P2 §1.5 Algebraic division

Pure mathematics Year 2

Table of contents

1. Algebraic division
2. Examples
3. Edexcel P2 Ch1 Exercise 1F
4. Improper fractions in partial fractions
5. Edexcel P2 Ch1 Exercise 1G

1. Algebraic division

An improper algebraic fraction is one whose numerator has a degree equal to or larger than the denominator.

• The degree of a polynomial is the largest exponent in the expression. For example, $x^3 + 5x - 9$ has degree 3.
• $\frac{ x^2 + 5x + 8 }{ x - 2 }$ is an example where the degree of the numerator is greater than the degree of the denominator.
• $\frac{ x^3 + 5x - 9 }{ x^3 - 4x^2 + 7x - 3 }$ is an example where the degrees of the numerator and denominator are equal.

 

An improper fraction must be converted to a mixed fraction before we can express it in partial fractions. We can use algebraic division: $$\begin{align} && \textrm F(x) &= \textrm{(Quotient)} \times \textrm{(Divisor)} + \textrm{(Remainder)} \\ \\ &\Leftrightarrow& \textrm F(x) &= \textrm Q(x) \textrm D(x) + \textrm R(x) \\ \\ &\Leftrightarrow& \frac{ \textrm F(x) }{ \textrm D(x) } &\equiv \textrm Q(x) + \frac{ \textrm R(x) }{ \textrm D(x) } \qquad (\textrm{a mixed fraction}) \end{align}$$

• Our first example above can be written as: $$\begin{align} x^2 + 5x + 8 &\equiv (x+7)(x-2) + 22 \\ \Leftrightarrow \qquad \frac{ x^2 + 5x + 8 }{ x - 2 } &\equiv x + 7 + \frac{ 22 }{ x - 2 } \end{align}$$
• The second example above can be written as: $$\begin{align} x^3 + 5x - 9 &\equiv 1 \times \left( x^3 - 4x^2 + 7x - 3 \right) + \underbrace{ \left( 4x^2 - 2x - 6 \right) }_{2(x+1)(2x-3)} \\ \Leftrightarrow \qquad \frac{ x^3 + 5x - 9 }{ x^3 - 4x^2 + 7x - 3 } &\equiv 1 + \frac{ 2(x+1)(2x-3) }{ x^3 - 4x^2 + 7x - 3 } \end{align}$$

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2. Examples

Example 1. By algebraic long division, or otherwise, show that $$x^3 + x^2 - 7 \equiv \left( Ax^2 + Bx + C \right) (x-3) + D$$ and that $$\frac{ x^3 + x^2 - 7 }{ x - 3 } \equiv Ax^2 + Bx + C + \frac{ D }{ x-3 },$$ where $A,B,C$ and $D$ are constants to be found.

 

(Edexcel 2017 Specifications, P2 Ch1 Examples 11 and 12.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} x^3 + x^2 - 7 &\equiv \left( x^2 + 4x + 12 \right) (x-3) + 29 \\ \\ \Leftrightarrow \qquad \frac{ x^3 + x^2 - 7 }{ x - 3 } &\equiv x^2 + 4x + 12 + \frac{ 29 }{ x-3 } \end{align}$$ Hence, $$\begin{align} A &= 1 \\ B &= 4 \\ C &= 12 \\ D &= 29 \end{align}$$

 

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Example 2. $$\textrm f(x) = \frac{ x^4 + x^3 + x - 10 }{ x^2 + 2x - 3 }$$ Show that $\textrm f(x)$ can be written as $$Ax^2 + Bx + C + \frac{ Dx + E }{ x^2 + 2x - 3 }$$ and find the values of $A,B,C,D$ and $E$.

 

(Edexcel 2017 Specifications, P2 Ch1 Example 13.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} x^4 + x^3 + x - 10 &\equiv \left( x^2 - x + 5 \right) \left( x^2 + 2x - 3 \right) + ( -12x + 5 ) \\ \\ \Leftrightarrow \qquad \textrm f(x) = \frac{ x^4 + x^3 + x - 10 }{ x^2 + 2x - 3 } &\equiv x^2 - x + 5 + \frac{ -12x + 5 }{ x^2 + 2x - 3 } \end{align}$$ Hence, $$\begin{align} A &= 1 \\ B &= -1 \\ C &= 5 \\ D &= -12 \\ E &= 5 \end{align}$$

 

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3. Edexcel P2 Ch1 Exercise 1F

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

 

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

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} x^3 + 2x^2 + 3x - 4 &\equiv \left( x^2 + x + 2 \right) (x+1) -6 \\ \\ \Leftrightarrow \qquad \frac{ x^3 + 2x^2 + 3x - 4 }{ x+1 } &\equiv x^2 + x + 2 + \frac{ -6 }{ x+1 } \end{align}$$ Hence, $$\begin{align} A &= 1 \\ B &= 1 \\ C &= 2 \\ D &= -6 \end{align}$$

 

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Question 2. (E) Given that $$\frac{ 2x^3 + 3x^2 - 4x + 5 }{ x+3 } \equiv ax^2 + bx + c + \frac{ d }{ x+3 },$$ find the values of $a,b,c$ and $d$. [4 marks]

