Solving Quadratics by Factorising GCSE Questions & Answers

How to solve a quadratic by factorising

Rearrange so that one side is $0$ : $x^2 = 4x + 21$ becomes $x^2 - 4x - 21 = 0$ .
Factorise the quadratic: $x^2 - 4x - 21 = (x - 7)(x + 3)$ .
Set each bracket equal to $0$ : $x - 7 = 0$ or $x + 3 = 0$ .
Solve each one: $x = 7$ or $x = -3$ . A quadratic usually has two solutions.
In a context question, check that each solution makes sense: a length can't be negative.

Key facts

If $AB = 0$ , then $A = 0$ or $B = 0$ . This only works when one side is $0$ .
$(x - p)(x - q) = 0$ has solutions $x = p$ and $x = q$ : the signs change.
$x^2 - 5x = 0$ factorises to $x(x - 5) = 0$ , so $x = 0$ or $x = 5$ .
The solutions of $x^2 + bx + c = 0$ are where the graph of $y = x^2 + bx + c$ crosses the $x$ -axis.

Common mistakes

Forgetting to change the sign: $(x - 2)(x - 5) = 0$ gives $x = 2$ and $x = 5$ , not $-2$ and $-5$ .
Factorising before making one side $0$ : $(x + 1)(x - 2) = 4$ does not mean $x + 1 = 4$ .
Dividing both sides by $x$ (or by a bracket), which loses a solution.
Giving only one solution, or keeping an impossible one in a context question.

Solving quadratics by factorising practice questions

Work through the levels in order. Try each question on paper first, then open the worked answer to check every step and see where the marks go.

Level 1: Warm-up

Equations that are already $= 0$ and ready to factorise. 12 marks

Solve $(x - 3)(x + 5) = 0$ . [2 marks]

Show worked answer

Answer $x = 3$ or $x = -5$

$\displaystyle x - 3 = 0 \quad\text{or}\quad x + 5 = 0$ If two brackets multiply to $0$ , one of them is $0$
$\displaystyle x = 3 \quad\text{or}\quad x = -5$

B1 $x = 3$
B1 $x = -5$

Solve $x^2 + 7x + 12 = 0$ . [3 marks]

Show worked answer

Answer $x = -3$ or $x = -4$

$\displaystyle (x + 3)(x + 4) = 0$ $3 \times 4 = 12$ and $3 + 4 = 7$
$\displaystyle x + 3 = 0 \quad\text{or}\quad x + 4 = 0$
$\displaystyle x = -3 \quad\text{or}\quad x = -4$

M1 $(x \pm 3)(x \pm 4)$
A1 $(x + 3)(x + 4) = 0$
A1 $x = -3$ and $x = -4$

Solve $x^2 - 6x + 8 = 0$ . [3 marks]

Show worked answer

Answer $x = 2$ or $x = 4$

$\displaystyle (x - 2)(x - 4) = 0$ $(-2) \times (-4) = 8$ and $(-2) + (-4) = -6$
$\displaystyle x = 2 \quad\text{or}\quad x = 4$ The signs change: $x - 2 = 0$ gives $x = 2$

M1 $(x \pm 2)(x \pm 4)$
A1 $(x - 2)(x - 4) = 0$
A1 $x = 2$ and $x = 4$

Solve $x^2 - 5x = 0$ . [2 marks]

Show worked answer

Answer $x = 0$ or $x = 5$

$\displaystyle x(x - 5) = 0$ Take out the common factor $x$ . Don't divide by $x$ : you would lose $x = 0$
$\displaystyle x = 0 \quad\text{or}\quad x = 5$

M1 $x(x - 5) = 0$
A1 $x = 0$ and $x = 5$

Solve $x^2 - 36 = 0$ . [2 marks]

Show worked answer

Answer $x = 6$ or $x = -6$

$\displaystyle (x + 6)(x - 6) = 0$ Difference of two squares
$\displaystyle x = -6 \quad\text{or}\quad x = 6$ Both $6^2$ and $(-6)^2$ are $36$

