P2 §1.2 Algebraic fractions

Pure mathematics Year 2

Table of contents

1. Algebraic fractions
2. Multiplication
3. Division
4. Edexcel P2 Ch1 Exercise 1B
5. Addition and subtraction
6. Edexcel P2 Ch1 Exercise 1C

1. Algebraic fractions

Algebraic fractions work in the same way as numeric fractions. We can simplify them by cancelling common factors for multiplication and division, and finding common denominators for addition and subtraction.

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2. Multiplication

To multiply fractions, cancel any common factors, then multiply the numerators together and multiply the denominators together. We illustrate this in Example 1.

Example 1. Simplify the following products: $$ \begin{align} \textbf{(a)} \qquad \frac{3}{5} \times \frac{5}{9} \qquad\qquad \textbf{(b)} \qquad \frac{a}{b} \times \frac{c}{a} \qquad\qquad \textbf{(c)} \qquad \frac{x+1}{2} \times \frac{3}{x^2-1} \end{align} $$

 

(Edexcel 2017 Specifications, P2 Ch1 Example 5.)

 

Solution.

$$ \begin{align} \textbf{(a)}&&& \frac{3}{5} \times \frac{5}{9} = \frac{1}{3} \qquad \checkmark \\ \\ \textbf{(b)}&&& \frac{a}{b} \times \frac{c}{a} = \frac{c}{b} \qquad \checkmark \\ \\ \textbf{(c)}&&& \frac{x+1}{2} \times \frac{3}{x^2-1} = \frac{x+1}{2} \times \frac{3}{(x+1)(x-1)} = \frac{3}{2(x-1)} \qquad \checkmark \end{align} $$

 

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3. Division

To divide two fractions, multiply the first fraction by the reciprocal of the second fraction. We illustrate this in Example 2.

Example 2. Simplify: $$ \begin{align} \textbf{(a)} \qquad \frac{a}{b} \div \frac{a}{c} \qquad\qquad \textbf{(b)} \qquad \frac{x+2}{x+4} \div \frac{3x+6}{x^2-16} \end{align} $$

 

(Edexcel 2017 Specifications, P2 Ch1 Example 6.)

 

Solution.

$$ \begin{align} \textbf{(a)}&&& \frac{a}{b} \div \frac{a}{c} = \frac{a}{b} \times \frac{c}{a} = \frac{c}{b} \qquad \checkmark \\ \\ \textbf{(b)}&&& \frac{x+2}{x+4} \div \frac{3x+6}{x^2-16} \\ &&&= \frac{x+2}{x+4} \times \frac{x^2-16}{3x+6} \\ &&&= \frac{x+2}{x+4} \times \frac{ (x+4)(x-4) }{3(x+2)} \\ &&&= \frac{x-4}{3} \qquad \checkmark \\ \\ \end{align} $$

 

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4. Edexcel P2 Ch1 Exercise 1B

Question 1. Simplify: $$ \begin{align} &\textbf{(a)} \qquad \frac{a}{d} \times \frac{a}{c} & &\textbf{(b)} \qquad \frac{a^2}{c} \times \frac{c}{a} & &\textbf{(c)} \qquad \frac{2}{x} \times \frac{x}{4} \\ \\ &\textbf{(d)} \qquad \frac{3}{x} \div \frac{6}{x} & &\textbf{(e)} \qquad \frac{4}{xy} \div \frac{x}{y} & &\textbf{(f)} \qquad \frac{2r^2}{5} \div \frac{4}{r^3} & \end{align} $$

 

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

 

Solution.

$$ \begin{align} &\textbf{(a)} \qquad \frac{a}{d} \times \frac{a}{c} = \frac{a^2}{cd} \qquad \checkmark \\ \\ &\textbf{(b)} \qquad \frac{a^2}{c} \times \frac{c}{a} = a \qquad \checkmark \\ \\ &\textbf{(c)} \qquad \frac{2}{x} \times \frac{x}{4} = \frac12 \qquad \checkmark \\ \\ &\textbf{(d)} \qquad \frac{3}{x} \div \frac{6}{x} = \frac{3}{x} \times \frac{x}{6} = \frac12 \qquad \checkmark \\ \\ &\textbf{(e)} \qquad \frac{4}{xy} \div \frac{x}{y} = \frac{4}{xy} \times \frac{y}{x} = \frac{4}{x^2} \qquad \checkmark \\ \\ &\textbf{(f)} \qquad \frac{2r^2}{5} \div \frac{4}{r^3} = \frac{2r^2}{5} \times \frac{r^3}{4} = \frac{ r^{2+3} }{5 \times 2} = \frac{ r^5 }{ 10 } \qquad \checkmark \end{align} $$

