Hard A-Level Polynomials and factor theorem Questions

Challenging, exam-style A-Level Polynomials and factor theorem questions with worked solutions. Stretch yourself on the hardest factorising, quadratic in disguise, difference of two squares, factor theorem problems.

factorisingquadratic in disguisedifference of two squaresfactor theoremalgebraic divisionsolving cubics
A-Level34 questionsStep-by-step solutions
Question 1
6 markschallenging
A monic cubic f(x)=x3+ax2+bx+cf(x)=x^3+ax^2+bx+c has three roots that are consecutive integers, and f(0)=−60f(0)=-60. Given the roots are positive, find f(x)f(x) in expanded form.
Show worked solution

Worked solution

  1. Name the consecutive roots

    n−1, n, n+1n-1,\ n,\ n+1

    Three consecutive integers can be written around a middle value n. Using this structure is the key insight.

  2. Write f(x) in factor form

    f(x)=(x−(n−1))(x−n)(x−(n+1))f(x)=(x-(n-1))(x-n)(x-(n+1))

    A monic cubic with those roots is the product of the three corresponding factors.

  3. Use the value at 0

    f(0)=−(n−1)⋅(−n)⋅−(n+1)f(0)=-(n-1)\cdot(-n)\cdot-(n+1)

    Substitute x = 0; each factor becomes the negative of a root.

  4. Simplify f(0)

    f(0)=−(n−1)n(n+1)f(0)=-(n-1)n(n+1)

    Three minus signs multiply to give one overall minus sign.

  5. Form an equation

    −(n−1)n(n+1)=−60-(n-1)n(n+1)=-60

    Set this equal to the given value f(0) = -60.

  6. Simplify

    (n−1)n(n+1)=60(n-1)n(n+1)=60

    Divide both sides by -1: the product of three consecutive integers is 60.

  7. Solve by inspection

    3×4×5=60⇒n=43\times4\times5=60\Rightarrow n=4

    Since 3 times 4 times 5 is 60, the middle integer is n = 4. The roots are positive, as required.

  8. State the roots

    x=3, 4, 5x=3,\ 4,\ 5

    The three consecutive integer roots are 3, 4 and 5.

  9. Multiply the first two factors

    (x−3)(x−4)=x2−7x+12(x-3)(x-4)=x^2-7x+12

    Expand two brackets at a time.

  10. Multiply by the third factor

    (x2−7x+12)(x−5)(x^2-7x+12)(x-5)

    Now multiply by the remaining factor (x - 5).

  11. Distribute

    x3−5x2−7x2+35x+12x−60x^3-5x^2-7x^2+35x+12x-60

    Multiply each term of the quadratic by x and then by -5.

  12. Collect like terms

    x3−12x2+47x−60x^3-12x^2+47x-60

    Combine x squared terms (-5 - 7 = -12) and x terms (35 + 12 = 47).

  13. Check the constant

    f(0)=−60 ✓f(0)=-60\ \checkmark

    The constant term is -60, matching the given condition, so the answer is confirmed.

  14. Explain why f(0) gives the product

    f(0)=(0−r1)(0−r2)(0−r3)=−r1r2r3f(0)=(0-r_1)(0-r_2)(0-r_3)=-r_1r_2r_3

    Substituting x = 0 into the factor form gives minus the product of the roots.

  15. Confirm the roots are positive

    3, 4, 5>0 ✓3,\ 4,\ 5>0\ \checkmark

    All three roots are positive, matching the condition stated in the question.

Answer
f(x)=x3−12x2+47x−60f(x)=x^3-12x^2+47x-60
Question 2
5 markschallenging
Identify the correct sketch of y=(x−2)2(x+1)y=(x-2)^2(x+1), showing the behaviour at each root and the yy-intercept.
Show worked solution

Worked solution

  1. Find the roots

    x=2 (repeated), x=−1 (simple)x=2\ (\text{repeated}),\ x=-1\ (\text{simple})

    Set each factor to zero. The squared factor gives a repeated root at x = 2.