M1 $(x + 6)(x - 6) = 0$ or $x^2 = 36$
A1 $x = 6$ and $x = -6$

Level 2: Build

Harder signs, repeated roots, and a first look at $ax^2 + bx + c = 0$ and rearranging. 16 marks

Solve $x^2 + 2x - 24 = 0$ . [3 marks]

Show worked answer

Answer $x = 4$ or $x = -6$

$\displaystyle (x + 6)(x - 4) = 0$ $6 \times (-4) = -24$ and $6 + (-4) = 2$
$\displaystyle x = -6 \quad\text{or}\quad x = 4$

M1 $(x \pm 6)(x \pm 4)$
A1 $(x + 6)(x - 4) = 0$
A1 $x = 4$ and $x = -6$

Solve $x^2 - 10x + 25 = 0$ . [2 marks]

Show worked answer

Answer $x = 5$

$\displaystyle (x - 5)(x - 5) = 0$ $(-5) \times (-5) = 25$ and $(-5) + (-5) = -10$
$\displaystyle x = 5$ Both brackets give the same answer, so there is only one solution

M1 $(x - 5)(x - 5) = 0$ or $(x - 5)^2 = 0$
A1 $x = 5$ (only)

Solve $x^2 + 9x = 0$ . [2 marks]

Show worked answer

Answer $x = 0$ or $x = -9$

$\displaystyle x(x + 9) = 0$ Common factor $x$
$\displaystyle x = 0 \quad\text{or}\quad x = -9$

M1 $x(x + 9) = 0$
A1 $x = 0$ and $x = -9$

Solve $x^2 - 3x - 40 = 0$ . [3 marks]

Show worked answer

Answer $x = 8$ or $x = -5$

$\displaystyle (x - 8)(x + 5) = 0$ $(-8) \times 5 = -40$ and $(-8) + 5 = -3$
$\displaystyle x = 8 \quad\text{or}\quad x = -5$

M1 $(x \pm 8)(x \pm 5)$
A1 $(x - 8)(x + 5) = 0$
A1 $x = 8$ and $x = -5$

HigherSolve $2x^2 + 5x - 3 = 0$ . [3 marks]

Show worked answer

Answer $x = \tfrac{1}{2}$ or $x = -3$

$\displaystyle ac = -6;\quad 6 \times (-1) = -6,\ 6 + (-1) = 5$ Two numbers that multiply to $ac$ and add to $b$
$\displaystyle 2x^2 + 6x - x - 3 = 0$
$\displaystyle 2x(x + 3) - 1(x + 3) = 0$
$\displaystyle (2x - 1)(x + 3) = 0$
$\displaystyle x = \tfrac{1}{2} \quad\text{or}\quad x = -3$ $2x - 1 = 0$ gives $x = \tfrac{1}{2}$

M1 $(2x \pm 1)(x \pm 3)$ , or a correct split of the middle term
A1 $(2x - 1)(x + 3) = 0$
A1 $x = \tfrac{1}{2}$ and $x = -3$ — allow $0.5$

HigherSolve $x^2 = 4x + 21$ . [3 marks]

Show worked answer

Answer $x = 7$ or $x = -3$

$\displaystyle x^2 - 4x - 21 = 0$ Make one side $0$ first
$\displaystyle (x - 7)(x + 3) = 0$
$\displaystyle x = 7 \quad\text{or}\quad x = -3$

M1 $x^2 - 4x - 21 = 0$
M1 $(x - 7)(x + 3) = 0$
A1 $x = 7$ and $x = -3$

Level 3: Exam-style

Show that, hence and graph questions in real GCSE style. 21 marks

(a) Factorise $x^2 - 2x - 15$ . [2 marks]

Show worked answer

Answer $(x - 5)(x + 3)$

$\displaystyle (-5) \times 3 = -15,\quad (-5) + 3 = -2$
$\displaystyle x^2 - 2x - 15 = (x - 5)(x + 3)$

M1 $(x \pm 5)(x \pm 3)$
A1 $(x - 5)(x + 3)$

(b) Hence solve $x^2 - 2x - 15 = 0$ . [1 mark]

Show worked answer

Answer $x = 5$ or $x = -3$

$\displaystyle (x - 5)(x + 3) = 0$ Use part (a)
$\displaystyle x = 5 \quad\text{or}\quad x = -3$

B1 $x = 5$ and $x = -3$ — ft their (a)

(a) A rectangle is $x$ cm wide and $(x + 5)$ cm long. Its area is $36\ \text{cm}^2$ . Show that $x^2 + 5x - 36 = 0$ . [2 marks]

Show worked answer

Answer $x(x + 5) = 36$ , so $x^2 + 5x = 36$ , so $x^2 + 5x - 36 = 0$ .

