P2 §1.3 Partial fractions

Pure mathematics Year 2

Table of contents

1. Partial fractions
2. Examples
3. Edexcel P2 Ch1 Exercise 1D

1. Partial fractions

In the previous section, P2 §1.2 Algebraic fractions, we learnt how to add and subtract two (or more) fractions, e.g. $$ \frac{2}{x+3} - \frac{1}{x+1} = \frac{x-1}{(x+3)(x+1)} $$ Now, we can perform the reverse. In other words, a single fraction with two distinct linear factors in the denominator can be split into two separate fractions with linear denominators. $$ \frac{x-1}{(x+3)(x+1)} = \frac{A}{x+3} + \frac{B}{x+1} $$ This reverse process is called 'splitting it into partial fractions'. Note:

• The left-hand side has two linear factors in the denominator.
• The right-hand side is written as the sum of two partial fractions.
• The constants $A$ and $B$ are to be found. There are two methods to find them: by substitution and by equating coefficients.

Note: Partial fractions are used for binomial expansions in P2 §4.3 Using partial fractions, and in integration in P2 §11.7 Partial fractions.

반응형

2. Examples

Example 1. Split $$ \frac{ 6x - 2 }{ (x-3)(x+1) } $$ into partial fractions by

(a) substitution

(a) equating coefficients.

 

(Edexcel 2017 Specifications, P2 Ch1 Example 8.)

 

Solution.

Let $$ \begin{align} \frac{ 6x - 2 }{ (x-3)(x+1) } &= \frac{ A }{ x - 3 } + \frac{ B }{ x + 1 } \\ &= \frac{ A (x+1) + B (x-3) }{ (x-3)(x+1) } \end{align} $$ Compare the numerators on both sides: $$ A(x+1) + B(x-3) = 6x - 2 $$ (a) By substitution: $$ \begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& -4B &= -8 \\ &&&&&\Rightarrow& B&=2 \\ \\ \textrm{(ii)}&& x &= 3 &&\Rightarrow& 4A &= 16 \\ &&&&&\Rightarrow& A&=4 \end{align} $$ and we find: $$ \begin{align} \Rightarrow \qquad \frac{ 6x - 2 }{ (x-3)(x+1) } &= \frac{ 4 }{ x - 3 } + \frac{ 2 }{ x + 1 } \qquad \checkmark \end{align} $$

 

(b) By equating coefficients: $$ A(x+1) + B(x-3) = (A+B)x + (A-3B) = 6x - 2 $$ gives $$ \begin{align} \left\{\begin{array}{lr} A+B = 6 & (1)\\ A-3B = -2 & (2) \end{array}\right. \end{align} $$ (1)-(2) gives $4B = 8$, i.e. $B=2$, and we also find $A=4$. Thus, we obtain: $$ \begin{align} \Rightarrow \qquad \frac{ 6x - 2 }{ (x-3)(x+1) } &= \frac{ 4 }{ x - 3 } + \frac{ 2 }{ x + 1 } \qquad \checkmark \end{align} $$

 

반응형

Example 2. Given that $$ \frac{ 6x^2 + 5x - 2 }{ x(x-1)(2x+1) } \equiv \frac{A}{x} + \frac{B}{x-1} + \frac{C}{2x+1}, $$ find the values of the constants $A,B$ and $C$.

 

(Edexcel 2017 Specifications, P2 Ch1 Example 9.)

 

Solution.

Use the common denominator for the right-hand side: $$ \frac{ 6x^2 + 5x - 2 }{ x(x-1)(2x+1) } \equiv \frac{ A(x-1)(2x+1) + Bx(2x+1) + Cx(x-1) }{ x(x-1)(2x+1) } $$ Compare the numerators on both sides: $$ A(x-1)(2x+1) + Bx(2x+1) + Cx(x-1) \equiv 6x^2 + 5x - 2. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& -A&=-2 \\ &&&&&\Rightarrow& A&=2 \\ \\ \textrm{(ii)}&& x &= 1 &&\Rightarrow& 3B &= 9 \\ &&&&&\Rightarrow& B&=3 \\ \\ \textrm{(iii)}&& x &= -\frac12 &&\Rightarrow& \frac34C = -3 \\ &&&&&\Rightarrow& C&=-4 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ 6x^2 + 5x - 2 }{ x(x-1)(2x+1) } \equiv \frac{2}{x} + \frac{3}{x-1} - \frac{4}{2x+1} \qquad \checkmark $$

