Solving monic quadratics by factorisation

(a) $x = -2$ or $x = -3$

1. Factorise.
   $$(x + 2)(x + 3) = 0$$
2. One of the brackets must be zero: $x + 2 = 0$ or $x + 3 = 0$, so $x = -2$ or $x = -3$.

• M1 Factorising to $(x + 2)(x + 3)$.
• A1 $x = -2$ and $x = -3$.

(b) $x = 4$ or $x = -4$

1. $(x + 4)(x - 4) = 0$, so $x = -4$ or $x = 4$.

• B1 Both $4$ and $-4$.

(a) $x = 0$ or $x = 7$

1. Factorise: $x(x - 7) = 0$. So $x = 0$ or $x = 7$.

• M1 Factorising to $x(x - 7)$.
• A1 Both $0$ and $7$.

(b) $x = 5$ or $x = -3$

1. $(x - 5)(x + 3) = 0$, so $x = 5$ or $x = -3$.

• M1 Factorising to $(x - 5)(x + 3)$.
• A1 $x = 5$ and $x = -3$.

(a) $5$ cm

1. Equate the product of the lengths to the area.
   $$x(x+3)=40$$
2. Factorise the quadratic after moving the area to the left.
   $$(x-5)(x+8)=0$$
3. Therefore $5$ cm.

• P1 Equate the product of the lengths to the area.
• P1 Factorise the quadratic after moving the area to the left.
• A1 Correct answer: $5$ cm

(b) The width is a positive length, so the negative root is rejected.

1. The width is a positive length, so the negative root is rejected.

• C1 Correct conclusion with supporting reasoning: The width is a positive length, so the negative root is rejected.

(a) x² − 4x = 21, and subtracting 21 from both sides gives x² − 4x - 21 = 0.

1. Write the words as an equation.
   $$x^{2}-4x=21$$
2. Therefore x² − 4x = 21, and subtracting 21 from both sides gives x² − 4x - 21 = 0.

• M1 Write the words as an equation.
• C1 Correct conclusion with supporting reasoning: x² − 4x = 21, and subtracting 21 from both sides gives x² − 4x - 21 = 0.

(b) $7, -3$

1. Factorise and set each factor equal to zero.
   $$(x-7)(x+3)=0$$
2. Therefore $7, -3$.

• M1 Factorise and set each factor equal to zero.
• A1 Correct answer: $7, -3$

(a) (10 + x)² − 100 = 44, so x² + 20x + 100 − 100 − 44 = 0, which gives x² + 20x - 44 = 0.

1. Subtract the lawn area from the large square.
   $$(10+x)^{2}-100=44$$
2. Therefore (10 + x)² − 100 = 44, so x² + 20x + 100 − 100 − 44 = 0, which gives x² + 20x - 44 = 0.

• P1 Subtract the lawn area from the large square.
• C1 Correct conclusion with supporting reasoning: (10 + x)² − 100 = 44, so x² + 20x + 100 − 100 − 44 = 0, which gives x² + 20x - 44 = 0.

(b) $2$ m

1. Factorise the quadratic.
   $$(x+22)(x-2)=0$$
2. Therefore $2$ m.

• P1 Factorise the quadratic.
• A1 Correct answer: $2$ m

(a) $x = 5$ or $x = -2$

1. Rearrange so one side is zero.
   $$x^2 - 3x - 10 = 0$$
2. Factorise.
   $$(x - 5)(x + 2) = 0 \Rightarrow x = 5 \text{ or } x = -2$$

• M1 Rearranging to $x^2 - 3x - 10 = 0$.
• M1 Factorising to $(x - 5)(x + 2)$.
• A1 $x = 5$ and $x = -2$.

(b) $5$ cm

1. Form an equation.
   $$x(x + 3) = 40 \Rightarrow x^2 + 3x - 40 = 0$$
2. Factorise.
   $$(x + 8)(x - 5) = 0 \Rightarrow x = -8 \text{ or } x = 5$$
3. A width cannot be negative, so the width is 5 cm.

• P1 Forming $x^2 + 3x - 40 = 0$.
• P1 Factorising to $(x + 8)(x - 5)$.
• A1 The correct answer, $5$ cm.

(a) $-9$

1. For a monic quadratic, the product of the roots is the constant term, 18.
2. The positive integer pair with product 18 and difference 3 is 3 and 6.
3. The factors are (x $-$ 3)(x $-$ 6) = $x^{2}$ $-$ 9x + 18, so k = $-9.$

• B1 Correct answer: $-9$

(a) $\frac{27}{4}$

1. $$4r=6$$
2. $$3\times (3/2)^{2}$$
3. Let the roots be r and 3r. Their sum is 6, so 4r = 6 and r = 3/2.
4. Their product is the constant term c.
5. Therefore c = (3/2)(9/2) = 27/4.

• P1 Establishing $4r=6$ or an equivalent valid method.
• P1 Establishing $3\times (3/2)^{2}$ or an equivalent valid method.
• A1 Correct answer: $\frac{27}{4}$