Factorising Quadratics GCSE Questions

Answer $(x + 2)(x + 5)$

1. $\displaystyle 2 \times 5 = 10,\quad 2 + 5 = 7$Find two numbers that multiply to $10$ and add to $7$
2. $\displaystyle x^2 + 7x + 10 = (x + 2)(x + 5)$

• M1 $(x \pm 2)(x \pm 5)$
• A1 $(x + 2)(x + 5)$

Answer $x(x + 5)$

1. $\displaystyle x^2 + 5x = x(x + 5)$Both terms have a common factor of $x$

• B1 $x(x + 5)$

Answer $(x - 4)(x - 5)$

1. $\displaystyle (-4) \times (-5) = 20,\quad (-4) + (-5) = -9$Both numbers are negative: they multiply to $+20$ and add to $-9$
2. $\displaystyle x^2 - 9x + 20 = (x - 4)(x - 5)$

• M1 $(x \pm 4)(x \pm 5)$
• A1 $(x - 4)(x - 5)$

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

1. $\displaystyle 5 \times (-3) = -15,\quad 5 + (-3) = 2$A negative product means one number is negative
2. $\displaystyle x^2 + 2x - 15 = (x + 5)(x - 3)$

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

Answer $(x + 4)(x + 6)$

1. $\displaystyle 4 \times 6 = 24,\quad 4 + 6 = 10$List factor pairs of $24$: $1, 24$; $2, 12$; $3, 8$; $4, 6$
2. $\displaystyle x^2 + 10x + 24 = (x + 4)(x + 6)$

• M1 $(x \pm 4)(x \pm 6)$
• A1 $(x + 4)(x + 6)$

Answer $(x - 7)(x + 4)$

1. $\displaystyle (-7) \times 4 = -28,\quad (-7) + 4 = -3$Opposite signs; the bigger number is negative because $b$ is negative
2. $\displaystyle x^2 - 3x - 28 = (x - 7)(x + 4)$

• M1 $(x \pm 7)(x \pm 4)$
• A1 $(x - 7)(x + 4)$

Answer $(x - 3)(x - 8)$

1. $\displaystyle (-3) \times (-8) = 24,\quad (-3) + (-8) = -11$Positive $c$ and negative $b$: both numbers are negative
2. $\displaystyle x^2 - 11x + 24 = (x - 3)(x - 8)$

• M1 $(x \pm 3)(x \pm 8)$
• A1 $(x - 3)(x - 8)$

Answer $(x + 8)(x - 8)$

1. $\displaystyle x^2 - 64 = x^2 - 8^2$Difference of two squares: $a^2 - b^2 = (a + b)(a - b)$
2. $\displaystyle = (x + 8)(x - 8)$

• B1 $(x + 8)(x - 8)$ — accept in either order

Answer $(x + 7)^2$

1. $\displaystyle 7 \times 7 = 49,\quad 7 + 7 = 14$The same number twice, so it is a perfect square
2. $\displaystyle x^2 + 14x + 49 = (x + 7)(x + 7) = (x + 7)^2$

• M1 $(x \pm 7)(x \pm 7)$
• A1 $(x + 7)^2$ — allow $(x + 7)(x + 7)$

Answer $(2x + 1)(x + 3)$

1. $\displaystyle ac = 2 \times 3 = 6;\quad 1 \times 6 = 6,\ 1 + 6 = 7$Find two numbers that multiply to $ac$ and add to $b$
2. $\displaystyle 2x^2 + x + 6x + 3$Split the middle term using $1$ and $6$
3. $\displaystyle = x(2x + 1) + 3(2x + 1)$Factorise each pair: the brackets match
4. $\displaystyle = (2x + 1)(x + 3)$

• M1 $(2x \pm 1)(x \pm 3)$, or a correct split of the middle term, e.g. $2x^2 + x + 6x + 3$
• A1 $(2x + 1)(x + 3)$

Answer $(3x - 4)(x - 2)$

1. $\displaystyle ac = 24;\quad (-4) \times (-6) = 24,\ (-4) + (-6) = -10$Both negative, because $b$ is negative and $ac$ is positive
2. $\displaystyle 3x^2 - 4x - 6x + 8$
3. $\displaystyle = x(3x - 4) - 2(3x - 4)$Take out $-2$ so the second bracket matches
4. $\displaystyle = (3x - 4)(x - 2)$