 

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Question 2. Simplify: $$ \begin{align} &\textbf{(a)} \qquad (x + 2) \times \frac{1}{ x^2 - 4 } \\ \\ &\textbf{(b)} \qquad \frac{1}{ a^2 + 6a + 9 } \times \frac{ a^2 - 9 }{ 2 } \\ \\ &\textbf{(c)} \qquad \frac{ x^2 - 3x }{ y^2 + y } \times \frac{ y + 1 }{ x } \\ \\ &\textbf{(d)} \qquad \frac{ y }{ y + 3 } \div \frac{ y^2 }{ y^2 + 4y + 3 } \\ \\ &\textbf{(e)} \qquad \frac{ x^2 }{ 3 } \div \frac{ 2x^3 - 6x^2 }{ x^2 - 3x } \\ \\ &\textbf{(f)} \qquad \frac{ 4x^2 - 25 }{ 4x - 10 } \div \frac{ 2x + 5 }{ 8 } \\ \\ &\textbf{(g)} \qquad \frac{ x + 3 }{ x^2 + 10x + 25 } \times \frac{ x^2 + 25 }{ x^2 + 3x } \\ \\ &\textbf{(h)} \qquad \frac{ 3y^2 + 4y - 4 }{ 10 } \div \frac{ 3y + 6 }{ 15 } \\ \\ &\textbf{(i)} \qquad \frac{ x^2 + 2xy + y^2 }{2} \times \frac{ 4 }{ (x - y)^2 } \end{align} $$

 

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

 

Solution.

$$ \begin{align} &\textbf{(a)} && (x + 2) \times \frac{1}{ x^2 - 4 } \\ &&&= (x + 2) \times \frac{1}{ (x + 2)(x - 2) } \\ &&&= \frac1{ x - 2 } \qquad \checkmark \\ \\ &\textbf{(b)} &&\frac{1}{ a^2 + 6a + 9 } \times \frac{ a^2 - 9 }{ 2 } \\ &&&= \frac{1}{ (a+3)^2 } \times \frac{ (a+3)(a-3) }{ 2 } \\ &&&= \frac{ a - 3 }{ 2(a+3) } \qquad \checkmark \\ \\ &\textbf{(c)} && \frac{ x^2 - 3x }{ y^2 + y } \times \frac{ y + 1 }{ x } \\ &&&= \frac{ x(x-3) }{ y(y+1) } \times \frac{ y + 1 }{ x } \\ &&&= \frac{ x - 3 }{ y } \qquad \checkmark \\ \\ &\textbf{(d)} && \frac{ y }{ y + 3 } \div \frac{ y^2 }{ y^2 + 4y + 3 } \\ &&&= \frac{ y }{ y + 3 } \times \frac{ (y+1)(y+3) }{ y^2 } \\ &&&= \frac{ y + 1 }{ y } \qquad \checkmark \\ \\ &\textbf{(e)} && \frac{ x^2 }{ 3 } \div \frac{ 2x^3 - 6x^2 }{ x^2 - 3x } \\ &&&= \frac{ x^2 }{ 3 } \times \frac{ x ( x - 3) }{ 2x^2 ( x - 3 ) } \\ &&&= \frac{ x }{ 6 } \qquad \checkmark \\ \\ &\textbf{(f)} && \frac{ 4x^2 - 25 }{ 4x - 10 } \div \frac{ 2x + 5 }{ 8 } \\ &&&= \frac{ (2x + 5)(2x - 5) }{ 2 (2x - 5) } \times \frac{ 8 }{ 2x + 5 } \\ &&&= 4 \qquad \checkmark \\ \\ &\textbf{(g)} && \frac{ x + 3 }{ x^2 + 10x + 25 } \times \frac{ x^2 + 25 }{ x^2 + 3x } \\ &&&= \frac{ x + 3 }{ (x + 5)^2 } \times \frac{ x ( x + 5 ) }{ x ( x + 3 ) } \\ &&&= \frac{1}{x+5} \qquad \checkmark \\ \\ &\textbf{(h)} && \frac{ 3y^2 + 4y - 4 }{ 10 } \div \frac{ 3y + 6 }{ 15 } \\ &&&= \frac{ (3y - 2)(y + 2) }{ 10 } \times \frac{ 15 }{ 3(y + 2) } \\ &&&= \frac{ 3y - 2 }{ 2 } \qquad \checkmark \\ \\ &\textbf{(i)} && \frac{ x^2 + 2xy + y^2 }{2} \times \frac{ 4 }{ (x - y)^2 } \\ &&&= \frac{ (x+y)^2 }{2} \times \frac{ 4 }{ (x - y)^2 } \\ &&&= \frac{ 2(x+y)^2 }{ (x - y)^2 } \qquad \checkmark \end{align} $$