 

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

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} 2x^3 + 3x^2 - 4x + 5 &\equiv \left( 2x^2 - 3x + 5 \right) (x+3) - 10 \\ \\ \Leftrightarrow \qquad \frac{ 2x^3 + 3x^2 - 4x + 5 }{ x+3 } &\equiv 2x^2 - 3x + 5 + \frac{ -10 }{ x+3 } \end{align}$$ Hence, $$\begin{align} a &= 2 \\ b &= -3 \\ c &= 5 \\ d &= -10 \end{align}$$

 

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Question 3. (E) $$\textrm f(x) = \frac{ x^3 - 8 }{ x-2 }$$ Show that $\textrm f(x)$ can be written in the form $$px^2 + qx + r$$ and find the values of $p,q$ and $r$. [4 marks]

 

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

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} x^3 - 8 &\equiv \left( x^2 + 2x + 4 \right) (x-1) \\ \\ \Leftrightarrow \qquad \frac{ x^3 - 8 }{ x-2 } &\equiv x^2 + 2x + 4 \end{align}$$ Hence, $$\begin{align} p &= 1 \\ q &= 2 \\ r &= 4 \end{align}$$

 

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Question 4. (E) Given that $$\frac{ 2x^2 + 4x + 5 }{ x^2 - 1 } \equiv m + \frac{ nx + p }{ x^2 - 1 },$$ find the values of $m,n$ and $p$. [4 marks]

 

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

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} 2x^2 + 4x + 5 &\equiv 2 \left( x^2 -1 \right) + 4x + 7 \\ \\ \Leftrightarrow \qquad \frac{ 2x^2 + 4x + 5 }{ x^2 - 1 } &\equiv 2 + \frac{ 4x + 7 }{ x^2 - 1 } \end{align}$$ Hence, $$\begin{align} m &= 2 \\ n &= 4 \\ p &= 7 \end{align}$$

 

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Question 5. (E) Find the values of the constants $A,B,C$ and $D$ in the following identity: $$8x^3 + 2x^2 + 5 \equiv (Ax+B) \left( 2x^2+2 \right) + Cx + D$$ [4 marks]

 

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

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} 8x^3 + 2x^2 + 5 &\equiv (4x+1) \left( 2x^2+2 \right) - 8x + 3 \\ \\ \Leftrightarrow \qquad \frac{ 8x^3 + 2x^2 + 5 }{ 2x^2+2 } &\equiv 4x+1 + \frac{ - 8x + 3 }{ 2x^2+2 } \end{align}$$ Hence, $$\begin{align} A &= 4 \\ B &= 1 \\ C &= -8 \\ D &= 3 \end{align}$$

 

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Question 6. (E) $$\frac{ 4x^3 - 5x^2 + 3x - 14 }{ x^2 + 2x - 1 } \equiv Ax + B + \frac{ Cx + D }{ x^2 + 2x - 1}.$$ Find the values of the constants $A,B,C$ and $D$. [4 marks]

 

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

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} 4x^3 - 5x^2 + 3x - 14 &\equiv ( 4x - 13 ) \left( x^2 + 2x - 1 \right) + 33x - 27 \\ \\ \Leftrightarrow \qquad \frac{ 4x^3 - 5x^2 + 3x - 14 }{ x^2 + 2x - 1 } &\equiv 4x - 13 + \frac{ 33x - 27 }{ x^2 + 2x - 1} \end{align}$$ Hence, $$\begin{align} A &= 4 \\ B &= - 13 \\ C &= 33 \\ D &= - 27 \end{align}$$

 

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Question 7. (E) $$\textrm g(x) = \frac{ x^4 + 3x^2 - 4 }{ x^2 + 1 }$$ Show that $\textrm g(x)$ can be written in the form $$px^2 + qx + r + \frac{ sx + t }{ x^2 + 1 }$$ and find the values of $p,q,r,s$ and $t$. [4 marks]

 

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

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} x^4 + 3x^2 - 4 &\equiv \left( x^2 + 2 \right) \left( x^2 + 1 \right) -6 \\ \\ \Leftrightarrow \qquad \textrm g(x) = \frac{ x^4 + 3x^2 - 4 }{ x^2 + 1 } &\equiv x^2 + 2 + \frac{ -6 }{ x^2 + 1 } \end{align}$$ Hence, $$\begin{align} p &= 1 \\ q &= 0 \\ r &= 2 \\ s &= 0 \\ t &= -6 \end{align}$$

 

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Question 8. (E) Given that $$\frac{ 2x^4 + 3x^3 - 2x^2 + 4x - 6 }{ x^2 + x - 2 } \equiv ax^2 + bx + c + \frac{ dx + e }{ x^2 + x - 2 },$$ find the values of $a,b,c,d$ and $e$. [5 marks]

 

