align}$$ Hence, we find: $$A = 2, \qquad B = -1, \qquad C = 6 \qquad \checkmark$$ and the partial fractions read: $$\Rightarrow \qquad \frac{ 6x^2 - x + 2 }{ 4x^3 - 4x^2 + x } = \frac{ 6x^2 - x + 2 }{ x (2x - 1)^2 } \equiv \frac{2}{x} - \frac{1}{ 2x-1 } + \frac{6}{ (2x-1)^2 } \qquad \checkmark$$