 

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Question 3. (E/P) Show that $$ \frac{ x^2 - 64 }{ x^2 - 36 } \div \frac{ 64 - x^2 }{ x^2 - 36 } = -1 $$ [4 marks]

 

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

 

Solution.

$$ \begin{align} \textrm{L.H.S.} &= \frac{ x^2 - 64 }{ x^2 - 36 } \div \frac{ 64 - x^2 }{ x^2 - 36 } \\ &= \frac{ x^2 - 64 }{ x^2 - 36 } \times \frac{ x^2 - 36 }{ - \left( x^2 - 64 \right) } \\ &= -1 \\ &= \textrm{R.H.S.} \qquad \square \end{align} $$

 

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Question 4. (E/P) Show that $$ \frac{ 2x^2 - 11x - 40 }{ x^2 - 4x - 32 } \times \frac{ x^2 + 8x + 16 }{ 6x^2 - 3x - 45 } \div \frac{ 8x^2 + 20x - 48 }{ 10x^2 - 45x + 45 } = \frac{ a }{ b } $$ and find the values of the constants $a$ and $b$, where $a$ and $b$ are integers. [4 marks]

 

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

 

Solution.

$$ \begin{align} \textrm{L.H.S.} &= \frac{ 2x^2 - 11x - 40 }{ x^2 - 4x - 32 } \times \frac{ x^2 + 8x + 16 }{ 6x^2 - 3x - 45 } \div \frac{ 8x^2 + 20x - 48 }{ 10x^2 - 45x + 45 } \\ &= \frac{ (2x + 5)(x - 8) }{ (x + 4)(x - 8) } \times \frac{ (x + 4)^2 }{ 3 (2x + 5)( x - 3) } \times \frac{ 5(2x - 3)(x - 3) }{ 4(2x - 3)(x + 4) } \\ &= \frac{5}{12} = \frac{a}{b} \end{align} $$ Hence, we find $a = 5$ and $ b = 12$ with no common factors.

 

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Question 5. (E/P)

(a) Simplify fully: [3 marks]

$$ \frac{ x^2 + 2x - 24 }{ 2x^2 + 10x } \times \frac{ x^2 - 3x }{ x^2 + 3x - 18 } $$

(b) Given that $$ \ln \left[ \left( x^2 + 2x - 24 \right) \left( x^2 - 3x \right) \right] = 2 + \ln \left[ \left( 2x^2 + 10x \right) \left( x^2 + 3x - 18 \right) \right] $$ find $x$ in terms of $\textrm e$. [4 marks]

[Hint: Simplify and then solve the logarithmic equation. See P1 §14.6 Solving equations using logarithms.]

 

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

 

Solution.

(a) $$ \begin{align} & \frac{ x^2 + 2x - 24 }{ 2x^2 + 10x } \times \frac{ x^2 - 3x }{ x^2 + 3x - 18 } \\ &= \frac{ (x+6)(x-4) }{ 2x(x+5) } \times \frac{ x(x-3) }{ (x+6)(x-3) } \\ &= \frac{ x - 4 }{ 2(x+5) } \qquad \checkmark \end{align} $$

 

