Hard GCSE Quadratics by factorising Questions

Challenging, exam-style GCSE Quadratics by factorising questions with worked solutions. Stretch yourself on the hardest common factor, null-factor law, zero root, difference of two squares problems.

common factornull-factor lawzero rootdifference of two squaresfactorising a quadraticrepeated root
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Two positive numbers differ by 55. Their product is 8484. Find the two numbers.
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Worked solution

  1. Name the numbers

    n and n+5n \text{ and } n + 5

    The numbers differ by 5, so call the smaller one n and the larger one n + 5.

  2. Form the equation

    n(n+5)=84n(n + 5) = 84

    Their product is 84.

  3. Expand and rearrange

    n2+5n=84⇒n2+5n−84=0n^2 + 5n = 84 \Rightarrow n^2 + 5n - 84 = 0

    Multiply out, then subtract 84 so the equation equals zero.

  4. Look for the pair of numbers

    product=−84,sum=5\text{product} = -84, \quad \text{sum} = 5

    For n2n^2 + bn + c the two numbers in the brackets must multiply to c=−84c = -84 and add to b=5b = 5.

  5. List the factor pairs of -84

    (−84)×1,(−42)×2,(−28)×3,(−21)×4,(−14)×6,(−12)×7,(−7)×12,(−6)×14,(−4)×21,(−3)×28,(−2)×42,(−1)×84(-84) \times 1,\quad (-42) \times 2,\quad (-28) \times 3,\quad (-21) \times 4,\quad (-14) \times 6,\quad (-12) \times 7,\quad (-7) \times 12,\quad (-6) \times 14,\quad (-4) \times 21,\quad (-3) \times 28,\quad (-2) \times 42,\quad (-1) \times 84

    Write out every pair of whole numbers whose product is -84, then test which pair also adds to 5.

  6. Choose the pair that works

    (−7)×12=−84,(−7)+12=5(-7) \times 12 = -84, \quad (-7) + 12 = 5

    -7 and 12 pass both tests, so they are the numbers that go in the brackets.

  7. Write the factorised equation

    (n−7)(n+12)=0(n - 7)(n + 12) = 0

    The product of the brackets is zero — this is the form the null-factor law needs.

  8. Check the factorisation by expanding

    (n−7)(n+12)=n2+5n−84(n - 7)(n + 12) = n^2 + 5n - 84

    Multiplying the brackets back out returns the original quadratic, so the factorisation is right.

  9. Apply the null-factor law

    n−7=0orn+12=0n - 7 = 0 \quad \text{or} \quad n + 12 = 0

    If several things multiply to give zero, at least one of them must be zero. So set each bracket equal to zero in turn.

  10. Solve bracket 1

    n−7=0⇒n=7n - 7 = 0 \Rightarrow n = 7

    Solve the little linear equation n−7=0n - 7 = 0 to get n=7n = 7.

  11. Solve bracket 2

    n+12=0⇒n=−12n + 12 = 0 \Rightarrow n = -12

    Solve the little linear equation n+12=0n + 12 = 0 to get n=−12n = -12.

  12. State the solutions

    n=7 or n=−12n = 7 \text{ or } n = -12

    Each bracket gives one solution, so a quadratic that factorises into two different brackets has two solutions.

  13. Reject the impossible solution

    n=−12 is rejected, so n=7n = -12 \text{ is rejected, so } n = 7

    Both numbers must be positive, so the negative root cannot be the answer.

  14. Answer the question that was asked

    7 and 127 \text{ and } 12

    The numbers are n=7n = 7 and n+5=12n + 5 = 12. Check: 12−7=512 - 7 = 5 and 7×12=847 \times 12 = 84.

  15. Check the solution n=7n = 7

    (7)2+5(7)−84=49+35−84=0(7)^2 + 5(7) - 84 = 49 + 35 - 84 = 0

    Substituting n=7n = 7 back into the original quadratic gives 0, so this solution is correct.

Answer
7 and 127 \text{ and } 12
Question 2
5 markschallenging
Solve 5x2=3x+25x^2 = 3x + 2 by factorising.
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Worked solution

  1. Rearrange so that one side is zero

    5x2−3x−2=05x^2 - 3x - 2 = 0

    Subtract 3x and 2 from both sides. Factorising can only solve the equation once it equals zero.

  2. Use the ac method

    a×c=5×(−2)=−10,sum=−3a \times c = 5 \times (-2) = -10, \quad \text{sum} = -3

    When the x2x^2 coefficient is not 11, look for two numbers that multiply to a×c=−10a \times c = -10 and add to b=−3b = -3.

  3. List the factor pairs of -10

    (−10)×1,(−5)×2,(−2)×5,(−1)×10(-10) \times 1,\quad (-5) \times 2,\quad (-2) \times 5,\quad (-1) \times 10

    Test each pair to see which one adds to -3.