  2. Interpret the repeated root

    (x−2)2⇒curve touches at x=2(x-2)^2\Rightarrow\text{curve touches at }x=2

    A squared factor means the curve touches and turns at x = 2 without crossing.

  3. Interpret the simple root

    (x+1)⇒curve crosses at x=−1(x+1)\Rightarrow\text{curve crosses at }x=-1

    A single factor means the curve passes straight through at x = -1.

  4. Find the y-intercept

    y=(−2)2(1)=4y=(-2)^2(1)=4

    Substitute x = 0 to find where the curve meets the y-axis.

  5. Determine the shape

    +x3⇒ up to the right+x^3\Rightarrow\ \text{up to the right}

    The leading term is positive x cubed, so the curve rises to the top-right and falls to the bottom-left.

  6. Expand the equation

    y=x3−3x2+4y=x^3-3x^2+4

    Expanding shows the standard cubic form, which is useful for finding turning points.

  7. State the y-intercept

    y(0)=4y(0)=4

    The curve meets the y-axis at 4.

  8. Differentiate for turning points

    y′=3x2−6x=3x(x−2)y'=3x^2-6x=3x(x-2)

    Setting the derivative to zero locates the turning points.

  9. Solve for turning points

    x=0 and x=2x=0\ \text{and}\ x=2

    The turning points occur at x = 0 and x = 2.

  10. Evaluate the local maximum

    y(0)=4y(0)=4

    At x = 0 there is a local maximum at height 4.

  11. Evaluate the local minimum

    y(2)=0y(2)=0

    At x = 2 there is a local minimum lying on the axis — this is the repeated root.

  12. Describe end behaviour

    x→+∞: y→+∞x\to+\infty:\ y\to+\infty

    A positive cubic rises to the top-right.

  13. Eliminate the reflected option

    reject falling-right curves\text{reject falling-right curves}

    The positive leading term rules out reflected shapes.

  14. Eliminate the negated-roots option

    touch must be at x=2\text{touch must be at }x=2

    The squared factor is (x - 2), so the touch is at 2, not at -2.

  15. Eliminate the quadratic option

    reject parabolas\text{reject parabolas}

    The equation is a cubic, not a quadratic.

Answer
Positive cubic crossing at x=−1 and touching at x=2, y-intercept (0,4)\text{Positive cubic crossing at }x=-1\text{ and touching at }x=2,\ y\text{-intercept }(0,4)
Question 3
6 markschallenging
Given that (x−1)2(x-1)^2 is a factor of f(x)=x3+ax+bf(x)=x^3+ax+b, find aa and bb and hence factorise f(x)f(x) completely.
Show worked solution

Worked solution

  1. Model the factorisation

    x3+ax+b=(x−1)2(x−r)x^3+ax+b=(x-1)^2(x-r)

    Since (x - 1) squared is a factor of a cubic, the remaining factor must be a single linear factor (x - r).

  2. Expand the square

    (x−1)2=x2−2x+1(x-1)^2=x^2-2x+1

    Square the repeated bracket first.

  3. Multiply by (x-r)

    (x2−2x+1)(x−r)(x^2-2x+1)(x-r)

    Now multiply by the unknown linear factor.

  4. Distribute

    x3−rx2−2x2+2rx+x−rx^3-rx^2-2x^2+2rx+x-r

    Multiply each term of the quadratic by x and then by -r.

  5. Collect like terms

    x3−(r+2)x2+(2r+1)x−rx^3-(r+2)x^2+(2r+1)x-r

    Group by power of x, ready to compare with f(x).

  6. Compare x-squared coefficients

    −(r+2)=0-(r+2)=0

    f(x) has no x squared term, so this coefficient must be zero.

  7. Solve for r

    r=−2r=-2

    So the third root is r = -2, giving the factor (x + 2).

  8. Find a from the x coefficient

    a=2r+1=2(−2)+1=−3a=2r+1=2(-2)+1=-3

    Match the coefficient of x to find a.