$\displaystyle x(x + 5) = 36$ Area $=$ width $\times$ length
$\displaystyle x^2 + 5x = 36$
$\displaystyle x^2 + 5x - 36 = 0$ Subtract $36$ from both sides

M1 $x(x + 5) = 36$
A1 Expands and rearranges correctly to $x^2 + 5x - 36 = 0$

(b) Solve $x^2 + 5x - 36 = 0$ . [3 marks]

Show worked answer

Answer $x = 4$ or $x = -9$

$\displaystyle (x + 9)(x - 4) = 0$ $9 \times (-4) = -36$ and $9 + (-4) = 5$
$\displaystyle x = -9 \quad\text{or}\quad x = 4$

M1 $(x \pm 9)(x \pm 4)$
A1 $(x + 9)(x - 4) = 0$
A1 $x = 4$ and $x = -9$

Show worked answer

Answer $4$ cm

$\displaystyle x = 4 \text{ cm}$ A width can't be negative, so reject $x = -9$

B1 $4$ cm — ft a positive solution from (b)

Ella solves $x^2 - 7x + 10 = 0$ . She writes $(x - 2)(x - 5) = 0$ , so $x = -2$ or $x = -5$ . Explain Ella's mistake, and write down the correct solutions. [2 marks]

Show worked answer

Answer $x = 2$ or $x = 5$ . Ella's brackets are right, but $x - 2 = 0$ gives $x = 2$ , not $-2$ .

$\displaystyle (x - 2)(x - 5) = 0$ Ella's factorising is correct
$\displaystyle x - 2 = 0 \;\Rightarrow\; x = 2$ Add $2$ to both sides: the sign changes
$\displaystyle x - 5 = 0 \;\Rightarrow\; x = 5$

B1 Explains that $x - 2 = 0$ gives $x = 2$ (the signs should change)
B1 $x = 2$ and $x = 5$

The graph of $y = x^2 - x - 12$ crosses the $x$ -axis at two points. Use algebra to find the coordinates of these points. [3 marks]

Show worked answer

Answer $(4,\ 0)$ and $(-3,\ 0)$

$\displaystyle x^2 - x - 12 = 0$ On the $x$ -axis, $y = 0$
$\displaystyle (x - 4)(x + 3) = 0$
$\displaystyle x = 4 \quad\text{or}\quad x = -3$
$\displaystyle (4,\ 0) \text{ and } (-3,\ 0)$ Give coordinates, not just the $x$ values

M1 $x^2 - x - 12 = 0$
M1 $(x - 4)(x + 3) = 0$
A1 $(4,\ 0)$ and $(-3,\ 0)$

HigherSolve $3x^2 - 11x - 4 = 0$ . [3 marks]

Show worked answer

Answer $x = 4$ or $x = -\tfrac{1}{3}$

$\displaystyle ac = -12;\quad (-12) \times 1 = -12,\ -12 + 1 = -11$
$\displaystyle 3x^2 - 12x + x - 4 = 0$
$\displaystyle 3x(x - 4) + 1(x - 4) = 0$
$\displaystyle (3x + 1)(x - 4) = 0$
$\displaystyle x = -\tfrac{1}{3} \quad\text{or}\quad x = 4$

M1 $(3x \pm 1)(x \pm 4)$ , or a correct split of the middle term
A1 $(3x + 1)(x - 4) = 0$
A1 $x = 4$ and $x = -\tfrac{1}{3}$

HigherSolve $(x + 2)(x - 1) = 4x + 2$ . [4 marks]