  4. Choose the pair that works

    (−5)×2=−10,(−5)+2=−3(-5) \times 2 = -10, \quad (-5) + 2 = -3

    -5 and 2 multiply to -10 and add to -3.

  5. Split the middle term

    5x2−5x+2x−2=05x^2 - 5x + 2x - 2 = 0

    Rewrite -3x as -5x + 2x. The expression is worth exactly the same, but it can now be factorised in pairs.

  6. Factorise in pairs

    5x(x−1)+2(x−1)=05x(x - 1) + 2(x - 1) = 0

    Factorise the first two terms and the last two terms separately. The identical bracket (x - 1) appears in both — that is the sign that the split was right. Now that bracket itself comes out as a factor.

  7. Write the factorised equation

    (x−1)(5x+2)=0(x - 1)(5x + 2) = 0

    The product of the brackets is zero — this is the form the null-factor law needs.

  8. Check the factorisation by expanding

    (x−1)(5x+2)=5x2−3x−2(x - 1)(5x + 2) = 5x^2 - 3x - 2

    Multiplying the brackets back out returns the original quadratic, so the factorisation is right.

  9. Apply the null-factor law

    x−1=0or5x+2=0x - 1 = 0 \quad \text{or} \quad 5x + 2 = 0

    If several things multiply to give zero, at least one of them must be zero. So set each bracket equal to zero in turn.

  10. Solve bracket 1

    x−1=0⇒x=1x - 1 = 0 \Rightarrow x = 1

    Solve the little linear equation x−1=0x - 1 = 0 to get x=1x = 1.

  11. Solve bracket 2

    5x+2=0⇒x=−255x + 2 = 0 \Rightarrow x = -\frac{2}{5}

    Solve the little linear equation 5x+2=05x + 2 = 0 to get x=−2/5x = -2/5.

  12. State the solutions

    x=1 or x=−25x = 1 \text{ or } x = -\frac{2}{5}

    Each bracket gives one solution, so a quadratic that factorises into two different brackets has two solutions.

  13. Check the solution x=1x = 1

    5(1)2−3(1)−2=5−3−2=05(1)^2 - 3(1) - 2 = 5 - 3 - 2 = 0

    Substituting x=1x = 1 back into the original quadratic gives 0, so this solution is correct.

  14. Check the solution x=−2/5x = -2/5

    5(−25)2−3(−25)−2=45+65−2=05\left(-\frac{2}{5}\right)^2 - 3\left(-\frac{2}{5}\right) - 2 = \frac{4}{5} + \frac{6}{5} - 2 = 0

    Substituting x=−2/5x = -2/5 back into the original quadratic gives 0, so this solution is correct.

  15. Check the symmetry of the roots

    x=1+−252=310x = \frac{1 + -\frac{2}{5}}{2} = \frac{3}{10}

    A parabola is symmetrical, so its two roots sit the same distance either side of the line of symmetry, here x=3/10x = 3/10. This is a quick sanity check on the pair of answers.

Answer
x=1 or x=−25x = 1 \text{ or } x = -\frac{2}{5}
Question 3
6 markschallenging
A rectangle has perimeter 2626 cm and area 4040 cm2^2. Find the length and the width of the rectangle.
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Worked solution

  1. Use the perimeter

    2(l+w)=26⇒l+w=132(l + w) = 26 \Rightarrow l + w = 13

    Halving the perimeter gives the sum of the length and the width.

  2. Introduce one unknown

    w=x,l=13−xw = x, \quad l = 13 - x

    If the width is x then the length must be 13 - x, because the two add to 13.

  3. Use the area

    x(13−x)=40x(13 - x) = 40

    Length × width = area.

  4. Expand and rearrange

    13x−x2=40⇒x2−13x+40=013x - x^2 = 40 \Rightarrow x^2 - 13x + 40 = 0

    Multiply out, then collect everything on one side so the x2x^2 term is positive and the equation equals zero.

  5. Look for the pair of numbers

    product=40,sum=−13\text{product} = 40, \quad \text{sum} = -13

    For x2x^2 + bx + c the two numbers in the brackets must multiply to c=40c = 40 and add to b=−13b = -13.

  6. List the factor pairs of 40

    (−40)×(−1),(−20)×(−2),(−10)×(−4),(−8)×(−5),1×40,2×20,4×10,5×8(-40) \times (-1),\quad (-20) \times (-2),\quad (-10) \times (-4),\quad (-8) \times (-5),\quad 1 \times 40,\quad 2 \times 20,\quad 4 \times 10,\quad 5 \times 8

    Write out every pair of whole numbers whose product is 40, then test which pair also adds to -13.

  7. Choose the pair that works

    (−8)×(−5)=40,(−8)+(−5)=−13(-8) \times (-5) = 40, \quad (-8) + (-5) = -13

    -8 and -5 pass both tests, so they are the numbers that go in the brackets.