(b) $$ \begin{align} && \ln \left[ \left( x^2 + 2x - 24 \right) \left( x^2 - 3x \right) \right] &= 2 + \ln \left[ \left( 2x^2 + 10x \right) \left( x^2 + 3x - 18 \right) \right] \\ & \Rightarrow & \ln \underbrace{ \left[ \frac{ \left( x^2 + 2x - 24 \right) \left( x^2 - 3x \right) }{ \left( 2x^2 + 10x \right) \left( x^2 + 3x - 18 \right) } \right]}_{ = \frac{ x - 4 }{ 2(x+5) } } &= 2 \\ & \Rightarrow & \ln \left[ \frac{ x - 4 }{ 2(x+5) } \right] &= 2 \\ & \Rightarrow & \frac{ x - 4 }{ 2(x+5) } &= \textrm e^2 \\ & \Rightarrow & x - 4 &= 2 \textrm e^2 (x+5) \\ & \Rightarrow & \left( 1 - 2 \textrm e^2 \right) x &= 10 \textrm e^2 + 4 = 2 \left( 5 \textrm e^2 + 2 \right) \\ \\ & \Rightarrow & x &= \frac{ 2 \left( 5 \textrm e^2 + 2 \right) }{ 1 - 2 \textrm e^2 } \qquad \checkmark \end{align} $$

 

Note: A similar question: P2 Ch1 Mixed exercise 1 Q4.

 

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Question 6. (E/P) $$ \textrm f(x) = \frac{ 2x^2 - 3x - 2 }{ 6x - 8 } \div \frac{ x - 2 }{ 3x^2 + 14x - 24 } $$

(a) Show that: [4 marks]

$$ \textrm f(x) = \frac{ 2x^2 + 13x + 6 }{ 2 } $$

(b) Hence differentiate $\textrm f(x)$ and find $ \textrm f'(4)$. [3 marks]

[Hint: Differentiate each term separately. See P1 §12.5. Differentiating functions with two or more terms.]

 

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

 

Solution.

(a) $$ \begin{align} \textrm f(x) &= \frac{ 2x^2 - 3x - 2 }{ 6x - 8 } \div \frac{ x - 2 }{ 3x^2 + 14x - 24 } \\ &= \frac{ (2x+1)(x-2) }{ 2(3x-4) } \times \frac{ (3x - 4)(x + 6) }{ x - 2 } \\ &= \frac{ (2x+1)(x+6) }{ 2 } \\ &= \frac{ 2x^2 + 13x + 6 }{ 2 } \qquad \square \end{align} $$

 

(b) $$ \begin{align} && \textrm f(x) &= \frac{ 2x^2 + 13x + 6 }{ 2 } = x^2 + \frac{13}{2} x + 3 \\ & \Rightarrow & \textrm f'(x) &= 2x + \frac{13}{2} \\ & \Rightarrow & \textrm f'(4) &= 8 + \frac{13}{2} = \frac{29}{2} = 14.5 \qquad \checkmark \end{align} $$

 

Note: A similar question: P2 Ch1 Mixed exercise 1 Q5.

 

 

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5. Addition and subtraction

To add or subtract two fractions, find a common denominator.

Example 1. Simplify the following: $$ \begin{align} &\textbf{(a)} \qquad \frac{1}{3} + \frac{3}{4} \\ \\ &\textbf{(b)} \qquad \frac{a}{2x} + \frac{b}{3x} \\ \\ &\textbf{(c)} \qquad \frac{2}{x+3} - \frac{1}{x+1} \\ \\ &\textbf{(d)} \qquad \frac{3}{x+1} - \frac{4x}{x^2-1} \end{align} $$

 

(Edexcel 2017 Specifications, P2 Ch1 Example 7.)

 

Solution.

$$ \begin{align} &\textbf{(a)} \qquad \frac{1}{3} + \frac{3}{4} = \frac{4}{12} + \frac{9}{12} = \frac{13}{12} \qquad \checkmark \\ \\ &\textbf{(b)} \qquad \frac{a}{2x} + \frac{b}{3x} = \frac{3a + 2b}{6x} \qquad \checkmark \\ \\ &\textbf{(c)} \qquad \frac{2}{x+3} - \frac{1}{x+1} = \frac{2(x+1)-(x+3)}{(x+3)(x+1)}=\frac{x-1}{(x+3)(x+1)} \qquad \checkmark \\ \\ &\textbf{(d)} \qquad \frac{3}{x+1} - \frac{4x}{x^2-1} = \frac{3(x-1)-4x}{(x+1)(x-1)} = \frac{-x-3}{(x+1)(x-1)} \qquad \checkmark \end{align} $$