Show worked answer

Answer $x = 4$ or $x = -1$

$\displaystyle x^2 + x - 2 = 4x + 2$ Expand the brackets first. You can't set each bracket to $4x + 2$
$\displaystyle x^2 - 3x - 4 = 0$ Make one side $0$
$\displaystyle (x - 4)(x + 1) = 0$
$\displaystyle x = 4 \quad\text{or}\quad x = -1$

M1 Expands $(x + 2)(x - 1)$ to $x^2 + x - 2$
M1 $x^2 - 3x - 4 = 0$
M1 $(x - 4)(x + 1) = 0$
A1 $x = 4$ and $x = -1$

Level 4: Challenge

Foundation: Grade 4–5 problem solving: forming and solving quadratic equations from words. Higher: Grade 8–9: fractions, unknown coefficients and contexts with conditions. 27 marks

(a) I think of a number, $x$ . I square it, then subtract $3$ times the number. The answer is $28$ . Show that $x^2 - 3x - 28 = 0$ . [1 mark]

Show worked answer

Answer $x^2 - 3x = 28$ , so $x^2 - 3x - 28 = 0$ .

$\displaystyle x^2 - 3x = 28$ Write the words as an equation
$\displaystyle x^2 - 3x - 28 = 0$

B1 $x^2 - 3x = 28$ rearranged to $x^2 - 3x - 28 = 0$

(b) Find the two possible values of the number. [3 marks]

Show worked answer

Answer $7$ or $-4$

$\displaystyle (x - 7)(x + 4) = 0$ $(-7) \times 4 = -28$ and $(-7) + 4 = -3$
$\displaystyle x = 7 \quad\text{or}\quad x = -4$ Check: $7^2 - 3 \times 7 = 28$ and $(-4)^2 - 3 \times (-4) = 28$

M1 $(x \pm 7)(x \pm 4)$
A1 $(x - 7)(x + 4) = 0$
A1 $7$ and $-4$

(a) The product of two consecutive positive whole numbers is $132$ . The smaller number is $x$ . Show that $x^2 + x - 132 = 0$ . [1 mark]

Show worked answer

Answer $x(x + 1) = 132$ , so $x^2 + x - 132 = 0$ .

$\displaystyle x(x + 1) = 132$ The next whole number is $x + 1$
$\displaystyle x^2 + x - 132 = 0$

B1 $x(x + 1) = 132$ expanded and rearranged to $x^2 + x - 132 = 0$

(b) Find the two numbers. [3 marks]

Show worked answer

Answer $11$ and $12$

$\displaystyle 12 \times (-11) = -132,\quad 12 + (-11) = 1$ Look for factors of $132$ that differ by $1$
$\displaystyle (x + 12)(x - 11) = 0$
$\displaystyle x = 11 \quad (\text{or } x = -12)$
$\displaystyle \text{The numbers are } 11 \text{ and } 12$ The numbers are positive, so reject $x = -12$

M1 $(x \pm 12)(x \pm 11)$
A1 $(x + 12)(x - 11) = 0$
A1 $11$ and $12$ (rejecting $x = -12$ )

A rectangle has length $(x + 3)$ cm and width $(x - 3)$ cm. Its area is $40\ \text{cm}^2$ . Work out the value of $x$ . [3 marks]

Show worked answer

Answer $x = 7$

$\displaystyle (x + 3)(x - 3) = 40$ Area $=$ length $\times$ width
$\displaystyle x^2 - 9 = 40$ Difference of two squares: $(x + 3)(x - 3) = x^2 - 9$
$\displaystyle x^2 = 49$
$\displaystyle x = 7$ $x = -7$ would make the width $-10$ cm, so reject it

M1 $(x + 3)(x - 3) = 40$
M1 $x^2 - 9 = 40$ or $x^2 = 49$
A1 $x = 7$ only

HigherSolve $\dfrac{5}{x + 2} + \dfrac{3}{x} = 2$ . [4 marks]