  8. Write the factorised equation

    (x−8)(x−5)=0(x - 8)(x - 5) = 0

    The product of the brackets is zero — this is the form the null-factor law needs.

  9. Check the factorisation by expanding

    (x−8)(x−5)=x2−13x+40(x - 8)(x - 5) = x^2 - 13x + 40

    Multiplying the brackets back out returns the original quadratic, so the factorisation is right.

  10. Apply the null-factor law

    x−8=0orx−5=0x - 8 = 0 \quad \text{or} \quad x - 5 = 0

    If several things multiply to give zero, at least one of them must be zero. So set each bracket equal to zero in turn.

  11. Solve bracket 1

    x−8=0⇒x=8x - 8 = 0 \Rightarrow x = 8

    Solve the little linear equation x−8=0x - 8 = 0 to get x=8x = 8.

  12. Solve bracket 2

    x−5=0⇒x=5x - 5 = 0 \Rightarrow x = 5

    Solve the little linear equation x−5=0x - 5 = 0 to get x=5x = 5.

  13. State the solutions

    x=8 or x=5x = 8 \text{ or } x = 5

    Each bracket gives one solution, so a quadratic that factorises into two different brackets has two solutions.

  14. Interpret the two solutions

    width 5 cm,length 8 cm\text{width } 5 \text{ cm}, \quad \text{length } 8 \text{ cm}

    The two roots are the width and the length: 5+8=135 + 8 = 13 (perimeter 2626 cm) and 5×8=40 cm25 \times 8 = 40\text{ cm}^2 (the area). Both roots are used, not rejected.

  15. Check the solution x=8x = 8

    (8)2−13(8)+40=64−104+40=0(8)^2 - 13(8) + 40 = 64 - 104 + 40 = 0

    Substituting x=8x = 8 back into the original quadratic gives 0, so this solution is correct.

Answer
length 8 cm,  width 5 cm\text{length } 8 \text{ cm}, \; \text{width } 5 \text{ cm}
Question 4
5 markschallenging
A student solves (x−3)(x+2)=14(x - 3)(x + 2) = 14 by writing x−3=14x - 3 = 14 or x+2=14x + 2 = 14. Explain what is wrong with this method and give the correct solutions.
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Worked solution

  1. State what the student did

    x−3=14orx+2=14x - 3 = 14 \quad \text{or} \quad x + 2 = 14

    The student used the null-factor law, but the product on the right is 14, not 0.

  2. Why that is wrong

    AB=0⇒A=0 or B=0AB = 0 \Rightarrow A = 0 \text{ or } B = 0

    The law only works for zero. Zero is special: it is the only number that forces one of the factors to take a particular value.

  3. A counterexample makes it obvious

    2×7=142 \times 7 = 14

    Two numbers can multiply to 14 without either of them being 14. So the brackets do not have to equal 14.

  4. The correct first move is to expand

    (x−3)(x+2)=x2−x−6(x - 3)(x + 2) = x^2 - x - 6

    Multiply out: x×x=x2x \times x = x^2, and 2x−3x=−x2x - 3x = -x, and −3×2=−6-3 \times 2 = -6.

  5. Write the full equation

    x2−x−6=14x^2 - x - 6 = 14

    Now the left-hand side is a single quadratic expression.

  6. Rearrange so that one side is zero

    x2−x−20=0x^2 - x - 20 = 0

    Subtract 14 from both sides. Only now can the null-factor law be used.

  7. Look for the pair of numbers

    product=−20,sum=−1\text{product} = -20, \quad \text{sum} = -1

    The two numbers in the brackets must multiply to -20 and add to -1.

  8. List the factor pairs of -20

    (−1)×20,(−2)×10,(−4)×5,(−5)×4,(−10)×2,(−20)×1(-1) \times 20, \quad (-2) \times 10, \quad (-4) \times 5, \quad (-5) \times 4, \quad (-10) \times 2, \quad (-20) \times 1

    One number must be negative because the product is negative.

  9. Choose the pair that works

    (−5)×4=−20,−5+4=−1(-5) \times 4 = -20, \quad -5 + 4 = -1

    -5 and 4 pass both tests.

  10. Write the factorised equation

    (x−5)(x+4)=0(x - 5)(x + 4) = 0

    The numbers -5 and 4 go straight into the brackets.

  11. Apply the null-factor law

    x−5=0orx+4=0x - 5 = 0 \quad \text{or} \quad x + 4 = 0

    Now the product really is zero, so the law applies.

  12. Solve each bracket

    x=5orx=−4x = 5 \quad \text{or} \quad x = -4

    Two brackets, two solutions.

  13. Check the first solution

    (5−3)(5+2)=2×7=14(5 - 3)(5 + 2) = 2 \times 7 = 14

    Substituting x=5x = 5 into the original equation gives 14, as required.

