Single brackets and common factors

(a) $5x + 20$

1. Multiply each term in the bracket by 5: $5 \times x + 5 \times 4 = 5x + 20$.

• B1 The correct answer, $5x + 20$.

(b) $10y - 3$

1. Expand, then collect like terms.
   $$6y - 3 + 4y = 10y - 3$$

• M1 Expanding to $6y - 3$.
• A1 The correct answer, $10y - 3$.

(c) $3(2a + 5)$

1. The highest common factor of 6 and 15 is 3: $6a + 15 = 3(2a + 5)$.

• B1 The correct answer, $3(2a + 5)$.

(a) $24 - 6x$

1. $$9-6x+15$$
2. Multiply both terms inside the brackets by $-3$: $-6x$ + 15.
3. Add the constant 9: 9 $-$ 6x + 15 = 24 $-$ 6x.

• M1 Establishing $9-6x+15$ or an equivalent valid method.
• A1 Correct answer: $24 - 6x$

(a) $x^2 + 7x$

1. $x \times x = x^2$ and $x \times 7 = 7x$.

• B1 The correct answer, $x^2 + 7x$.

(b) $y(y - 4)$

1. Both terms contain $y$: $y^2 - 4y = y(y - 4)$.

• B1 The correct answer, $y(y - 4)$.

(a) $10x + 18$

1. Expand each bracket. The $-2$ multiplies both $x$ and $-5$.
   $$12x + 8 - 2x + 10$$
2. Collect like terms.
   $$10x + 18$$

• M1 Expanding one bracket correctly: $12x + 8$ or $-2x + 10$.
• A1 The correct answer, $10x + 18$.

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

1. The highest common factor of $8x^2$ and $12x$ is $4x$.
2. $$8x^2 - 12x = 4x(2x - 3)$$

• M1 Taking out a common factor: $4(2x^2 - 3x)$, $x(8x - 12)$ or $2x(4x - 6)$.
• A1 The correct answer, $4x(2x - 3)$.

(a) $12-6x$

1. Distribute minus 3 to both terms inside the bracket.
   $$6-6x+6$$
2. Therefore $12-6x$.

• M1 Distribute minus 3 to both terms inside the bracket.
• A1 Correct answer: $12-6x$

(b) $2$

1. Set the simplified expression equal to zero and isolate x.
   $$6x=12$$
2. Therefore $2$.

• M1 Set the simplified expression equal to zero and isolate x.
• A1 Correct answer: $2$

(a) $9(2x+3)$

1. Identify the greatest common numerical factor.
   $$18=9\times 2$$
2. Therefore $9(2x+3)$.

• M1 Identify the greatest common numerical factor.
• A1 Correct answer: $9(2x+3)$

(b) $2x+3$

1. Divide area by the given length.
   $$\frac{18x+27}{9}$$
2. Therefore $2x+3$.

• M1 Divide area by the given length.
• A1 Correct answer: $2x+3$

(a) $5x+2$

1. Subtract both bracketed sides from the perimeter.
   $$10x+5-(2x+6)-(3x-3)$$
2. Therefore $5x+2$.

• M1 Subtract both bracketed sides from the perimeter.
• A1 Correct answer: $5x+2$

(b) $5(2x+1)$

1. Therefore $5(2x+1)$.

• B1 Correct answer: $5(2x+1)$

(c) $3$

1. Form and solve an equation from the third side.
   $$5x+2=17$$
2. Therefore $3$.

• M1 Form and solve an equation from the third side.
• A1 Correct answer: $3$

(a) It simplifies to $17$.

1. Expand both brackets.
   $$6n + 15 - 6n + 2$$
2. The $n$ terms cancel, leaving $17$, which does not depend on $n$.

• M1 Expanding both brackets correctly, including $-2 \times (-1) = +2$.
• A1 Reaching $17$ and stating that it is the same for every $n$.

(b) $6ab(3a - 2b)$

1. The highest common factor is $6ab$.
2. $$18a^2b - 12ab^2 = 6ab(3a - 2b)$$

• M1 Taking out a partial common factor such as $6a(3ab - 2b^2)$ or $2ab(9a - 6b)$.
• A1 The correct answer, $6ab(3a - 2b)$.

(c) $(14x + 6)$ cm

1. Factorise the area to find the length.
   $$10x^2 + 15x = 5x(2x + 3)$$
2. So the length is $2x + 3$.
3. $$\text{perimeter} = 2(5x) + 2(2x + 3) = 10x + 4x + 6 = 14x + 6$$

• P1 Factorising to $5x(2x + 3)$, or dividing to find the length $2x + 3$.
• P1 Adding all four sides: $2 \times 5x + 2 \times (2x + 3)$.
• A1 The correct answer, $14x + 6$.