Show worked answer

Answer $x = 3$ or $x = -1$

$\displaystyle 5x + 3(x + 2) = 2x(x + 2)$ Multiply every term by $x(x + 2)$
$\displaystyle 8x + 6 = 2x^2 + 4x$
$\displaystyle 2x^2 - 4x - 6 = 0$
$\displaystyle x^2 - 2x - 3 = 0$ Divide by $2$
$\displaystyle (x - 3)(x + 1) = 0$
$\displaystyle x = 3 \quad\text{or}\quad x = -1$

M1 Multiplies by $x(x + 2)$ : $5x + 3(x + 2) = 2x(x + 2)$
M1 $2x^2 - 4x - 6 = 0$ or $x^2 - 2x - 3 = 0$
M1 $(x - 3)(x + 1) = 0$ — ft their quadratic
A1 $x = 3$ and $x = -1$

HigherThe equation $x^2 + bx + c = 0$ has solutions $x = 4$ and $x = -7$ . Find the value of $b$ and the value of $c$ . [3 marks]

Show worked answer

Answer $b = 3,\ c = -28$

$\displaystyle (x - 4)(x + 7) = 0$ Work backwards: $x = 4$ comes from the bracket $(x - 4)$
$\displaystyle x^2 + 7x - 4x - 28 = x^2 + 3x - 28$
$\displaystyle b = 3,\quad c = -28$

M1 $(x - 4)(x + 7) = 0$
M1 Expands to $x^2 + 3x - 28$
A1 $b = 3$ and $c = -28$

Higher(a) A triangle has base $(x + 1)$ cm and perpendicular height $(2x - 3)$ cm. Its area is $9\ \text{cm}^2$ . Show that $2x^2 - x - 21 = 0$ . [3 marks]

Show worked answer

Answer $\tfrac{1}{2}(x + 1)(2x - 3) = 9$ , so $2x^2 - x - 3 = 18$ , so $2x^2 - x - 21 = 0$ .

$\displaystyle \tfrac{1}{2}(x + 1)(2x - 3) = 9$ Area of a triangle $= \tfrac{1}{2} \times$ base $\times$ height
$\displaystyle (x + 1)(2x - 3) = 18$ Multiply both sides by $2$
$\displaystyle 2x^2 - 3x + 2x - 3 = 18$
$\displaystyle 2x^2 - x - 21 = 0$

M1 $\tfrac{1}{2}(x + 1)(2x - 3) = 9$
M1 $(x + 1)(2x - 3) = 18$ , or $2x^2 - x - 3$ seen
A1 Fully correct working to $2x^2 - x - 21 = 0$

(b) Work out the length of the base of the triangle. [3 marks]

Show worked answer

Answer $4.5$ cm

$\displaystyle ac = -42;\quad (-7) \times 6 = -42,\ -7 + 6 = -1$
$\displaystyle 2x^2 - 7x + 6x - 21 = 0$
$\displaystyle x(2x - 7) + 3(2x - 7) = 0$
$\displaystyle (2x - 7)(x + 3) = 0$
$\displaystyle x = 3.5$ $x = -3$ would make the base and height negative
$\displaystyle \text{base} = 3.5 + 1 = 4.5 \text{ cm}$ The question asks for the base, not $x$

M1 $(2x - 7)(x + 3) = 0$
A1 $x = 3.5$ , rejecting $x = -3$
A1 $4.5$ cm

HigherSolve $(x - 3)^2 = 2(x - 3)$ . [3 marks]

Show worked answer

Answer $x = 3$ or $x = 5$

$\displaystyle (x - 3)^2 - 2(x - 3) = 0$ Don't divide by $(x - 3)$ : it could be $0$ , and you would lose a solution
$\displaystyle (x - 3)\big[(x - 3) - 2\big] = 0$ Take out the common factor $(x - 3)$
$\displaystyle (x - 3)(x - 5) = 0$
$\displaystyle x = 3 \quad\text{or}\quad x = 5$

M1 $(x - 3)^2 - 2(x - 3) = 0$ , or $x^2 - 8x + 15 = 0$
M1 $(x - 3)(x - 5) = 0$
A1 $x = 3$ and $x = 5$

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