• M1 $(3x \pm 4)(x \pm 2)$, or a correct split of the middle term
• A1 $(3x - 4)(x - 2)$

Answer $3(x + 5)(x - 1)$

1. $\displaystyle 3x^2 + 12x - 15 = 3(x^2 + 4x - 5)$Take out the common factor $3$ first
2. $\displaystyle 5 \times (-1) = -5,\quad 5 + (-1) = 4$
3. $\displaystyle = 3(x + 5)(x - 1)$

• M1 $3(x^2 + 4x - 5)$, or $(3x \pm 15)(x \pm 1)$, or $(3x \pm 3)(x \pm 5)$
• A1 $3(x + 5)(x - 1)$

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

1. $\displaystyle x^2 - 9 = x^2 - 3^2 = (x + 3)(x - 3)$Difference of two squares

• B1 $(x + 3)(x - 3)$

Answer $10\,600$

1. $\displaystyle 103^2 - 9 = (103 + 3)(103 - 3)$Use part (a) with $x = 103$
2. $\displaystyle = 106 \times 100$
3. $\displaystyle = 10\,600$

• M1 $(103 + 3)(103 - 3)$ or $106 \times 100$
• A1 $10\,600$

Answer $(x - 7)(x + 2)$. Sam's numbers multiply to $+14$, not $-14$.

1. $\displaystyle (-7) \times (-2) = +14 \ne -14$Sam's brackets give the wrong sign for $c$
2. $\displaystyle (-7) \times 2 = -14,\quad (-7) + 2 = -5$A negative $c$ needs opposite signs
3. $\displaystyle x^2 - 5x - 14 = (x - 7)(x + 2)$

• B1 Explains that $(-7) \times (-2) = +14$, or that the numbers need opposite signs
• B1 $(x - 7)(x + 2)$

Answer $x + 7$ cm

1. $\displaystyle x^2 + 11x + 28 = (x + 4)(x + 7)$Factorise the area: one bracket must be the width
2. $\displaystyle \text{length} = x + 7$Area $=$ width $\times$ length

• M1 $(x + 4)(x + 7)$
• A1 $x + 7$

Answer $4x + 22$ cm

1. $\displaystyle 2(x + 4) + 2(x + 7)$Two widths and two lengths
2. $\displaystyle = 2x + 8 + 2x + 14$
3. $\displaystyle = 4x + 22$

• M1 $2(x + 4) + 2(x + 7)$ oe — ft their length
• A1 $4x + 22$

Answer $(2x - 1)(3x + 2)$

1. $\displaystyle ac = -12;\quad 4 \times (-3) = -12,\ 4 + (-3) = 1$Two numbers that multiply to $-12$ and add to $1$
2. $\displaystyle 6x^2 + 4x - 3x - 2$
3. $\displaystyle = 2x(3x + 2) - 1(3x + 2)$
4. $\displaystyle = (2x - 1)(3x + 2)$

• M1 $(2x \pm 1)(3x \pm 2)$, or a correct split of the middle term, e.g. $6x^2 + 4x - 3x - 2$
• A1 $(2x - 1)(3x + 2)$

Answer $3(2x + 3)(2x - 3)$

1. $\displaystyle 12x^2 - 27 = 3(4x^2 - 9)$Common factor $3$ first
2. $\displaystyle = 3\left((2x)^2 - 3^2\right)$$4x^2 - 9$ is a difference of two squares
3. $\displaystyle = 3(2x + 3)(2x - 3)$

• M1 $3(4x^2 - 9)$, or $(6x \pm 9)(2x \pm 3)$, or $(2x \pm 3)(6x \pm 9)$
• A1 $3(2x + 3)(2x - 3)$

Answer $7400$

1. $\displaystyle 87^2 - 13^2 = (87 + 13)(87 - 13)$Difference of two squares: $a^2 - b^2 = (a + b)(a - b)$
2. $\displaystyle = 100 \times 74$
3. $\displaystyle = 7400$

• M1 $(87 + 13)(87 - 13)$ or $100 \times 74$
• A1 $7400$

Answer $(x + 3)^2$

1. $\displaystyle 3 \times 3 = 9,\quad 3 + 3 = 6$
2. $\displaystyle x^2 + 6x + 9 = (x + 3)^2$

• M1 $(x \pm 3)(x \pm 3)$
• A1 $(x + 3)^2$ — allow $(x + 3)(x + 3)$

Answer $x^2 + 6x + 9 = (x + 3)^2$, and a square is never negative, so $x^2 + 6x + 9 \ge 0$.