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

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} 2x^4 + 3x^3 - 2x^2 + 4x - 6 &\equiv \left( 2x^2 + x + 1 \right) \left( x^2 + x - 2 \right) + 5x - 4 \\ \\ \Leftrightarrow \qquad \frac{ 2x^4 + 3x^3 - 2x^2 + 4x - 6 }{ x^2 + x - 2 } &\equiv 2x^2 + x + 1 + \frac{ 5x - 4 }{ x^2 + x - 2 } \end{align}$$ Hence, $$\begin{align} a &= 2 \\ b &= 1 \\ c &= 1 \\ d &= 5 \\ e &= -4 \end{align}$$

 

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Question 9. (E) Find the values of the constants $A,B,C,D$ and $E$ in the following identity: $$3x^4 - 4x^3 - 8x^2 + 16x - 2 \equiv \left( Ax^2 + Bx + C \right) \left( x^2 - 3 \right) + Dx + E$$ [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1F Q9.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details),

$$\begin{align} 3x^4 - 4x^3 - 8x^2 + 16x - 2 &\equiv \left( 3x^2 - 4x + 1 \right) \left( x^2 - 3 \right) + 4x + 1 \\ \\ \Leftrightarrow \qquad \frac{ 3x^4 - 4x^3 - 8x^2 + 16x - 2 }{ x^2 - 3 } &\equiv 3x^2 - 4x + 1 + \frac{ 4x + 1 }{ x^2 - 3 } \end{align}$$ Hence, $$\begin{align} A &= 3 \\ B &= -4 \\ C &= 1 \\ D &= 4 \\ E &= 1 \end{align}$$

 

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Question 10. (E/P)

(a) Fully factorise the expression $x^4 - 1$. [2 marks]

(b) Hence, or otherwise, write the algebraic fraction $$\frac{ x^4 - 1 }{ x + 1 }$$ in the form $$(ax + b) \left( cx^2 + dx + e \right)$$ and find the values of $a,b,c,d$ and $e$. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1F Q10.)

 

Solution.

(a) We have: $$\begin{align} x^4 - 1 &= \left( x^2 + 1 \right) \left( x^2 - 1 \right) \\ &= \left( x^2 + 1 \right) (x+1)(x-1) \end{align}$$

 

(b) This gives $$\begin{align} \frac{ x^4 - 1 }{ x + 1 } &= \frac{ \left( x^2 + 1 \right) (x+1)(x-1) }{ x+1 } \\ &= (x-1) \left( x^2 + 1 \right) \end{align}$$ so we find: $$\begin{align} A &= 1 \\ B &= -1 \\ C &= 1 \\ D &= 0 \\ E &= 1 \end{align}$$

 

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4. Improper fractions in partial fractions

In order to express an improper algebraic fraction in partial fractions, we need to divide the numerator by the denominator first. Recall that an improper algebraic fraction is one where the degree of the numerator is greater than or equal to the degree of the denominator.

 

Example. Given that $$\frac{ 3x^2 - 3x - 2 }{ (x-1)(x-2)} \equiv A + \frac{B}{x-1} + \frac{C}{x-2},$$ find the values of $A,B$ and $C$.

 

(Edexcel 2017 Specifications, P2 Ch1 Example 14.)

 

Solution.

(i) Step 1: Algebraic division.

We note that $$(x-1)(x-2) = x^2 - 3x + 2.$$ By algebraic long division (see the YouTube clip below for more details), $$\begin{align} 3x^2 - 3x - 2 &\equiv 3 \left( x^2 - 3x + 2 + 3x - 2 \right) - 3x - 2 \\ &\equiv 3 \left( x^2 - 3x + 2 \right) + 6x - 8 \\ \\ \Leftrightarrow \qquad \frac{ 3x^2 - 3x - 2 }{ (x-1)(x-2) } &\equiv 3 + \frac{ 6x - 8 }{ (x-1)(x-2) } \end{align}$$

 

(ii) Step 2: Partial fractions.

Let $$\begin{align} \frac{ 6x - 8 }{ (x-1)(x-2) } &\equiv \frac{ B }{ x-1 } + \frac{ C }{ x-2 } \\ &\equiv \frac{ B(x-2) + C(x-1) }{ (x-1)(x-2) } \end{align}$$ Compare the numerators on both sides: $$B(x-2) + C(x-1) \equiv 6x - 8$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 1 &&\Rightarrow& -B&=-2 \\ &&&&&\Rightarrow& B&=2 \\ \\ \textrm{(ii)}&& x &= 2 &&\Rightarrow& C&= 12-8=4 \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ 6x - 8 }{ (x-1)(x-2) } \equiv \frac{ 2 }{ x-1 } + \frac{ 4 }{ x-2 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ 3x^2 - 3x - 2 }{ (x-1)(x-2)} &\equiv 3 + \frac{ 6x - 8 }{ (x-1)(x-2) } \\ &\equiv 3 + \frac{ 2 }{x-1} + \frac{4}{x-2} \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 3 \\ B &= 2 \\ C &= 4 \end{align}$$

 

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5. Edexcel P2 Ch1 Exercise 1G

Question 1. (E) $$\textrm g(x) = \frac{ x^2 + 3x - 2 }{ (x-1)(x-2)}.$$ Show that $\textrm g(x)$ can be written in the form $$A + \frac{ B }{ x - 1 } + \frac{ C }{ x - 2 }$$ and find the values of the constants $A,B$ and $C$. [4 marks]

 

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

 

Solution.