  9. Find b from the constant

    b=−r=−(−2)=2b=-r=-(-2)=2

    Match the constant term to find b.

  10. State a and b

    a=−3, b=2a=-3,\ b=2

    So f(x) = x cubed - 3x + 2.

  11. Write the full factorisation

    f(x)=(x−1)2(x+2)f(x)=(x-1)^2(x+2)

    Combine the repeated factor with (x + 2).

  12. Verify

    (x−1)2(x+2)=x3−3x+2 ✓(x-1)^2(x+2)=x^3-3x+2\ \checkmark

    Expanding confirms the cubic matches f(x) with a = -3, b = 2.

  13. Explain the repeated-factor model

    (x−1)2(x−r) for a cubic(x-1)^2(x-r)\ \text{for a cubic}

    A cubic with a squared factor must have exactly one more linear factor.

  14. Cross-check a

    a=2r+1=2(−2)+1=−3a=2r+1=2(-2)+1=-3

    Using the x-coefficient gives the same value a = -3, a reassuring check.

  15. Describe the graph

    touches at x=1, crosses at x=−2\text{touches at }x=1,\ \text{crosses at }x=-2

    The double root gives a touch at x = 1; the simple root gives a crossing at x = -2.

Answer
a=−3, b=2,  f(x)=(x−1)2(x+2)a=-3,\ b=2,\ \ f(x)=(x-1)^2(x+2)
Question 4
6 markschallenging
Express x3x^3 in the form A(x−1)3+B(x−1)2+C(x−1)+DA(x-1)^3+B(x-1)^2+C(x-1)+D, stating the values of A,B,C,DA,B,C,D.
Show worked solution

Worked solution

  1. Introduce a substitution

    let t=x−1⇒x=t+1\text{let }t=x-1\Rightarrow x=t+1

    Rewriting in terms of t = x - 1 turns the problem into expanding a cube. This is a neat change of variable.

  2. Rewrite the target

    x3=(t+1)3x^3=(t+1)^3

    Since x = t + 1, we just need to expand (t + 1) cubed.

  3. Expand the cube

    (t+1)3=t3+3t2+3t+1(t+1)^3=t^3+3t^2+3t+1

    Use the binomial expansion of (t + 1) cubed.

  4. Return to x-1

    =(x−1)3+3(x−1)2+3(x−1)+1=(x-1)^3+3(x-1)^2+3(x-1)+1

    Replace t with (x - 1) to write the expression in the required form.

  5. Read off A

    A=1A=1

    The coefficient of (x - 1) cubed is 1.

  6. Read off B

    B=3B=3

    The coefficient of (x - 1) squared is 3.

  7. Read off C

    C=3C=3

    The coefficient of (x - 1) is 3.

  8. Read off D

    D=1D=1

    The constant term is 1.

  9. Verify at x = 2

    1(1)+3(1)+3(1)+1=8=23 ✓1(1)+3(1)+3(1)+1=8=2^3\ \checkmark

    Substituting x = 2 gives 8 on both sides, confirming the identity.

  10. Motivate the method

    expand about x=1\text{expand about }x=1

    Rewriting in powers of (x - 1) is like performing a Taylor-style expansion around x = 1.

  11. Expand the square first

    (t+1)2=t2+2t+1(t+1)^2=t^2+2t+1

    Building the cube from the square keeps the algebra manageable.

  12. Multiply by (t+1)

    (t2+2t+1)(t+1)=t3+3t2+3t+1(t^2+2t+1)(t+1)=t^3+3t^2+3t+1

    Multiplying out confirms the cube expansion.

  13. Verify at x = 1

    RHS=0+0+0+1=1=13 ✓\text{RHS}=0+0+0+1=1=1^3\ \checkmark

    At x = 1 both sides equal 1.

  14. Verify at x = 0

    −1+3−3+1=0=03 ✓-1+3-3+1=0=0^3\ \checkmark

    At x = 0 both sides equal 0.