 

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6. Edexcel P2 Ch1 Exercise 1C

Question 1. Write as a single fraction: $$ \begin{align} &\textbf{(a)} \qquad \frac13 + \frac14 \\ \\ &\textbf{(b)} \qquad \frac34 - \frac25 \\ \\ &\textbf{(c)} \qquad \frac1p + \frac1q \\ \\ &\textbf{(d)} \qquad \frac{3}{4x} + \frac{1}{8x} \\ \\ &\textbf{(e)} \qquad \frac{3}{x^2} - \frac{1}{x} \\ \\ &\textbf{(f)} \qquad \frac{a}{5b} - \frac{3}{2b} \end{align} $$

 

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

 

Solution.

$$ \begin{align} &\textbf{(a)} \qquad \frac13 + \frac14 = \frac{4+3}{12} = \frac{7}{12} \qquad \checkmark \\ \\ &\textbf{(b)} \qquad \frac34 - \frac25 = \frac{15 - 8}{20} = \frac{7}{20} \qquad \checkmark \\ \\ &\textbf{(c)} \qquad \frac1p + \frac1q = \frac{p+q}{pq} \qquad \checkmark \\ \\ &\textbf{(d)} \qquad \frac{3}{4x} + \frac{1}{8x} = \frac{6 + 1 }{8x} = \frac{7}{8x} \qquad \checkmark \\ \\ &\textbf{(e)} \qquad \frac{3}{x^2} - \frac{1}{x} = \frac{3-x}{x^2} \qquad \checkmark \\ \\ &\textbf{(f)} \qquad \frac{a}{5b} - \frac{3}{2b} = \frac{2a-15}{10b} \qquad \checkmark \end{align} $$

 

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Question 2. Write as a single fraction: $$ \begin{align} &\textbf{(a)} \qquad \frac{3}{x} - \frac{2}{x+1} \\ \\ &\textbf{(b)} \qquad \frac{2}{x-1} - \frac{3}{x+2} \\ \\ &\textbf{(c)} \qquad \frac{4}{2x+1} + \frac{2}{x-1} \\ \\ &\textbf{(d)} \qquad \frac13 (x+2) - \frac12 (x+3) \\ \\ &\textbf{(e)} \qquad \frac{3x}{ (x+4)^2 } - \frac{1}{x+4} \\ \\ &\textbf{(f)} \qquad \frac{5}{ 2(x+3) } + \frac{4}{ 3(x-1) } \end{align} $$

 

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

 

Solution.

$$ \begin{align} &\textbf{(a)} \qquad \frac{3}{x} - \frac{2}{x+1} = \frac{ 3(x+1) - 2x }{ x ( x+1) } = \frac{ x + 3 }{ x ( x+1) } \qquad \checkmark \\ \\ &\textbf{(b)} \qquad \frac{2}{x-1} - \frac{3}{x+2} = \frac{ 2(x+2) - 3(x-1) }{ (x-1)(x+2) } = \frac{ -x + 7 }{ (x-1)(x+2) } \qquad \checkmark \\ \\ &\textbf{(c)} \qquad \frac{4}{2x+1} + \frac{2}{x-1} = \frac{ 4(x-1) + 2(2x+1) }{ (2x+1)(x-1) } = \frac{ 8x -2 }{ (2x+1)(x-1) } = \frac{ 2 (4x - 1) }{ (2x+1)(x-1) } \qquad \checkmark \\ \\ &\textbf{(d)} \qquad \frac13 (x+2) - \frac12 (x+3) = \frac{ 2(x+2) - 3(x+3) }{6} = \frac{ -x - 5 }{6} \qquad \checkmark \\ \\ &\textbf{(e)} \qquad \frac{3x}{ (x+4)^2 } - \frac{1}{x+4} = \frac{ 3x - (x+4) }{ (x+4)^2 } = \frac{ 2x - 4 }{ (x+4)^2 } = \frac{ 2(x-2) }{ (x+4)^2 } \qquad \checkmark \\ \\ &\textbf{(f)} \qquad \frac{5}{ 2(x+3) } + \frac{4}{ 3(x-1) } = \frac{ 15(x-1) + 8(x+3) }{ 6(x+3)(x-1) } = \frac{ 23x + 9 }{ 6(x+3)(x-1) } \qquad \checkmark \end{align} $$