(i) Step 1: Algebraic division.

We note that $$(x-1)(x-2) = x^2 - 3x + 2.$$ By algebraic long division (see the YouTube clip below for more details), $$\begin{align} x^2 + 3x - 2 &\equiv \left( x^2 - 3x + 2 + 3x - 2 \right) + 3x - 2 \\ &\equiv \left( x^2 - 3x + 2 \right) + 6x - 4 \\ \\ \Leftrightarrow \qquad \frac{ x^2 - 3x - 2 }{ (x-1)(x-2) } &\equiv 1 + \frac{ 6x - 4 }{ (x-1)(x-2) } \end{align}$$

 

(ii) Step 2: Partial fractions.

Let $$\begin{align} \frac{ 6x - 4 }{ (x-1)(x-2) } &\equiv \frac{ B }{ x-1 } + \frac{ C }{ x-2 } \\ &\equiv \frac{ B(x-2) + C(x-1) }{ (x-1)(x-2) } \end{align}$$ Compare the numerators on both sides: $$B(x-2) + C(x-1) \equiv 6x - 4$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 1 &&\Rightarrow& -B&=2 \\ &&&&&\Rightarrow& B&=-2 \\ \\ \textrm{(ii)}&& x &= 2 &&\Rightarrow& C&= 12-4=8 \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ 6x - 4 }{ (x-1)(x-2) } \equiv - \frac{ 2 }{ x-1 } + \frac{ 8 }{ x-2 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ 3x^2 - 3x - 2 }{ (x-1)(x-2)} &\equiv 1 + \frac{ 6x - 4 }{ (x-1)(x-2) } \\ &\equiv 1 - \frac{ 2 }{x-1} + \frac{ 8 }{x-2} \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 1 \\ B &= -2 \\ C &= 8 \end{align}$$

 

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Question 2. (E) Given that $$\frac{ x^2 - 10 }{ (x-2)(x+1) } \equiv A + \frac{ B }{ x-2 } + \frac{ C }{x+1},$$ find the values of the constants $A,B$ and $C$. [4 marks]

 

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

 

Solution.

(i) Step 1: Algebraic division.

We note that $$(x-2)(x+1) = x^2 - x - 2.$$ By algebraic long division (see the YouTube clip below for more details), $$\begin{align} x^2 - 10 &\equiv \left( x^2 - x - 2 + x + 2 \right) - 10 \\ &\equiv \left( x^2 - x -2 \right) + x - 8 \\ \\ \Leftrightarrow \qquad \frac{ x^2 - 10 }{ (x-2)(x+1) } &\equiv 1 + \frac{ x - 8 }{ (x-2)(x+1) } \end{align}$$

 

(ii) Step 2: Partial fractions.

Let $$\begin{align} \frac{ x - 8 }{ (x-2)(x+1) } &\equiv \frac{ B }{ x-2 } + \frac{ C }{ x+1 } \\ &\equiv \frac{ B(x+1) + C(x-2) }{ (x-1)(x-2) } \end{align}$$ Compare the numerators on both sides: $$B(x+1) + C(x-2) \equiv x - 8$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 2 &&\Rightarrow& 3B & = -6 \\ &&&&&\Rightarrow& B&=-2 \\ \\ \textrm{(ii)}&& x &= -1 &&\Rightarrow& -3C &= - 1 - 8 = - 9 \\ &&&&&\Rightarrow& C &= 3 \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ x - 8 }{ (x-2)(x+1) } \equiv - \frac{ 2 }{ x-2 } + \frac{ 3 }{ x+1 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ x^2 - 10 }{ (x-2)(x+1) } &\equiv 1 + \frac{ x - 8 }{ (x-2)(x+1) } \\ &\equiv 1 - \frac{ 2 }{ x-2 } + \frac{ 3 }{ x+1 } \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 1 \\ B &= -2 \\ C &= 3 \end{align}$$

 

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Question 3. (E) Find the values of the constants $A,B,C$ and $D$ in the following identity: $$\frac{ x^3 - x^2 - x - 3 }{ x(x-1) } \equiv Ax + B + \frac{ C }{ x } + \frac{ D }{ x-1 }$$ [5 marks]

 

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

 

Solution.

(i) Step 1: Algebraic division.

We note that $$x(x-1) = x^2 - x.$$ By algebraic long division (see the YouTube clip below for more details), $$\begin{align} x^3 - x^2 - x - 3 &\equiv x \left( x^2 - x \right) - x - 3 \\ \\ \Leftrightarrow \qquad \frac{ x^3 - x^2 - x - 3 }{ x(x-1) } &\equiv x + \frac{ - x - 3 }{ x(x-1) } \end{align}$$

 

(ii) Step 2: Partial fractions.