  14. Check the second solution

    (−4−3)(−4+2)=(−7)×(−2)=14(-4 - 3)(-4 + 2) = (-7) \times (-2) = 14

    Substituting x=−4x = -4 also gives 14, so both solutions are correct.

  15. Summarise the error

    x=5 or x=−4x = 5 \text{ or } x = -4

    The student's method only works when the product is zero. Expanding, rearranging to = 0 and then factorising gives the correct solutions x=5x = 5 and x=−4x = -4.

Answer
x=5 or x=−4x = 5 \text{ or } x = -4
Question 5
5 markschallenging
The graph of y=x2−x−4y = x^2 - x - 4 crosses the xx-axis at two points. Which pair of estimates gives the solutions of x2−x−4=0x^2 - x - 4 = 0 correct to 1 decimal place?
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Worked solution

  1. Check whether it factorises

    x2−x−4=0x^2 - x - 4 = 0

    Two whole numbers would have to multiply to -4 and add to -1. The pairs are 1 × (-4) (sum -3), (-1) × 4 (sum 3) and (-2) × 2 (sum 0). None gives -1, so this does not factorise.

  2. Use the graph instead

    y=x2−x−4y = x^2 - x - 4

    The solutions of x2−x−4=0x^2 - x - 4 = 0 are the x-values where this curve meets the x-axis, so read them off a sketch.

  3. Build a table of values

    x: −3, −2, −1, 0, 1, 2, 3, 4y: 8, 2, −2, −4, −4, −2, 2, 8x:\ -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4 \quad y:\ 8,\ 2,\ -2,\ -4,\ -4,\ -2,\ 2,\ 8

    For example, when x=−2x = -2, y=4+2−4=2y = 4 + 2 - 4 = 2.

  4. Plot the points and draw the curve

    y=x2−x−4y = x^2 - x - 4

    The values repeat symmetrically, so the line of symmetry is x=0.5x = 0.5.

  5. Find where the curve changes sign

    −2<x<−1and2<x<3-2 < x < -1 \quad \text{and} \quad 2 < x < 3

    The curve goes from y=2y = 2 to y=−2y = -2 between x=−2x = -2 and x=−1x = -1, so it crosses the axis there; the same happens between x=2x = 2 and x=3x = 3.

  6. Read the left-hand root

    x≈−1.6x \approx -1.6

    The crossing is a little past x=−1.5x = -1.5.

  7. Test the left-hand estimate

    (−1.6)2−(−1.6)−4=2.56+1.6−4=0.16(-1.6)^2 - (-1.6) - 4 = 2.56 + 1.6 - 4 = 0.16

    Close to zero, so -1.6 is right to 1 decimal place.

  8. Test the neighbouring value

    (−1.5)2−(−1.5)−4=2.25+1.5−4=−0.25(-1.5)^2 - (-1.5) - 4 = 2.25 + 1.5 - 4 = -0.25

    This is further from zero than -1.6 was, and it has the opposite sign, so the root lies between -1.6 and -1.5 — and to 1 decimal place it reads as -1.6.

  9. Read the right-hand root

    x≈2.6x \approx 2.6

    The curve rises through the axis just after x=2.5x = 2.5.

  10. Test the right-hand estimate

    2.62−2.6−4=6.76−2.6−4=0.162.6^2 - 2.6 - 4 = 6.76 - 2.6 - 4 = 0.16

    Close to zero, so 2.6 is right to 1 decimal place.

  11. Test the neighbouring value

    2.52−2.5−4=6.25−2.5−4=−0.252.5^2 - 2.5 - 4 = 6.25 - 2.5 - 4 = -0.25

    Opposite sign again, so the root is between 2.5 and 2.6.

  12. Check with the line of symmetry

    −1.6+2.62=0.5\frac{-1.6 + 2.6}{2} = 0.5

    The midpoint of the two roots is x=0.5x = 0.5, matching the symmetry of the table.

  13. Compare with the exact answer

    x=1±172=−1.5615…,  2.5615…x = \frac{1 \pm \sqrt{17}}{2} = -1.5615\ldots, \; 2.5615\ldots

    The exact roots round to -1.6 and 2.6, so the graphical estimates are correct.

  14. Reject the tempting wrong readings

    x=−4 is the y-interceptx = -4 \text{ is the } y\text{-intercept}

    The number -4 in the equation is where the curve meets the y-axis, not a solution.

  15. Choose the correct pair

    x≈−1.6andx≈2.6x \approx -1.6 \quad \text{and} \quad x \approx 2.6

    Both estimates pass the substitution test and the symmetry test.

Answer
x=−1.6 or x=2.6 (to 1 dp)x = -1.6 \text{ or } x = 2.6 \text{ (to 1 dp)}

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