 

반응형

3. Edexcel P2 Ch1 Exercise 1D

Question 1. Express the following as partial fractions: $$ \begin{align} &\textbf{(a)} \qquad \frac{ 6x - 2 }{ (x - 2)(x + 3) } \\ \\ &\textbf{(b)} \qquad \frac{ 2x + 11 }{ (x + 1)(x + 4) } \\ \\ &\textbf{(c)} \qquad \frac{ -7x - 12 }{ 2x(x - 4) } \\ \\ &\textbf{(d)} \qquad \frac{ 2x - 13 }{ (2x + 1)(x - 3) } \\ \\ &\textbf{(e)} \qquad \frac{ 6x + 6 }{ (x^2 - 9 } \\ \\ &\textbf{(f)} \qquad \frac{ 7 - 3x }{ x^2 - 3x - 4 } \\ \\ &\textbf{(g)} \qquad \frac{ 8 - x }{ x^2 + 4x } \\ \\ &\textbf{(h)} \qquad \frac{ 2x - 14 }{ x^2 + 2x - 15 } \end{align} $$ [Hint: Factorise the denominator first.]

 

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

 

Solution.

(a) Let $$ \begin{align} \frac{ 6x - 2 }{ (x - 2)(x + 3) } &\equiv \frac{ A }{ x-2 } + \frac{ B }{ x+3 } \\ &= \frac{ A(x+3) + B(x-2) }{ (x - 2)(x + 3) } \end{align} $$ Compare the numerators on both sides: $$ A(x+3) + B(x-2) \equiv 6x - 2. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 2 &&\Rightarrow& 5A&=10 \\ &&&&&\Rightarrow& A&=2 \\ \\ \textrm{(ii)}&& x &= -3 &&\Rightarrow& -5B&= -20 \\ &&&&&\Rightarrow& B&=4 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ 6x - 2 }{ (x - 2)(x + 3) } \equiv \frac{ 2 }{ x-2 } + \frac{ 4 }{ x+3 } \qquad \checkmark $$

 

(b) Let $$ \begin{align} \frac{ 2x + 11 }{ (x + 1)(x + 4) } &\equiv \frac{ A }{ x+1 } + \frac{ B }{ x+4 } \\ &= \frac{ A(x+4) + B(x+1) }{ (x + 1)(x + 4) } \end{align} $$ Compare the numerators on both sides: $$ A(x+4) + B(x+1) \equiv 2x + 11. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& 3A&=9 \\ &&&&&\Rightarrow& A&=3 \\ \\ \textrm{(ii)}&& x &= -4 &&\Rightarrow& -3B&= 3 \\ &&&&&\Rightarrow& B&=-1 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ 2x + 11 }{ (x + 1)(x + 4) } \equiv \frac{ 3 }{ x+1 } - \frac{ 1 }{ x+4 } \qquad \checkmark $$

 

(c) Let $$ \begin{align} \frac{ -7x - 12 }{ 2x(x-4) } &\equiv \frac{ A }{ 2x } + \frac{ B }{ x-4 } \\ &= \frac{ A(x-4) + 2Bx }{ 2x(x-4) } \end{align} $$ Compare the numerators on both sides: $$ A(x-4) + 2Bx \equiv -7x - 12. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& -4A&=-12 \\ &&&&&\Rightarrow& A&=3 \\ \\ \textrm{(ii)}&& x &= 4 &&\Rightarrow& 8B&=-40 \\ &&&&&\Rightarrow& B&=-5 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ -7x - 12 }{ 2x(x-4) } \equiv \frac{ 3 }{ 2x } - \frac{ 5 }{ x-4 } \qquad \checkmark $$

 