  15. State the coefficients together

    (A,B,C,D)=(1,3,3,1)(A,B,C,D)=(1,3,3,1)

    The coefficients are the binomial numbers 1, 3, 3, 1.

Answer
A=1, B=3, C=3, D=1A=1,\ B=3,\ C=3,\ D=1
Question 5
6 markschallenging
Solve the equation 2x3+3x2−1=02x^3+3x^2-1=0, stating clearly any repeated root.
Show worked solution

Worked solution

  1. List candidate roots

    x=±1,±12x=\pm1,\pm\tfrac12

    Rational roots are factors of the constant 1 over factors of the leading 2. Test these.

  2. Test x = -1

    2(−1)3+3(−1)2−1=−2+3−1=02(-1)^3+3(-1)^2-1=-2+3-1=0

    Substituting x = -1 gives zero, so (x + 1) is a factor.

  3. Divide by (x+1)

    2x3+3x2−1=(x+1)(2x2+x−1)2x^3+3x^2-1=(x+1)(2x^2+x-1)

    Include the missing x term as 0 when dividing. The quotient is 2x squared + x - 1.

  4. Factorise the quadratic

    2x2+x−1=(2x−1)(x+1)2x^2+x-1=(2x-1)(x+1)

    Split the middle term using +2 and -1, which multiply to -2 and add to +1.

  5. Notice the repeated factor

    (x+1)(2x−1)(x+1)=(x+1)2(2x−1)(x+1)(2x-1)(x+1)=(x+1)^2(2x-1)

    The factor (x + 1) appears twice, so it is a repeated (double) factor.

  6. Set factors to zero

    (x+1)2=0 or 2x−1=0(x+1)^2=0\ \text{or}\ 2x-1=0

    Solve each distinct factor for x.

  7. Solve for the roots

    x=−1 (repeated), x=12x=-1\ (\text{repeated}),\ x=\tfrac12

    x = -1 is a repeated root, and x = 1/2 is a simple root.

  8. Interpret the repeated root

    curve touches the x-axis at x=−1\text{curve touches the }x\text{-axis at }x=-1

    A double root means the graph just touches the axis at x = -1 rather than crossing it.

  9. Explain the candidate roots

    ±pq, p∣1, q∣2\pm\tfrac{p}{q},\ p\mid1,\ q\mid2

    The rational root theorem restricts the possible roots to these few values.

  10. Interpret the first factor

    f(−1)=0⇒(x+1)∣ff(-1)=0\Rightarrow(x+1)\mid f

    The zero value gives the first factor.

  11. Note the missing term

    2x3+3x2+0x−12x^3+3x^2+0x-1

    There is no x term, so we use a 0 placeholder when dividing.

  12. Confirm the repeated root

    f′(−1)=6−6=0f'(-1)=6-6=0

    The derivative is also zero at x = -1, confirming it is a repeated root.

  13. Verify the simple root

    f(12)=14+34−1=0 ✓f\left(\tfrac12\right)=\tfrac14+\tfrac34-1=0\ \checkmark

    Substituting x = 1/2 gives zero, confirming the simple root.

  14. Describe the graph

    touches at x=−1, crosses at x=12\text{touches at }x=-1,\ \text{crosses at }x=\tfrac12

    The double root means the curve touches the axis at -1 and crosses at 1/2.

  15. State the solution set

    x=−1 (repeated), x=12x=-1\ (\text{repeated}),\ x=\tfrac12

    The complete solution, noting the repeated root.

Answer
x=−1 (repeated), x=12x=-1\ (\text{repeated}),\ x=\tfrac12

Unlock 29 more Polynomials and factor theorem questions

Create a free account to work through every A-Level Polynomials and factor theorem question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Polynomials and factor theorem practice

Practice QuestionsWorked Solutions

Related Pure Maths topics

Areas and further applicationsMathematical argument and notationArithmetic sequences and seriesBinomial expansion (rational powers)Binomial expansionCircles