Let $$\begin{align} \frac{ - x - 3 }{ x(x-1) } &\equiv \frac{ C }{ x } + \frac{ D }{ x-1 } \\ &\equiv \frac{ C(x-1) + D x }{ x(x-1) } \end{align}$$ Compare the numerators on both sides: $$C(x-1) + D x \equiv - x - 3$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& -C & = -3 \\ &&&&&\Rightarrow& C &= 3 \\ \\ \textrm{(ii)}&& x &= 1 &&\Rightarrow& D &= - 1 - 3 = - 4 \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ - x - 3 }{ x(x-1) } \equiv \frac{ 3 }{ x } - \frac{ 4 }{ x-1 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ x^3 - x^2 - x - 3 }{ x(x-1) } &\equiv x + \frac{ - x - 3 }{ x(x-1) } \\ &\equiv x + \frac{ 3 }{ x } - \frac{ 4 }{ x-1 } \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 1 \\ B &= 0 \\ C &= 3 \\ D &= -4 \end{align}$$

 

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Question 4. (E) Show that $$\frac{ -3x^3 - 4x^2 + 19x + 8 }{ x^2 + 2x - 3 }$$ can be expressed in the form $$A + Bx + \frac{ C }{(x-1)} + \frac{D}{(x+3)},$$ where $A,B,C$ and $D$ are constants to be found. [5 marks]

 

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

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$\begin{align} -3x^3 - 4x^2 + 19x + 8 &\equiv ( - 3x + 2 ) \left( x^2 + 2x - 3 \right) + 6x + 14 \\ \\ \Leftrightarrow \qquad \frac{ -3x^3 - 4x^2 + 19x + 8 }{ x^2 + 2x - 3 } &\equiv -3x + 2 + \frac{ 6x + 14 }{ x^2 + 2x - 3 } \end{align}$$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$x^2 + 2x - 3 = (x-1) (x+3)$$ Let $$\begin{align} \frac{ 6x + 14 }{ (x-1)(x+3) } &\equiv \frac{ C }{ x-1 } + \frac{ D }{ x+3 } \\ &\equiv \frac{ C (x+3) + D (x-1) }{ (x-1)(x+3) } \end{align}$$ Compare the numerators on both sides: $$C (x+3) + D (x-1) \equiv 6x + 14$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 1 &&\Rightarrow& 4C & = 6+14 = 20 \\ &&&&&\Rightarrow& C &= 5 \\ \\ \textrm{(ii)}&& x &= -3 &&\Rightarrow& -4D &= - 18 + 14 = -4 \\ &&&&&\Rightarrow& D &= 1 \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ 6x + 14 }{ (x-1)(x+3) } \equiv \frac{ 5 }{ x-1 } + \frac{ 1 }{ x+3 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ -3x^3 - 4x^2 + 19x + 8 }{ x^2 + 2x - 3 } &\equiv 2 -3x + \frac{ 6x + 14 }{ x^2 + 2x - 3 } \\ &\equiv 2 -3x + \frac{ 5 }{ x-1 } + \frac{ 1 }{ x+3 } \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 2 \\ B &= -3 \\ C &= 5 \\ D &= 1 \end{align}$$

 

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Question 5. (E) $$\textrm p(x) \equiv \frac{ 4x^2 + 25 }{ 4x^2 - 25 }$$ Show that $\textrm p(x)$ can be written in the form $$A + \frac{ B }{ 2x - 5 } + \frac{ C }{ 2x + 5 },$$ where $A,B$ and $C$ are constants to be found. [4 marks]

 

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

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$\begin{align} \textrm p(x) \equiv \frac{ 4x^2 + 25 }{ 4x^2 - 25 } &\equiv \frac{ \left( 4x^2 - 25 \right) + 50 }{ 4x^2 - 25 } \\ &\equiv 1 + \frac{ 50 }{ 4x^2 - 25 } \end{align}$$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$4x^2 - 25 = (2x - 5) (2x + 5)$$ Let $$\begin{align} \frac{ 50 }{ (2x - 5) (2x + 5) } &\equiv \frac{ B }{ 2x - 5 } + \frac{ C }{ 2x + 5 } \\ &\equiv \frac{ B (2x + 5) + C (2x - 5) }{ (2x - 5) (2x + 5) } \end{align}$$ Compare the numerators on both sides: $$B (2x + 5) + C (2x - 5) \equiv 50$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= \frac52 &&\Rightarrow& 10 B & = 50 \\ &&&&&\Rightarrow& B &= 5 \\ \\ \textrm{(ii)}&& x &= \frac52 &&\Rightarrow& - 10 C &= 50 \\ &&&&&\Rightarrow& C &= -5 \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ 50 }{ (2x - 5) (2x + 5) } \equiv \frac{ 5 }{ 2x - 5 } - \frac{ 5 }{ 2x + 5 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \textrm p(x) \equiv \frac{ 4x^2 + 25 }{ 4x^2 - 25 } &\equiv 1 + \frac{ 50 }{ 4x^2 - 25 } \\ &\equiv 1 + \frac{ 5 }{ 2x - 5 } - \frac{ 5 }{ 2x + 5 } \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 1 \\ B &= 5 \\ C &= -5 \end{align}$$