(d) Let $$ \begin{align} \frac{ 2x - 13 }{ (2x+1)(x-3) } &\equiv \frac{ A }{ 2x+1 } + \frac{ B }{ x-3 } \\ &= \frac{ A(x-3) + B(2x+1) }{ (2x+1)(x-3) } \end{align} $$ Compare the numerators on both sides: $$ A(x-3) + B(2x+1) \equiv 2x - 13. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -\frac12 &&\Rightarrow& -\frac{7}{2}A&=-14 \\ &&&&&\Rightarrow& A&=4 \\ \\ \textrm{(ii)}&& x &= 3 &&\Rightarrow& 7B&=-7 \\ &&&&&\Rightarrow& B&=-1 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ 2x - 13 }{ (2x+1)(x-3) } \equiv \frac{ 4 }{ 2x+1 } - \frac{ 1 }{ x-3 } \qquad \checkmark $$

 

(e) Let $$ \begin{align} \frac{ 6x + 6 }{ x^2 - 9 } &\equiv \frac{ 6x+6 }{ (x+3)(x-3) } \\ &\equiv \frac{ A }{ x+3 } + \frac{ B }{ x-3 } \\ &= \frac{ A(x-3) + B(x+3) }{ (x+3)(x-3) } \end{align} $$ Compare the numerators on both sides: $$ A(x-3) + B(x+3) \equiv 6x + 6. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -3 &&\Rightarrow& -6A&=-12 \\ &&&&&\Rightarrow& A&=2 \\ \\ \textrm{(ii)}&& x &= 3 &&\Rightarrow& 6B&=24 \\ &&&&&\Rightarrow& B&=4 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ 6x + 6 }{ x^2 - 9 } \equiv \frac{ 6x + 6 }{ (x+3)(x-3) } \equiv \frac{ 2 }{ x+3 } + \frac{ 4 }{ x-3 } \qquad \checkmark $$

 

(f) Let $$ \begin{align} \frac{ 7 - 3x }{ x^2 - 3x - 4 } &\equiv \frac{ 7 - 3x }{ (x-4)(x+1) } \\ &\equiv \frac{ A }{ x-4 } + \frac{ B }{ x+1 } \\ &= \frac{ A(x+1) + B(x-4) }{ (x+3)(x-3) } \end{align} $$ Compare the numerators on both sides: $$ A(x+1) + B(x-4) \equiv 7 - 3x. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 4 &&\Rightarrow& 5A&=-5 \\ &&&&&\Rightarrow& A&=-1 \\ \\ \textrm{(ii)}&& x &= -1 &&\Rightarrow& -5B&=10 \\ &&&&&\Rightarrow& B&=-2 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ 7 - 3x }{ x^2 - 3x - 4 } \equiv \frac{ 7 - 3x }{ (x-4)(x+1) } \equiv -\frac{ 1 }{ x-4 } - \frac{ 2 }{ x+1 } \qquad \checkmark $$

 

(g) Let $$ \begin{align} \frac{ 8 - x }{ x^2 + 4x } &\equiv \frac{ 8 - x }{ x(x+4) } \\ &\equiv \frac{ A }{ x } + \frac{ B }{ x+4 } \\ &= \frac{ A(x+4) + Bx }{ x(x+4) } \end{align} $$ Compare the numerators on both sides: $$ A(x+4) + Bx \equiv 8 - x. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& 4A&=8 \\ &&&&&\Rightarrow& A&=2 \\ \\ \textrm{(ii)}&& x &= -4 &&\Rightarrow& -4B&=12 \\ &&&&&\Rightarrow& B&=-3 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ 8 - x }{ x^2 + 4x } \equiv \frac{ 8 - x }{ x(x+4) } \equiv \frac{ 2 }{ x } - \frac{ 3 }{ x+4 } \qquad \checkmark $$

 