 

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Question 3. Write as a single fraction: $$ \begin{align} &\textbf{(a)} \qquad \frac{ 2 }{ x^2 + 2x + 1 } + \frac{ 1 }{ x + 1 } \\ \\ &\textbf{(b)} \qquad \frac{ 7 }{ x^2 - 4 } + \frac{ 3 }{ x + 2 } \\ \\ &\textbf{(c)} \qquad \frac{ 2 }{ x^2 + 6x + 9 } - \frac{ 3 }{ x^2 + 4x + 3 } \\ \\ &\textbf{(d)} \qquad \frac{ 2 }{ y^2 - x^2 } + \frac{ 3 }{ y - x } \\ \\ &\textbf{(e)} \qquad \frac{ 3 }{ x^2 +3x + 2 } - \frac{ 1 }{ x^2 + 4x + 4 } \\ \\ &\textbf{(f)} \qquad \frac{ x + 2 }{ x^2 - x - 12 } - \frac{ x + 1 }{ x^2 +5x + 6 } \end{align} $$

 

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

 

Solution.

(a) $$ \begin{align} \frac{ 2 }{ x^2 + 2x + 1 } + \frac{ 1 }{ x + 1 } &= \frac{ 2 }{ (x + 1)^2 } + \frac{ 1 }{ x + 1 } \\ &= \frac{ 2 + (x+1) }{ (x + 1)^2 } \\ &= \frac{ x + 3 }{ (x + 1)^2 } \qquad \checkmark \end{align} $$ (b) $$ \begin{align} \frac{ 7 }{ x^2 - 4 } + \frac{ 3 }{ x + 2 } &= \frac{ 7 }{ (x+2) (x-2) } + \frac{ 3 }{ x + 2 } \\ &= \frac{ 7 + 3 (x-2) }{ (x+2) (x-2) } \\ &= \frac{ 3x + 1 }{ (x+2) (x-2) } \qquad \checkmark \end{align} $$ (c) $$ \begin{align} \frac{ 2 }{ x^2 + 6x + 9 } - \frac{ 3 }{ x^2 + 4x + 3 } &= \frac{ 2 }{ (x+3)^2 } - \frac{ 3 }{ (x+1)(x+3) } \\ &= \frac{ 2 (x+1) - 3 (x+3) }{ (x+1)(x+3)^2 } \\ &= \frac{ - x - 7 }{ (x+1)(x+3)^2 } \qquad \checkmark \end{align} $$ (d) $$ \begin{align} \frac{ 2 }{ y^2 - x^2 } + \frac{ 3 }{ y - x } &= \frac{ 2 }{ (y-x)(y+x) } + \frac{ 3 }{ y - x } \\ &= \frac{ 2 + 3 (x+y) }{ (y-x)(y+x) } \\ &= \frac{ 3x + 3y + 2 }{ (y-x)(y+x) } \qquad \checkmark \end{align} $$ (e) $$ \begin{align} \frac{ 3 }{ x^2 +3x + 2 } - \frac{ 1 }{ x^2 + 4x + 4 } &= \frac{ 3 }{ (x+1)(x+2) } - \frac{ 1 }{ (x+2)^2 } \\ &= \frac{ 3 (x+2) - (x+1) }{ (x+1)(x+2)^2 } \\ &= \frac{ (3x+6) - (x+1) }{ (x+1)(x+2)^2 } \\ &= \frac{ 2x + 5 }{ (x+1)(x+2)^2 } \qquad \checkmark \end{align} $$ (f) $$ \begin{align} \frac{ x + 2 }{ x^2 - x - 12 } - \frac{ x + 1 }{ x^2 +5x + 6 } &= \frac{ x + 2 }{ (x+3)(x-4) } - \frac{ x + 1 }{ (x+2)(x+3) } \\ &= \frac{ (x+2)^2 - (x+1)(x-4) }{ (x+2) (x+3) (x-4) } \\ &= \frac{ \left( x^2 + 4x + 4 \right) - \left( x^2 - 3x - 4 \right) }{ (x+2) (x+3) (x-4) } \\ &= \frac{ 7x + 8 }{ (x+2) (x+3) (x-4) } \qquad \checkmark \end{align} $$