 

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Question 6. (E) Given that $$\frac{ 2x^2 - 1 }{ x^2 + 2x + 1 } \equiv A + \frac{ B }{ x+1 } + \frac{ C }{ (x+1)^2 },$$ find the values of the constants $A,B$ and $C$. [4 marks]

 

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

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$\begin{align} \frac{ 2x^2 - 1 }{ x^2 + 2x + 1 } &\equiv \frac{ 2 \left( x^2 + 2x + 1 - 2x - 1 \right) - 1 }{ x^2 + 2x + 1 } \\ &\equiv \frac{ 2 \left( x^2 + 2x + 1 \right) - 4x - 3 }{ x^2 + 2x + 1 } \\ &\equiv 2 + \frac{ - 4x - 3 }{ x^2 + 2x + 1 } \end{align}$$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$x^2 + 2x + 1 = (x+1)^2$$ and this gives partial fractions with repeated factors. Let $$\begin{align} \frac{ - 4x - 3 }{ (x+1)^2 } &\equiv \frac{ B }{ x+1 } + \frac{ C }{ (x+1)^2 } \\ &\equiv \frac{ B (x+1) + C }{ (x+1)^2 } \end{align}$$ Compare the numerators on both sides: $$B (x+1) + C \equiv -4x - 3$$ By equating coefficients: $$\begin{align} \left\{ \begin{array}{l} B = - 4 \\ B + C = - 3 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} B = - 4 \\ C = 1 \end{array} \right. \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ - 4x - 3 }{ (x+1)^2 } \equiv - \frac{ 4 }{ x+1 } + \frac{ 1 }{ (x+1)^2 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ 2x^2 - 1 }{ x^2 + 2x + 1 } &\equiv 2 + \frac{ - 4x - 3 }{ x^2 + 2x + 1 } \\ &\equiv 2 - \frac{ 4 }{ x+1 } + \frac{ 1 }{ (x+1)^2 } \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 2 \\ B &= -4 \\ C &= 1 \end{align}$$

 

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Question 7. (P) By factorising the denominator, express the following as partial fractions: $$\begin{align} &\textbf{(a)} \qquad \frac{ 4x^2 + 17x - 11 }{ x^2 + 3x - 4 } \\ \\ &\textbf{(b)} \qquad \frac{ x^4 - 4x^3 + 9x^2 - 17x + 12 }{ x^3 - 4x^2 + 4x } \end{align}$$

 

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

 

Solution.

(a) (i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$\begin{align} \frac{ 4x^2 + 17x - 11 }{ x^2 + 3x - 4 } &\equiv \frac{ 4 \left( x^2 + 3x - 4 - 3x + 4 \right) + 17x - 11 }{ x^2 + 3x - 4 } \\ &\equiv \frac{ 4 \left( x^2 + 3x - 4 \right) + 5x + 5 }{ x^2 + 3x - 4 } \\ &\equiv 4 + \frac{ 5x + 5 }{ x^2 + 3x - 4 } \end{align}$$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$x^2 + 3x - 4 = (x-1)(x+4)$$ Let $$\begin{align} \frac{ 5x + 5 }{ (x-1)(x+4) } &\equiv \frac{ B }{ x-1 } + \frac{ C }{ x+4 } \\ &\equiv \frac{ B (x+4) + C (x-1) }{ (x+1)^2 } \end{align}$$ Compare the numerators on both sides: $$B (x+4) + C (x-1) \equiv 5x + 5$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 1 &&\Rightarrow& 5 B & = 5+5 = 10 \\ &&&&&\Rightarrow& B &= 2 \\ \\ \textrm{(ii)}&& x &= -4 &&\Rightarrow& -5 C &= - 20 + 5 = -15 \\ &&&&&\Rightarrow& C &= 3 \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ 5x + 5 }{ (x-1)(x+4) } \equiv \frac{ 2 }{ x-1 } + \frac{ 3 }{ x+4 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ 4x^2 + 17x - 11 }{ x^2 + 3x - 4 } &\equiv 4 + \frac{ 5x + 5 }{ x^2 + 3x - 4 } \\ &\equiv 4 + \frac{ 2 }{ x-1 } + \frac{ 3 }{ x+4 } \qquad \checkmark \end{align}$$