(h) Let $$ \begin{align} \frac{ 2x - 14 }{ x^2 + 2x - 15 } &\equiv \frac{ 2x - 14 }{ (x+5)(x-3) } \\ &\equiv \frac{ A }{ x+5 } + \frac{ B }{ x-3 } \\ &= \frac{ A(x-3) + B(x+5) }{ (x+5)(x-3) } \end{align} $$ Compare the numerators on both sides: $$ A(x-3) + B(x+5) \equiv 2x - 14. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -5 &&\Rightarrow& -8A&=-24 \\ &&&&&\Rightarrow& A&=3 \\ \\ \textrm{(ii)}&& x &= 3 &&\Rightarrow& 8B&=-8 \\ &&&&&\Rightarrow& B&=-1 \end{align} $$ and we obtain: $$ \Rightarrow \qquad \frac{ 2x - 14 }{ x^2 + 2x - 15 } \equiv \frac{ 2x - 14 }{ (x+5)(x-3) } \equiv \frac{ 3 }{ x+5 } - \frac{ 1 }{ x-3 } \qquad \checkmark $$

 

 

 

 

 

 

 

 

 

반응형

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

 

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

 

Solution.

We have $$ \begin{align} \frac{ -2x - 5 }{ (4 + x)(2 - x) } &\equiv \frac{A}{4+x} + \frac{B}{2-x} \\ &= \frac{ A(2-x) + B(4+x) }{ (4+x)(2-x) } \end{align} $$ Compare the numerators on both sides: $$ A(2-x) + B(4+x) \equiv -2x - 5. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -4 &&\Rightarrow& 6A&=3 \\ &&&&&\Rightarrow& A&=\frac12 \\ \\ \textrm{(ii)}&& x &= 2 &&\Rightarrow& 6B&=-9 \\ &&&&&\Rightarrow& B&=-\frac32 \end{align} $$ Hence, we find: $$ A = \frac12, \qquad B = - \frac32 \qquad \checkmark $$ Aside. Partial fractions read: $$ \Rightarrow \qquad \frac{ -2x - 5 }{ (4 + x)(2 - x) } \equiv \frac{1}{2(4+x)} - \frac{3}{2(2-x)} \qquad \checkmark $$

 

반응형

Question 3. (P) The expression $$ \frac{A}{ (x-4)(x+8) } $$ can be written in partial fractions as $$ \frac{2}{x-4} + \frac{B}{x+8}. $$ Find the values of the constants $A$ and $B$.

 

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

 

Solution.

Let $$ \begin{align} \frac{A}{ (x-4)(x+8) } &\equiv \frac{2}{x-4} + \frac{B}{x+8} \\ &= \frac{ 2(x+8) + B(x-4) }{ (x-4)(x+8) } \end{align} $$ Compare the numerators on both sides: $$ A \equiv 2(x+8) + B(x-4). $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 4 &&\Rightarrow& A&=2(4+8)=24 \\ \\ \textrm{(ii)}&& x &= -8 &&\Rightarrow& -12B&=24 \\ &&&&&\Rightarrow& B&=-2 \end{align} $$ Hence, we find: $$ A = 24, \qquad B = - 2 \qquad \checkmark $$ Aside. Partial fractions read: $$ \Rightarrow \qquad \frac{24}{ (x-4)(x+8) } \equiv \frac{2}{x-4} - \frac{2}{x+8} \qquad \checkmark $$

 

반응형

Question 4. (E) $$ \textrm h(x) = \frac{ 2x^2 - 12x - 26 }{ (x+1) (x-2) (x+5) }, \qquad x > 2 $$ Given that $ \textrm h(x) $ can be expressed in the form $$ \frac{ A }{ x + 1 } + \frac{ B }{ x - 2 } + \frac{ C }{ x + 5 }, $$ find the values of $A,B$ and $C$. [4 marks]

 

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

 

Solution.

It is given that: $$ \begin{align} \textrm h(x) = \frac{ 2x^2 - 12x - 26 }{ (x+1) (x-2) (x+5) } &= \frac{ A }{ x + 1 } + \frac{ B }{ x - 2 } + \frac{ C }{ x + 5 } \\ &= \frac{ A (x-2)(x+5) + B (x+1)(x+5) + C (x+1)(x-2) }{ (x+1) (x-2) (x+5) } \end{align} $$ Compare the numerators on both sides: $$ A (x-2)(x+5) + B (x+1)(x+5) + C (x+1)(x-2) \equiv 2x^2 - 12x - 26. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& (-3)(4)A&=2(-1)^2 - 12(-1) - 26 \\ &&&&&\Rightarrow& - 12 A &= -12 \\ &&&&&\Rightarrow& A &= 1 \\ \\ \textrm{(ii)}&& x &= 2 &&\Rightarrow& (3)(7)B&=2(2)^2-12(2)-26 \\ &&&&&\Rightarrow& 21 B &= -42 \\ &&&&&\Rightarrow& B &= -2 \\ \\ \textrm{(iii)}&& x &= -5 &&\Rightarrow& (-4)(-7)C&=2(-5)^2 - 12(-5) - 26 \\ &&&&&\Rightarrow& 28 C &= 84 \\ &&&&&\Rightarrow& C &= 3 \end{align} $$ Hence, we find: $$ A = 1, \qquad B = - 2, \qquad C = 3 \qquad \checkmark $$