 

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Question 4. (E) Express $$ \frac{ 6x + 1 }{ x^2 + 2x - 15 } - \frac{ 4 }{ x - 3 } $$ as a single fraction in its simplest form. [4 marks]

 

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

 

Solution.

$$ \begin{align} \frac{ 6x + 1 }{ x^2 + 2x - 15 } - \frac{ 4 }{ x - 3 } &= \frac{ 6x + 1 }{ (x-3)(x+5) } - \frac{ 4 }{ x - 3 } \\ &= \frac{ (6x + 1) - 4(x+5) }{ (x-3)(x+5) } \\ &= \frac{ (6x + 1) - (4x+20) }{ (x-3)(x+5) } \\ &= \frac{ 2x - 19 }{ (x-3)(x+5) } \qquad \checkmark \end{align} $$

 

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Question 5. Express each of the following as a fraction in its simplest form. $$ \begin{align} &\textbf{(a)} \qquad \frac{ 3 }{ x } + \frac{ 2 }{ x + 1 } + \frac{ 1 }{ x + 2 } \\ \\ &\textbf{(b)} \qquad \frac{ 4 }{ 3x } - \frac{ 2 }{ x - 2 } + \frac{ 1 }{ 2x + 1 } \\ \\ &\textbf{(c)} \qquad \frac{ 3 }{ x - 1 } + \frac{ 2 }{ x + 1 } + \frac{ 4 }{ x - 3 } \end{align} $$

 

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

 

Solution.

(a) $$ \begin{align} \frac{ 3 }{ x } + \frac{ 2 }{ x + 1 } + \frac{ 1 }{ x + 2 } &= \frac{ 3 (x+1) (x+2) + 2 x (x+2) + x(x+1) }{ x (x+1) (x+2) } \\ &= \frac{ 3 \left( x^2 + 3x + 2 \right) + 2 \left( x^2 + 2x \right) + \left( x^2 + x \right) }{ x (x+1) (x+2) } \\ &= \frac{ \left( 3x^2 + 9x + 6 \right) + \left( 2x^2 + 4x \right) + \left( x^2 + x \right) }{ x (x+1) (x+2) } \\ &= \frac{ 6x^2 + 14x + 6 }{ x (x+1) (x+2) } \qquad \checkmark \end{align} $$ (b) $$ \begin{align} \frac{ 4 }{ 3x } - \frac{ 2 }{ x - 2 } + \frac{ 1 }{ 2x + 1 } &= \frac{ 4 (x-2) (2x+1) - 6x (2x+1) + 3x (x-2) }{ 3x (x - 2) (2x + 1) } \\ &= \frac{ 4 \left( 2x^2 - 3x - 2 \right) - \left( 12x^2 + 6x \right) + \left( 3x^2 - 6x \right) }{ 3x (x - 2) (2x + 1) } \\ &= \frac{ \left( 8x^2 - 12x - 8 \right) - \left( 12x^2 + 6x \right) + \left( 3x^2 - 6x \right) }{ 3x (x - 2) (2x + 1) } \\ &= \frac{ -x^2 - 24x - 8 }{ 3x (x - 2) (2x + 1) } \qquad \checkmark \end{align} $$ (c) $$ \begin{align} \frac{ 3 }{ x - 1 } + \frac{ 2 }{ x + 1 } + \frac{ 4 }{ x - 3 } &= \frac{ 3 (x+1) (x-3) + 2 (x-1) (x-3) + 4 (x-1) (x+1) }{ (x - 1) (x + 1) (x - 3) } \\ &= \frac{ 3 \left( x^2 - 2x - 3 \right) + 2 \left( x^2 - 4x + 3 \right) + 4 \left( x^2 - 1 \right) }{ (x - 1) (x + 1) (x - 3) } \\ &= \frac{ \left( 3x^2 - 6x - 9 \right) + \left( 2x^2 - 8x + 6 \right) + \left( 4x^2 - 4 \right) }{ (x - 1) (x + 1) (x - 3) } \\ &= \frac{ 9x^2 - 14x - 7 }{ (x - 1) (x + 1) (x - 3) } \qquad \checkmark \end{align} $$

 

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Question 6. (E) Express $$ \frac{ 4 (2x - 1) }{ 36x^2 - 1 } + \frac{ 7 }{ 6x - 1 } $$ as a single fraction in its simplest form. [4 marks]