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(b) (i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$\begin{align} x^4 - 4x^3 + 9x^2 - 17x + 12 &\equiv x \left( x^3 - 4x^2 + 4x \right) + 5x^2 - 17x + 12 \\ \\ \Leftrightarrow \qquad \frac{ x^4 - 4x^3 + 9x^2 - 17x + 12 }{ x^3 - 4x^2 + 4x } &\equiv x + \frac{ 5x^2 - 17x + 12 }{ x^3 - 4x^2 + 4x } \end{align}$$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$\begin{align} x^3 - 4x^2 + 4x &= x \left( x^2 - 4x + 4 \right) \\ &= x ( x - 2 )^2 \end{align}$$ and this gives partial fractions with repeated factors. Let $$\begin{align} \frac{ 5x^2 - 17x + 12 }{ x^3 - 4x^2 + 4x } &\equiv \frac{ 5x^2 - 17x + 12 }{ x ( x-2 )^2 } \\ &\equiv \frac{ B }{ x } + \frac{ C }{ x-2 } + \frac{ D }{ (x-2)^2 } \\ &\equiv \frac{ B (x-2)^2 + C x (x-2) + D x }{ x (x-2)^2 } \end{align}$$ Compare the numerators on both sides: $$B (x-2)^2 + C x (x-2) + D x \equiv 5x^2 - 17x + 12$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& 4 B & = 12 \\ &&&&&\Rightarrow& B &= 3 \\ \\ \textrm{(ii)}&& x &= 2 &&\Rightarrow& 2 D &= 5(2)^2 - 17(2) + 12 \\ &&&&&&&= 20 - 34 + 12 \\ &&&&&&&= -2 \\ &&&&&\Rightarrow& D &= -1 \\ \\ \textrm{(iii)}&& x &= 1 &&\Rightarrow& B - C + D & = 5 - 17 + 12 = 0 \\ &&&&&\Rightarrow& C &= B + D = 2 \end{align}$$ and we obtain: $$\Rightarrow \qquad \frac{ 5x^2 - 17x + 12 }{ x^3 - 4x^2 + 4x } \equiv \frac{ 5x^2 - 17x + 12 }{ x ( x-2 )^2 } \equiv \frac{ 3 }{ x } + \frac{ 2 }{ x-2 } - \frac{ 1 }{ (x-2)^2 }$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ x^4 - 4x^3 + 9x^2 - 17x + 12 }{ x^3 - 4x^2 + 4x } &\equiv x + \frac{ 5x^2 - 17x + 12 }{ x^3 - 4x^2 + 4x } \\ &\equiv x + \frac{ 3 }{ x } + \frac{ 2 }{ x-2 } - \frac{ 1 }{ (x-2)^2 } \qquad \checkmark \end{align}$$

 

 

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Question 8. (E) Given that $$\frac{ 6x^3 - 7x^2 + 3 }{ 3x^2 + x - 10 } \equiv Ax + B + \frac{ C }{ 3x - 5 } + \frac{ D }{ x+2 },$$ find the values of the constants $A,B,C$ and $D$. [6 marks]

 

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

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$\begin{align} 6x^3 - 7x^2 + 3 &\equiv (2x-3) \left( 3x^2 + x - 10 \right) + 23x - 27 \\ \\ \Leftrightarrow \qquad \frac{ 6x^3 - 7x^2 + 3 }{ 3x^2 + x - 10 } &\equiv 2x - 3 + \frac{ 23x - 27 }{ 3x^2 + x - 10 } \end{align}$$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$\begin{align} 3x^2 + x - 10 = (3x - 5) (x + 2) \end{align}$$ Let $$\begin{align} \frac{ 23x - 27 }{ 3x^2 + x - 10 } &\equiv \frac{ 23x - 27 }{ (3x - 5) (x + 2) } \\ &\equiv \frac{ C }{ 3x-5 } + \frac{ D }{ x+2 } \\ &\equiv \frac{ C (x+2) + D (3x-5) }{ (3x - 5) (x + 2) } \end{align}$$ Compare the numerators on both sides: $$C (x+2) + D (3x-5) \equiv 23x - 27$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= \frac53 &&\Rightarrow& \left( \frac53 + 2 \right) C & = 23 \times \frac{5}{3} - 27 \\ &&&&&\Rightarrow& \frac{11}{3} C &= \frac{115}{3} - \frac{81}{3} = \frac{34}{3} \\ &&&&&\Rightarrow& C &= \frac{34}{11} \\ \\ \textrm{(ii)}&& x &= -2 &&\Rightarrow& - 11D &= -46-27 = -73 \\ &&&&&\Rightarrow& D &= \frac{73}{11} \end{align}$$ and we obtain: $$\begin{align} \Rightarrow \qquad \frac{ 23x - 27 }{ 3x^2 + x - 10 } &\equiv \frac{ 23x - 27 }{ (3x - 5) (x + 2) } \\ &\equiv \frac{ \frac{34}{11} }{ 3x-5 } + \frac{ \frac{73}{11} }{ x+2 } \\ &\equiv \frac{ 34 }{ 11( 3x-5 ) } + \frac{ 73 }{ 11(x+2) } \\ &\equiv \frac1{11} \left( \frac{ 34 }{ 3x-5 } + \frac{ 73 }{ x+2 } \right) \end{align}$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ 6x^3 - 7x^2 + 3 }{ 3x^2 + x - 10 } &\equiv 2x - 3 + \frac{ 23x - 27 }{ 3x^2 + x - 10 } \\ &\equiv 2x - 3 + \frac1{11} \left( \frac{ 34 }{ 3x-5 } + \frac{ 73 }{ x+2 } \right) \qquad \checkmark \end{align}$$

 

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Question 9. (E) $$\textrm q(x) = \frac{ 8x^3 + 1 }{ 4x^2 - 4x + 1 }$$ Show that $\textrm q(x)$ can be written in the form $$Ax + B + \frac{ C }{ 2x - 1 } + \frac{ D }{ (2x - 1)^2 }$$ and find the values of the constants $A,B,C$ and $D$. [6 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1G Q9.)

