A-Level Algebraic and partial fractions Practice Questions

Free A-Level Algebraic and partial fractions practice questions with full step-by-step worked solutions. Covers algebraic-fractions, simplify, combine, partial-fractions. Practise exam-style problems and check your method.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Simplify x2+5x+6x2+x−2\frac{x^{2} + 5 x + 6}{x^{2} + x - 2}.
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Worked solution

  1. Write the fraction

    x2+5x+6x2+x−2\frac{x^{2} + 5 x + 6}{x^{2} + x - 2}

    Start from the given algebraic fraction.

  2. Factorise and cancel the common factor

    (x+2)(x+3)(x−1)(x+2)=x+3x−1\frac{\left(x + 2\right) \left(x + 3\right)}{\left(x - 1\right) \left(x + 2\right)}=\frac{x + 3}{x - 1}

    Cancel the factor common to numerator and denominator.

  3. State the simplified fraction

    x+3x−1\frac{x + 3}{x - 1}

    This is the fraction in its simplest form.

Answer
x+3x−1\frac{x + 3}{x - 1}
Question 2
2 markseasy
Which of the following is the correct form of the partial fraction decomposition of x+1x2+3x−4\frac{x + 1}{x^{2} + 3 x - 4}?
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Worked solution

  1. Write the denominator

    x2+3x−4x^{2} + 3 x - 4

    Look at the denominator to decide the form.

  2. Factorise the denominator

    x2+3x−4=(x−1)(x+4)x^{2} + 3 x - 4=\left(x - 1\right) \left(x + 4\right)

    Each linear factor needs its own partial fraction.

  3. State the correct form of the decomposition

    Ax+4+Bx−1\frac{A}{x + 4}+\frac{B}{x - 1}

    This form matches every factor of the denominator.

Answer
Ax+4+Bx−1\frac{A}{x + 4}+\frac{B}{x - 1}
Question 3
3 marksintermediate
Which of the following is the correct partial fraction decomposition of 4x+7x2+7x+10\frac{4 x + 7}{x^{2} + 7 x + 10}?
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Worked solution

  1. Write the fraction to be decomposed

    4x+7x2+7x+10\frac{4 x + 7}{x^{2} + 7 x + 10}

    We start from the given rational function.

  2. Factorise the denominator

    x2+7x+10=(x+2)(x+5)x^{2} + 7 x + 10=\left(x + 2\right) \left(x + 5\right)

    Factorising shows which linear factors appear.

  3. Write the partial fraction form with unknown constants

    4x+7(x+2)(x+5)=Ax+2+Bx+5\frac{4 x + 7}{\left(x + 2\right) \left(x + 5\right)}=\frac{A}{x + 2}+\frac{B}{x + 5}

    Each distinct factor (and each power of a repeated factor) needs its own term.

  4. Cover up x + 2 and substitute x=-2

    −1=A (3) ⇒ A=−13-1=A\,(3)\ \Rightarrow\ A=- \frac{1}{3}

    Multiplying by the factor and substituting its root isolates one constant.

  5. Cover up x + 5 and substitute x=-5

    −13=B (−3) ⇒ B=133-13=B\,(-3)\ \Rightarrow\ B=\frac{13}{3}

    Multiplying by the factor and substituting its root isolates one constant.

  6. Select the correct decomposition

    4x+7(x+2)(x+5)=133(x+5)−13(x+2)\frac{4 x + 7}{\left(x + 2\right) \left(x + 5\right)}=\frac{13}{3 \left(x + 5\right)} - \frac{1}{3 \left(x + 2\right)}

    This is the fully correct partial fraction decomposition.

Answer
133(x+5)−13(x+2)\frac{13}{3 \left(x + 5\right)} - \frac{1}{3 \left(x + 2\right)}
Question 4
5 markshard
Which of the following is the correct partial fraction decomposition of 3x+5x2+2x+1\frac{3 x + 5}{x^{2} + 2 x + 1}?
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Worked solution

  1. Write the fraction to be decomposed

    3x+5x2+2x+1\frac{3 x + 5}{x^{2} + 2 x + 1}

    We start from the given rational function.

  2. Factorise the denominator

    x2+2x+1=(x+1)2x^{2} + 2 x + 1=\left(x + 1\right)^{2}

    Factorising shows which linear factors appear.

  3. Write the partial fraction form with unknown constants

    3x+5(x+1)2=Ax+1+B(x+1)2\frac{3 x + 5}{\left(x + 1\right)^{2}}=\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}

    Each distinct factor (and each power of a repeated factor) needs its own term.

  4. Cover up \left(x + 1\right)^{2} and substitute x=-1

    2=B (1) ⇒ B=22=B\,(1)\ \Rightarrow\ B=2

    Multiplying by the factor and substituting its root isolates one constant.

  5. Substitute x=0 to find A

    x=0: A=3x=0:\ A=3

    With the other constants known, any convenient value gives the last one.

  6. Multiply through by the denominator to form an identity

    3x+5=A(x+1)+B3 x + 5=A \left(x + 1\right) + B

    Clearing the fractions gives an identity true for all x.

  7. Expand the right-hand side

    3x+5=Ax+A+B3 x + 5=A x + A + B

    Expanding lets us compare like terms.