 

Aside 1. Partial fractions read: $$ \Rightarrow \qquad \textrm h(x) = \frac{ 2x^2 - 12x - 26 }{ (x+1) (x-2) (x+5) } \equiv \frac{ 1 }{ x + 1 } - \frac{ 2 }{ x - 2 } + \frac{ 3 }{ x + 5 } \qquad \checkmark $$

 

Aside 2. One may wonder whether it is valid to use $x=-1,-5$ since it is given that $x>2$. This condition is really there to avoid singularities, i.e. the largest domain would be $$ \{ x \in \mathbb R \; | \; x \ne -1, \; x \ne 2, \; x \ne -5 \}. $$ So the given domain, $x>2$, provides a safe region. Also, the method of substitution works because we only compare the numerators on both sides. If still in doubt, one could try to find $A,B$ and $C$ by equating coefficients as follows. $$ \begin{align} \textrm h(x) = \frac{ 2x^2 - 12x - 26 }{ (x+1) (x-2) (x+5) } &= \frac{ A }{ x + 1 } + \frac{ B }{ x - 2 } + \frac{ C }{ x + 5 } \\ &= \frac{ A (x-2)(x+5) + B (x+1)(x+5) + C (x+1)(x-2) }{ (x+1) (x-2) (x+5) } \\ &= \frac{ A \left( x^2 + 3x - 10 \right) + B \left( x^2 + 6x + 5 \right) + C \left( x^2 - x - 2 \right) }{ (x+1) (x-2) (x+5) } \\ &= \frac{ (A+B+C) x^2 + (3A+6B-C)x + (-10A + 5B - 2C) }{ (x+1) (x-2) (x+5) } \end{align} $$ This yields a set of simultaneous equations: $$ \begin{align} \left\{\begin{array}{l} A+B+C=2 \\ 3A+6B-C=-12 \\ -10A + 5B - 2C=-26 \end{array}\right. \qquad \Rightarrow \qquad \left\{\begin{array}{l} A=1 \\ B=-2 \\ C=3 \end{array}\right. \end{align} $$ so we find the same result.

 

반응형

Question 5. (E) Given that, for $ x < -1 $, $$ \frac{ -10x^2 - 8x + 2 }{ x (2x + 1) (3x - 2) } \equiv \frac{ D }{ x } + \frac{ E }{ 2x + 1 } + \frac{ F }{ 3x - 2 }, $$ where $D,E$ and $F$ are constants. Find the values of $D,E$ and $F$. [4 marks]

 

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

 

Solution.

It is given that: $$ \begin{align} \frac{ -10x^2 - 8x + 2 }{ x (2x + 1) (3x - 2) } &\equiv \frac{ D }{ x } + \frac{ E }{ 2x + 1 } + \frac{ F }{ 3x - 2 } \\ &\equiv \frac{ D (2x + 1) (3x - 2) + E x (3x - 2) + F x (2x + 1) }{ x (2x + 1) (3x - 2) } \end{align} $$ Compare the numerators on both sides: $$ D (2x + 1) (3x - 2) + E x (3x - 2) + F x (2x + 1) \equiv -10x^2 - 8x + 2. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& (1)(-2)D&=2 \\ &&&&&\Rightarrow& D &= -1 \\ \\ \textrm{(ii)}&& x &= -\frac12 &&\Rightarrow& \left( -\frac12 \right) \left( -\frac32 - 2 \right) E&=-10\left( -\frac12 \right)^2 - 8\left( -\frac12 \right) + 2 \\ &&&&&\Rightarrow& \frac74 E &= -\frac{5}{2}+4+2=\frac{7}{2} \\ &&&&&\Rightarrow& E &= 2 \\ \\ \textrm{(iii)}&& x &= \frac23 &&\Rightarrow& \left( \frac23 \right) \left( \frac43 +1 \right) F&=-10\left( \frac23 \right)^2 - 8\left( \frac23 \right) + 2 \\ &&&&&\Rightarrow& \frac{14}{9} F &= - \frac{40}{9} - \frac{16}{3} + 2 = \frac{-40-48+18}{9} = -\frac{70}{9} \\ &&&&&\Rightarrow& F &= -5 \end{align} $$ Hence, we find: $$ D = -1, \qquad E = 2, \qquad F = -5 \qquad \checkmark $$

