A-Level Polynomials and factor theorem Practice Questions

Free A-Level Polynomials and factor theorem practice questions with full step-by-step worked solutions. Covers expanding, double brackets, polynomial product, perfect square. Practise exam-style problems and check your method.

expandingdouble bracketspolynomial productperfect squarefactorisingquadratic
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Expand and simplify (x+3)(x+5)(x+3)(x+5).
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Worked solution

  1. Write out the product

    (x+3)(x+5)(x+3)(x+5)

    We must multiply everything in the first bracket by everything in the second bracket. This is the same double-bracket skill you met with quadratics.

  2. Multiply each pair of terms

    x⋅x+x⋅5+3⋅x+3⋅5x\cdot x + x\cdot 5 + 3\cdot x + 3\cdot 5

    Take each term in the first bracket and multiply it by each term in the second (often called FOIL). This gives four separate products.

  3. Simplify each product

    x2+5x+3x+15x^2 + 5x + 3x + 15

    Work out each multiplication. Remember x times x is x squared.

  4. Collect like terms

    x2+8x+15x^2 + 8x + 15

    The two x-terms, 5x and 3x, add to give 8x. Nothing else can be combined.

Answer
x2+8x+15x^2 + 8x + 15
Question 2
3 markseasy
Factorise x3+4x2+3xx^3 + 4x^2 + 3x completely.
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Worked solution

  1. Take out the common factor

    x3+4x2+3x=x(x2+4x+3)x^3+4x^2+3x=x(x^2+4x+3)

    Every term has at least one x, so x is a common factor. Taking it out leaves a quadratic inside.

  2. Factorise the quadratic

    x2+4x+3=(x+1)(x+3)x^2+4x+3=(x+1)(x+3)

    Find two numbers multiplying to 3 and adding to 4: they are 1 and 3.

  3. Combine all factors

    x(x+1)(x+3)x(x+1)(x+3)

    Bring the common factor and the two brackets together. 'Completely' means three linear factors here.

Answer
x(x+1)(x+3)x(x+1)(x+3)
Question 3
4 marksintermediate
An open-topped box has a square base of side xx cm and height (x+3)(x+3) cm; a second cuboid has dimensions xx, (x+1)(x+1) and (x+3)(x+3) cm. For the second cuboid, expand the volume V=x(x+1)(x+3)V=x(x+1)(x+3) and find VV when x=2x=2.
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Worked solution

  1. Multiply the first two factors

    x(x+1)=x2+xx(x+1)=x^2+x

    Start by combining two of the three dimensions. Multiply x through the bracket.

  2. Multiply by the third factor

    (x2+x)(x+3)(x^2+x)(x+3)

    Now multiply the result by the remaining dimension (x + 3).

  3. Distribute

    x3+3x2+x2+3xx^3+3x^2+x^2+3x

    Multiply each term by x and then by 3.

  4. Collect like terms

    V=x3+4x2+3xV=x^3+4x^2+3x

    Combine the x-squared terms to get the expanded volume.

  5. Substitute x = 2

    V=23+4(2)2+3(2)V=2^3+4(2)^2+3(2)

    Now put x = 2 into the expanded expression to find the volume.

  6. Evaluate

    V=8+16+6=30V=8+16+6=30

    The volume is 30 cubic centimetres.

Answer
V=x3+4x2+3x,  V(2)=30 cm3V=x^3+4x^2+3x,\ \ V(2)=30\text{ cm}^3
Question 4
5 markshard
The cubic x3+px2+qx+18x^3+px^2+qx+18 has (x−2)(x-2) and (x−3)(x-3) as factors. Find the third linear factor.
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Worked solution

  1. Name the third factor

    x3+px2+qx+18=(x−2)(x−3)(x−r)x^3+px^2+qx+18=(x-2)(x-3)(x-r)

    A monic cubic with these two factors must have a third linear factor (x - r) for some root r.

  2. Compare constant terms

    (−2)(−3)(−r)=18(-2)(-3)(-r)=18

    Multiplying the three constant terms in the brackets must give the constant 18 of the cubic.

  3. Simplify the left side

    −6r=18-6r=18

    Minus 2 times minus 3 is 6, times minus r gives -6r.

  4. Solve for r

    r=−3r=-3

    Divide both sides by -6 to find r = -3.

  5. Write the third factor

    (x−(−3))=(x+3)(x-(-3))=(x+3)

    The factor is (x - r) with r = -3, which is (x + 3).

  6. Verify by expanding

    (x−2)(x−3)(x+3)=x3−2x2−9x+18(x-2)(x-3)(x+3)=x^3-2x^2-9x+18

    Expanding confirms the constant is 18, matching the original cubic (with p = -2, q = -9).

  7. Recall the product-of-roots rule

    r1r2r3=−constantleadingr_1r_2r_3=-\tfrac{\text{constant}}{\text{leading}}

    For a monic cubic, multiplying the roots gives minus the constant term. This underlies the comparison.

  8. Compute the product of known roots

    2×3=62\times3=6

    The two known roots multiply to 6.

  9. Find the third root

    6×r3=−18⇒r3=−36\times r_3=-18\Rightarrow r_3=-3

    Since all three roots multiply to -18, the third root must be -3.

  10. State the factor

    (x+3)(x+3)

    A root of -3 corresponds to the linear factor (x + 3).

Answer
(x+3)(x+3)
Question 5
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.
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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

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