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$\begin{align} 8x^3 + 1 &\equiv (2x+2) \left( 4x^2 - 4x + 1 \right) + 6x - 1 \\ \\ \Leftrightarrow \qquad \frac{ 8x^3 + 1 }{ 4x^2 - 4x + 1 } &\equiv 2x + 2 + \frac{ 6x - 1 }{ 4x^2 - 4x + 1 } \end{align}$$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$\begin{align} 4x^2 - 4x + 1 = (2x - 1)^2 \end{align}$$ and this gives partial fractions with repeated factors. Let $$\begin{align} \frac{ 6x - 1 }{ 4x^2 - 4x + 1 } &\equiv \frac{ 6x - 1 }{ (2x - 1)^2 } \\ &\equiv \frac{ C }{ 2x - 1 } + \frac{ D }{ (2x - 1)^2 } \\ &\equiv \frac{ C (2x - 1) + D }{ 2x - 1 } \end{align}$$ Compare the numerators on both sides: $$C (2x - 1) + D \equiv 6x - 1$$ By equating coefficients: $$\begin{align} \left\{ \begin{array}{l} 2C = 6 \\ - C + D = - 1 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} C = 3 \\ D = 2 \end{array} \right. \end{align}$$ and we obtain: $$\begin{align} \Rightarrow \qquad \frac{ 6x - 1 }{ 4x^2 - 4x + 1 } \equiv \frac{ 6x - 1 }{ (2x - 1)^2 } \equiv \frac{ 3 }{ 2x - 1 } + \frac{ 2 }{ (2x - 1)^2 } \end{align}$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ 8x^3 + 1 }{ 4x^2 - 4x + 1 } &\equiv 2x + 2 + \frac{ 6x - 1 }{ 4x^2 - 4x + 1 } \\ &\equiv 2x + 2 + \frac{ 3 }{ 2x - 1 } + \frac{ 2 }{ (2x - 1)^2 } \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 2 \\ B &= 2 \\ C &= 3 \\ D &= 2 \end{align}$$

 

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Question 10. (E) $$\textrm h(x) = \frac{ x^4 + 2x^2 - 3x + 8 }{ x^2 + x - 2 }$$ Show that $\textrm h(x)$ can be written as $$Ax^2 + Bx + C + \frac{ D }{ x+2 } + \frac{ E }{ x-1 }$$ and find the values of $A,B,C,D$ and $E$. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Exercise 1G Q10.)

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$\begin{align} x^4 + 2x^2 - 3x + 8 &\equiv \left( x^2 - x + 5 \right) \left( x^2 + x - 2 \right) - 10 x + 18 \\ \\ \Leftrightarrow \qquad \frac{ x^4 + 2x^2 - 3x + 8 }{ x^2 + x - 2 } &\equiv x^2 - x + 5 + \frac{ - 10x + 18 }{ x^2 + x - 2 } \end{align}$$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$\begin{align} x^2 + x - 2 = (x + 2) (x - 1) \end{align}$$ Let $$\begin{align} \frac{ - 10x + 18 }{ x^2 + x - 2 } &\equiv \frac{ - 10x + 18 }{ (x + 2) (x - 1) } \\ &\equiv \frac{ D }{ x+2 } + \frac{ E }{ x-1 } \\ &\equiv \frac{ D (x-1) + E (x+2) }{ (x + 2) (x - 1) } \end{align}$$ Compare the numerators on both sides: $$D (x-1) + E (x+2) \equiv - 10x + 18$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= -2 &&\Rightarrow& -3 D & = 20 + 18 = 38 \\ &&&&&\Rightarrow& D &= - \frac{38}{3} \\ \\ \textrm{(ii)}&& x &= 1 &&\Rightarrow& 3 E &= - 10 + 18 = 8 \\ &&&&&\Rightarrow& E &= \frac{8}{3} \end{align}$$ and we obtain: $$\begin{align} \Rightarrow \qquad \frac{ - 10x + 18 }{ x^2 + x - 2 } &\equiv \frac{ - 10x + 18 }{ (x + 2) (x - 1) } \\ &\equiv - \frac{ \frac{38}{3} }{ x+2 } + \frac{ \frac{8}{3} }{ x-1 } \\ &\equiv \frac1{3} \left( \frac{ - 38 }{ x+2 } + \frac{ 8 }{ x-1 } \right) \end{align}$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ x^4 + 2x^2 - 3x + 8 }{ x^2 + x - 2 } &\equiv x^2 - x + 5 + \frac{ - 10x + 18 }{ x^2 + x - 2 } \\ &\equiv x^2 - x + 5 + \frac1{3} \left( \frac{ - 38 }{ x+2 } + \frac{ 8 }{ x-1 } \right) \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 1 \\ B &= -1 \\ C &= 5 \\ D &= -\frac{38}{3} \\ E &= \frac{8}{3} \end{align}$$