 

Aside 1. Partial fractions read: $$ \Rightarrow \qquad \frac{ -10x^2 - 8x + 2 }{ x (2x + 1) (3x - 2) } \equiv -\frac{ 1 }{ x } + \frac{ 2 }{ 2x + 1 } - \frac{ 5 }{ 3x - 2 } \qquad \checkmark $$

 

Aside 2. One may wonder whether it is valid to use $x=0,-\frac12,\frac23$ since it is given that $x<-1$. This condition is really there to avoid singularities, i.e. the largest domain would be $$ \left\{ x \in \mathbb R \; \bigg| \; x \ne 0, \; x \ne -\frac12, \; x \ne \frac23 \right\}. $$ So the given domain, $x>2$, provides a safe region. Also, the method of substitution works because we only compare the numerators on both sides. If still in doubt, one could try to find $D,E$ and $F$ by equating coefficients as follows. $$ \begin{align} \frac{ -10x^2 - 8x + 2 }{ x (2x + 1) (3x - 2) } &\equiv \frac{ D }{ x } + \frac{ E }{ 2x + 1 } + \frac{ F }{ 3x - 2 } \\ &\equiv \frac{ D (2x + 1) (3x - 2) + E x (3x - 2) + F x (2x + 1) }{ x (2x + 1) (3x - 2) } \\ &\equiv \frac{ D \left( 6x^2 - x - 2 \right) + E \left( 3x^2 - 2x \right) + F \left( 2x^2 + x \right) }{ (x+1) (x-2) (x+5) } \\ &\equiv \frac{ (6D+3E+2F) x^2 + (-D-2E+F)x -2D }{ (x+1) (x-2) (x+5) } \end{align} $$ This yields a set of simultaneous equations: $$ \begin{align} \left\{\begin{array}{l} 6D + 3E + 2F = -10 \\ -D - 2E + F = -8 \\ -2D = 2 \end{array}\right. \qquad \Rightarrow \qquad \left\{\begin{array}{l} D = -1 \\ E = 2 \\ F = -5 \end{array}\right. \end{align} $$ so we find the same result.

 

반응형

Question 6. Express the following as partial fractions: $$ \begin{align} \frac{ -5x^2 - 19x - 32 }{ (x+1) (x+2) (x-5) } \end{align} $$

 

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

 

Solution.

Let $$ \begin{align} \frac{ -5x^2 - 19x - 32 }{ (x+1) (x+2) (x-5) } &\equiv \frac{ A }{ x+1 } + \frac{ B }{ x+2 } + \frac{ C }{ x-5 } \\ &\equiv \frac{ A (x+2)(x-5) + B(x+1)(x-5) + C(x+1)(x+2) }{ (x+1) (x+2) (x-5) } \end{align} $$ Compare the numerators on both sides: $$ A (x+2)(x-5) + B(x+1)(x-5) + C(x+1)(x+2) \equiv -5x^2 - 19x - 32. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& (1)(-6) A &= -5(-1)^2 - 19(-1) - 32 \\ &&&&&\Rightarrow& -6A &= -5 + 19 - 32 = -18 \\ &&&&&\Rightarrow& A &= 3 \\ \\ \textrm{(ii)}&& x &= -2 &&\Rightarrow& (-1)(-7) B &= -5(-2)^2 - 19(-2) - 32 \\ &&&&&\Rightarrow& 7 B &= - 20 + 38 - 32 = -14 \\ &&&&&\Rightarrow& B &= -2 \\ \\ \textrm{(iii)}&& x &= 5 &&\Rightarrow& (6)(7) C &= -5(5)^2 - 19(5) - 32 \\ &&&&&\Rightarrow& 42 C &= -125 - 95 - 32 = - 252 \\ &&&&&\Rightarrow& C &= - 6 \end{align} $$ Hence, we obtain: $$ \Rightarrow \qquad \frac{ -5x^2 - 19x - 32 }{ (x+1) (x+2) (x-5) } \equiv \frac{ 3 }{ x+1 } - \frac{ 2 }{ x+2 } - \frac{ 6 }{ x-5 } \qquad \checkmark $$