  8. Compare coefficients of like powers of x

    A=3, A+B=5A=3,\ A + B=5

    Matching coefficients confirms the values of the constants.

  9. Express the assumed form over a common denominator

    Ax+1+B(x+1)2=A(x+1)+Bx2+2x+1\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}=\frac{A \left(x + 1\right) + B}{x^{2} + 2 x + 1}

    Recombining shows the numerators must match.

  10. Select the correct decomposition

    3x+5(x+1)2=3x+1+2(x+1)2\frac{3 x + 5}{\left(x + 1\right)^{2}}=\frac{3}{x + 1} + \frac{2}{\left(x + 1\right)^{2}}

    This is the fully correct partial fraction decomposition.

Answer
3x+1+2(x+1)2\frac{3}{x + 1} + \frac{2}{\left(x + 1\right)^{2}}
Question 5
8 markschallenging
Which of the following is the correct partial fraction decomposition of 2x+3x3−3x−2\frac{2 x + 3}{x^{3} - 3 x - 2}?
Show worked solution

Worked solution

  1. Write the fraction to be decomposed

    2x+3x3−3x−2\frac{2 x + 3}{x^{3} - 3 x - 2}

    We start from the given rational function.

  2. Factorise the denominator

    x3−3x−2=(x−2)(x+1)2x^{3} - 3 x - 2=\left(x - 2\right) \left(x + 1\right)^{2}

    Factorising shows which linear factors appear.

  3. Write the partial fraction form with unknown constants

    2x+3(x−2)(x+1)2=Ax+1+B(x+1)2+Cx−2\frac{2 x + 3}{\left(x - 2\right) \left(x + 1\right)^{2}}=\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}+\frac{C}{x - 2}

    Each distinct factor (and each power of a repeated factor) needs its own term.

  4. Cover up \left(x + 1\right)^{2} and substitute x=-1

    1=B (−3) ⇒ B=−131=B\,(-3)\ \Rightarrow\ B=- \frac{1}{3}

    Multiplying by the factor and substituting its root isolates one constant.

  5. Cover up x - 2 and substitute x=2

    7=C (9) ⇒ C=797=C\,(9)\ \Rightarrow\ C=\frac{7}{9}

    Multiplying by the factor and substituting its root isolates one constant.

  6. Substitute x=0 to find A

    x=0: A=−79x=0:\ A=- \frac{7}{9}

    With the other constants known, any convenient value gives the last one.

  7. Multiply through by the denominator to form an identity

    2x+3=A(x2−x−2)+B(x−2)+C(x2+2x+1)2 x + 3=A \left(x^{2} - x - 2\right) + B \left(x - 2\right) + C \left(x^{2} + 2 x + 1\right)

    Clearing the fractions gives an identity true for all x.

  8. Expand the right-hand side

    2x+3=Ax2−Ax−2A+Bx−2B+Cx2+2Cx+C2 x + 3=A x^{2} - A x - 2 A + B x - 2 B + C x^{2} + 2 C x + C

    Expanding lets us compare like terms.

  9. Compare coefficients of like powers of x

    A+C=0, −A+B+2C=2, −2A−2B+C=3A + C=0,\ - A + B + 2 C=2,\ - 2 A - 2 B + C=3

    Matching coefficients confirms the values of the constants.

  10. Express the assumed form over a common denominator

    Ax+1+B(x+1)2+Cx−2=A(x2−x−2)+B(x−2)+C(x2+2x+1)x3−3x−2\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}+\frac{C}{x - 2}=\frac{A \left(x^{2} - x - 2\right) + B \left(x - 2\right) + C \left(x^{2} + 2 x + 1\right)}{x^{3} - 3 x - 2}

    Recombining shows the numerators must match.

  11. Verify by recombining the partial fractions

    −79(x+1)−13(x+1)2+79(x−2)=2x+3(x−2)(x+1)2- \frac{7}{9 \left(x + 1\right)} - \frac{1}{3 \left(x + 1\right)^{2}} + \frac{7}{9 \left(x - 2\right)}=\frac{2 x + 3}{\left(x - 2\right) \left(x + 1\right)^{2}}

    Adding the fractions back returns the original expression.

  12. Check the result at a test value

    x=0: −32=−32x=0:\ - \frac{3}{2}=- \frac{3}{2}

    Both sides agree, confirming the decomposition.

  13. Note the degrees involved

    deg⁡N=1, deg⁡D=3\deg N=1,\ \deg D=3

    The degree comparison determines whether division is needed.

  14. Recall the cover-up method

    Multiply by a factor, then substitute its root.\text{Multiply by a factor, then substitute its root.}

    The cover-up method is the quickest route to each constant.

  15. Select the correct decomposition

    2x+3(x−2)(x+1)2=−79(x+1)−13(x+1)2+79(x−2)\frac{2 x + 3}{\left(x - 2\right) \left(x + 1\right)^{2}}=- \frac{7}{9 \left(x + 1\right)} - \frac{1}{3 \left(x + 1\right)^{2}} + \frac{7}{9 \left(x - 2\right)}

    This is the fully correct partial fraction decomposition.

Answer
−79(x+1)−13(x+1)2+79(x−2)- \frac{7}{9 \left(x + 1\right)} - \frac{1}{3 \left(x + 1\right)^{2}} + \frac{7}{9 \left(x - 2\right)}

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