1. $\displaystyle x^2 + 6x + 9 = (x + 3)^2$From part (a)
2. $\displaystyle (x + 3)^2 \ge 0$Squaring any number, positive or negative, never gives a negative answer

• B1 States that $(x + 3)^2$ is a square, so it is always $\ge 0$ — ft their (a) if it is a square

Answer $a = 3,\ b = 7$

1. $\displaystyle (-3) \times (-7) = 21,\quad (-3) + (-7) = -10$
2. $\displaystyle x^2 - 10x + 21 = (x - 3)(x - 7)$
3. $\displaystyle a = 3,\quad b = 7$The question says $a < b$

• M1 $(x - 3)(x - 7)$
• A1 $a = 3$ and $b = 7$

Answer $10\,400$

1. $\displaystyle (107 - 3)(107 - 7)$Substitute into the factorised form: much easier than $107^2$
2. $\displaystyle = 104 \times 100$
3. $\displaystyle = 10\,400$

• M1 $(107 - 3)(107 - 7)$ or $104 \times 100$ — ft their $a$ and $b$
• A1 $10\,400$

Answer $8(x + 1)$

1. $\displaystyle = \big[(x + 3) + (x - 1)\big]\big[(x + 3) - (x - 1)\big]$Difference of two squares with $a = x + 3$ and $b = x - 1$
2. $\displaystyle = (2x + 2)(4)$Careful: $-(x - 1) = -x + 1$
3. $\displaystyle = 8x + 8 = 8(x + 1)$"Fully" means take out the common factor $8$

• M1 $\big[(x + 3) + (x - 1)\big]\big[(x + 3) - (x - 1)\big]$, or both brackets expanded correctly
• M1 $(2x + 2) \times 4$ or $8x + 8$
• A1 $8(x + 1)$

Answer $(2n + 1)^2 - (2n - 1)^2 = 8n$, which is a multiple of $8$.

1. $\displaystyle \text{Let the numbers be } 2n - 1 \text{ and } 2n + 1$$n$ is any whole number; odd numbers are $2$ apart
2. $\displaystyle (2n + 1)^2 - (2n - 1)^2 = \big[(2n + 1) + (2n - 1)\big]\big[(2n + 1) - (2n - 1)\big]$Difference of two squares
3. $\displaystyle = 4n \times 2 = 8n$
4. $\displaystyle 8n = 8 \times n, \text{ a multiple of } 8$State the conclusion: it earns the final mark

• B1 Consecutive odd numbers written as $2n - 1$ and $2n + 1$ (or $2n + 1$ and $2n + 3$)
• M1 $\big[(2n + 1) + (2n - 1)\big]\big[(2n + 1) - (2n - 1)\big]$, or both squares expanded correctly
• A1 $8n$ with a conclusion that it is a multiple of $8$

Answer $b = -1,\ c = -2$

1. $\displaystyle (3x + 5)(x + c) = 3x^2 + (3c + 5)x + 5c$Expand the right-hand side
2. $\displaystyle 5c = -10 \;\Rightarrow\; c = -2$Compare the constant terms
3. $\displaystyle b = 3c + 5 = -6 + 5 = -1$Compare the $x$ terms

• M1 Expands $(3x + 5)(x + c)$ to $3x^2 + 3cx + 5x + 5c$
• M1 $5c = -10$
• A1 $c = -2$ and $b = -1$

Answer $(2x + 5)(2x - 5)$

1. $\displaystyle 4x^2 - 25 = (2x)^2 - 5^2 = (2x + 5)(2x - 5)$Difference of two squares with $a = 2x$

• B1 $(2x + 5)(2x - 5)$

Answer $x^2(2x + 5)(2x - 5)$

1. $\displaystyle 4x^4 - 25x^2 = x^2(4x^2 - 25)$Take out the common factor $x^2$ first
2. $\displaystyle = x^2(2x + 5)(2x - 5)$Use part (a)

• M1 $x^2(4x^2 - 25)$
• A1 $x^2(2x + 5)(2x - 5)$