 

반응형

Question 7. (P) Express the following as partial fractions: $$ \begin{align} &\textbf{(a)} \qquad \frac{ 6x^2 + 7x - 3 }{ x^3 - x } \\ \\ &\textbf{(b)} \qquad \frac{ 8x + 9 }{ 10x^2 + 3x - 4 } \end{align} $$ [Hint: Factorise the denominator first.]

 

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

 

Solution.

(a) We factorise the denominator first: $$ \begin{align} x^3 - x &= x \left( x^2 - 1 \right) \\ &= x ( x + 1 ) ( x - 1 ) \end{align} $$ and we have: $$ \begin{align} \frac{ 6x^2 + 7x - 3 }{ x^3 - x } &= \frac{ 6x^2 + 7x - 3 }{ x ( x + 1 ) ( x - 1 ) } \\ &= \frac{ A }{ x } + \frac{ B }{ x + 1 } + \frac{ C }{ x - 1 } \\ &= \frac{ A (x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) }{ x ( x + 1 ) ( x - 1 ) } \end{align} $$ Compare the numerators on both sides: $$ A (x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) \equiv 6x^2 + 7x - 3. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& (1)(-1) A &= -3 \\ &&&&&\Rightarrow& A &= 3 \\ \\ \textrm{(ii)}&& x &= -1 &&\Rightarrow& (-1)(-2) B &= 6(-1)^2 + 7(-1) - 3 \\ &&&&&\Rightarrow& 2 B &= 6 - 7 - 3 = -4 \\ &&&&&\Rightarrow& B &= -2 \\ \\ \textrm{(iii)}&& x &= 1 &&\Rightarrow& (1)(2) C &= 6 + 7 - 3 = 10 \\ &&&&&\Rightarrow& C &= 5 \end{align} $$ Hence, we obtain: $$ \Rightarrow \qquad \frac{ 6x^2 + 7x - 3 }{ x^3 - x } = \frac{ 6x^2 + 7x - 3 }{ x ( x + 1 ) ( x - 1 ) } = \frac{ 3 }{ x } - \frac{ 2 }{ x + 1 } + \frac{ 5 }{ x - 1 } \qquad \checkmark $$

 

(b) We factorise the denominator first: $$ \begin{align} 10x^2 + 3x - 4 &= ( 5x + 4) ( 2x - 1) \end{align} $$ and we have: $$ \begin{align} \frac{ 8x + 9 }{ 10x^2 + 3x - 4 } &= \frac{ 8x + 9 }{ ( 5x + 4) ( 2x - 1) } \\ &= \frac{ A }{ 5x + 4 } + \frac{ B }{ 2x - 1 } \\ &= \frac{ A(2x-1) + B(5x+4) }{ ( 5x + 4) ( 2x - 1) } \end{align} $$ Compare the numerators on both sides: $$ A(2x-1) + B(5x+4) \equiv 8x + 9. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -\frac45 &&\Rightarrow& \left( - \frac85 - 1 \right) A &= 8 \left( -\frac45 \right) + 9 \\ &&&&&\Rightarrow& - \frac{13}{5} A &= - \frac{32}{5} + 9 = \frac{13}{5} \\ &&&&&\Rightarrow& A &= -1 \\ \\ \textrm{(ii)}&& x &= \frac12 &&\Rightarrow& \left( \frac52 + 4 \right) B &= 8 \left( \frac12 \right) + 9 \\ &&&&&\Rightarrow& \frac{13}{2} B &= 13 \\ &&&&&\Rightarrow& B &= 2 \end{align} $$ Hence, we obtain: $$ \Rightarrow \qquad \frac{ 8x + 9 }{ 10x^2 + 3x - 4 } = \frac{ 8x + 9 }{ ( 5x + 4) ( 2x - 1) } \\ = - \frac{ 1 }{ 5x + 4 } + \frac{ 2 }{ 2x - 1 } \qquad \checkmark $$