 

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

 

Solution.

$$ \begin{align} \frac{ 4 (2x - 1) }{ 36x^2 - 1 } + \frac{ 7 }{ 6x - 1 } &= \frac{ 4 (2x - 1) }{ (6x + 1)(6x - 1) } + \frac{ 7 }{ 6x - 1 } \\ &= \frac{ 4 (2x - 1) + 7 (6x + 1) }{ (6x + 1)(6x - 1) } \\ &= \frac{ 8x - 4 + 42x + 7 }{ (6x + 1)(6x - 1) } \\ &= \frac{ 50x + 3 }{ (6x + 1)(6x - 1) } \qquad \checkmark \end{align}$$

 

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Question 7. (E/P) $$ \textrm g(x) = x + \frac{ 6 }{ x + 2 } + \frac{ 36 }{ x^2 - 2x - 8 }, \qquad x \in \mathbb R, \quad x \ne -2, \quad x \ne 4 $$

(a) Show that $$ \textrm g(x) = \frac{ x^3 - 2x^2 - 2x + 12 }{ (x + 2)(x - 4) } $$ [4 marks]

 

(b) Using algebraic long division, or otherwise, further show that $$ \textrm g(x) = \frac{ x^2 - 4x + 6 } { x - 4 } $$ [4 marks]

 

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

 

Solution.

(a) $$ \begin{align} \textrm g(x) &= x + \frac{ 6 }{ x + 2 } + \frac{ 36 }{ x^2 - 2x - 8 } \\ &= x + \frac{ 6 }{ x + 2 } + \frac{ 36 }{ (x + 2)(x - 4) } \\ &= \frac{ x \left( x^2 - 2x - 8 \right) + 6 (x - 4) + 36 }{ (x + 2)(x - 4) } \\ &= \frac{ \left( x^3 - 2x^2 - 8x \right) + (6x - 24) + 36 }{ (x + 2)(x - 4) } \\ &= \frac{ x^3 - 2x^2 - 2x + 12 }{ (x + 2)(x - 4) }. \qquad \square \end{align} $$

 

(b) Let the numerator be $ \textrm f(x) $, i.e. $$ \textrm f(x) = x^3 - 2x^2 - 2x + 12 $$ We notice that (by looking at the final result, the factor $(x+2)$ seems to have been cancelled) $$ \begin{align} \textrm f(-2) &= (-2)^3 - 2(-2)^2 - 2(-2) + 12 \\ &= -8 - 8 + 4 + 12 \\ &= 0 \end{align} $$ and, by factor theorem, we find $$ \begin{align} \textrm f(x) &= x^3 - 2x^2 - 2x + 12 = (x+2) \left( x^2 -4x + 6 \right) \\ \\ \Rightarrow \qquad \textrm g(x) &= \frac{ \textrm f(x) }{ (x + 2)(x - 4) } = \frac{ (x+2) \left( x^2 -4x + 6 \right) }{ (x + 2)(x - 4) } = \frac{ x^2 - 4x + 6 } { x - 4 }. \qquad \square \end{align} $$

 

Aside 1. Alternatively, by long division, we find: $$ \begin{align} \frac{ x^3 - 2x^2 - 2x + 12 }{ x + 2 } &= x^2 - 4x + 6 \\ \\ \Rightarrow \qquad \textrm g(x) = \frac{ x^3 - 2x^2 - 2x + 12 }{ (x + 2)(x - 4) } &= \frac{ x^2 - 4x + 6 } { x - 4 } \qquad \checkmark \end{align} $$ However, this approach still anticipates that $(x+2)$ would be a factor of the numerator, $ \textrm f(x) $. For a more general approach with long division, see aside 2 below.

 

Aside 2. More generally, long division for $ \textrm g(x) $ gives $$ \begin{align} \textrm g(x) = \frac{ x^3 - 2x^2 - 2x + 12 }{ x^2 - 2x - 8 } &= x + \frac{ 6 (x + 2) }{ (x+2)(x-4) } \\ &= x + \frac{ 6 }{ x - 4 } \\ &= \frac{ x (x-4) + 6 }{ x - 4 } \\ &= \frac{ x^2 - 4x + 6 }{ x - 4 } \qquad \checkmark \end{align} $$