 

 

 

반응형

Challenge. Express $$ \frac{ 5x^2 - 15x - 8 }{ x^3 - 4x^2 + x + 6 } $$ as a sum of fractions with linear denominators.

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1D Challenge.)

 

Solution.

We factorise the denominator first. It is a cubic expression, so we use the factor theorem. Let $$ \begin{align} \textrm f(x) &= x^3 - 4x^2 + x + 6 \end{align} $$ We try the factors of the constant term, 6, and they are $ \pm 1, \pm 2, \pm 3 $ and $ \pm 6$. $$ \begin{align} \textrm f(1) &= 1 - 4 + 1 + 6 \ne 0 \\ \textrm f(-1) &= -1 - 4 - 1 + 6 = 0 \qquad \checkmark \end{align} $$ hence $(x+1)$ is a factor of $\textrm f(x)$. $$ \begin{align} \Rightarrow \qquad \textrm f(x) &= x^3 - 4x^2 + x + 6 \\ &= (x+1) \left( x^2 -5x +6\right) \\ &= (x+1) (x-2) (x-3) \end{align} $$

 

Aside. In this case, we could have obtained all the factors by using factor theorem: $$ \begin{align} \textrm f(2) &= 8 - 16 + 2 + 6 = 0 \qquad \checkmark \\ \textrm f(-2) &= -8 - 16 - 2 + 6 \ne 0 \\ \textrm f(3) &= 27 - 36 + 3 + 6 = 0 \qquad \checkmark \\ \textrm f(-3) &= -27 - 36 - 3 + 6 \ne 0 \end{align} $$

 

We then have: $$ \begin{align} \frac{ 5x^2 - 15x - 8 }{ x^3 - 4x^2 + x + 6 } &= \frac{ 5x^2 - 15x - 8 }{ (x+1) (x-2) (x-3) } \\ &= \frac{ A }{ x+1 } + \frac{ B }{ x-2 } + \frac{ C }{ x-3 } \\ &= \frac{ A (x-2)(x-3) + B (x+1)(x-3) + C (x+1)(x-2) }{ (x-2) (x-3) (x+1) } \end{align} $$ Compare the numerators on both sides: $$ A (x-2)(x-3) + B (x+1)(x-3) + C (x+1)(x-2) \equiv 5x^2 - 15x - 8. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& (-3)(-4) A &= 5(-1)^2 - 15(-1) - 8 \\ &&&&&\Rightarrow& 12 A &= 5 + 15 - 8 = 12 \\ &&&&&\Rightarrow& A &= 1 \\ \\ \textrm{(ii)}&& x &= 2 &&\Rightarrow& (3)(-1) B &= 5(2)^2 - 15(2) - 8 \\ &&&&&\Rightarrow& -3 B &= 20 - 30 - 8 = - 18 \\ &&&&&\Rightarrow& B &= 6 \\ \\ \textrm{(iii)}&& x &= 3 &&\Rightarrow& (4)(1) C &= 5(3)^2 - 15(3) - 8 \\ &&&&&\Rightarrow& 4 C &= 45 - 45 - 8 = -8 \\ &&&&&\Rightarrow& C &= -2 \end{align} $$ Hence, we obtain: $$ \Rightarrow \qquad \frac{ 5x^2 - 15x - 8 }{ x^3 - 4x^2 + x + 6 } = \frac{ 5x^2 - 15x - 8 }{ (x+1) (x-2) (x-3) } = \frac{ 1 }{ x+1 } + \frac{ 6 }{ x-2 } - \frac{ 2 }{ x-3 } \qquad \checkmark $$