Three binomials and non-monic factorisation

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

1. $2 \times 3 = 6$: find two numbers that multiply to 6 and add to 7: $6$ and $1$.
2. Split the middle term and group.
   $$2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$$

• M1 Splitting to $2x^2 + 6x + x + 3$, or brackets $(2x \pm a)(x \pm b)$ with $ab = 3$.
• A1 $(2x + 1)(x + 3)$.

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

1. $6 \times (-2) = -12$: numbers that multiply to $-12$ and add to $-1$ are $-4$ and $3$.
2. $$6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (3x - 2)(2x + 1)$$

• M1 Splitting the middle term as $-4x + 3x$, or a pair of brackets with the right $x^2$ and constant terms.
• A1 $(3x - 2)(2x + 1)$.

(a) $(3x - 4)(2x + 3)$

1. $$6x^{2}+9x-8x-12$$
2. Find two numbers with product 6 $\times$ ($-12$) = $-72$ and sum 1: 9 and $-8.$
3. Split the middle term: $6x^{2}$ + 9x $-$ 8x $-$ 12.
4. Factorise in groups: 3x(2x + 3) $-$ 4(2x + 3) = (3x $-$ 4)(2x + 3).

• M1 Establishing $6x^{2}+9x-8x-12$ or an equivalent valid method.
• A1 Correct answer: $(3x - 4)(2x + 3)$

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

1. Take out the common factor 3.
   $$3(4x^2 - 9)$$
2. Difference of two squares.
   $$3(2x + 3)(2x - 3)$$

• M1 Taking out 3: $3(4x^2 - 9)$, or $(6x + 9)(2x - 3)$ or similar.
• A1 $3(2x + 3)(2x - 3)$.

(b) $x = 2$ or $x = -\frac{1}{3}$

1. Factorise.
   $$(3x + 1)(x - 2) = 0$$
2. $3x + 1 = 0$ gives $x = -\frac{1}{3}$; $x - 2 = 0$ gives $x = 2$.

• M1 Factorising to $(3x + 1)(x - 2)$.
• A1 $x = 2$ and $x = -\frac{1}{3}$.

(a) $(x^2 + x - 2)(2x + 3) = 2x^3 + 5x^2 - x - 6$

1. $$(x + 2)(x - 1) = x^2 + x - 2$$
2. $$(x^2 + x - 2)(2x + 3) = 2x^3 + 3x^2 + 2x^2 + 3x - 4x - 6$$
3. $$= 2x^3 + 5x^2 - x - 6$$

• M1 Expanding two brackets correctly.
• M1 Six terms from the second multiplication, at least four correct.
• A1 Collecting to $2x^3 + 5x^2 - x - 6$ with working shown.

(b) $90\ \text{cm}^3$

1. $5 \times 2 \times 9 = 90$.

• B1 The correct answer, $90$.

(a) $2x^{2}-5x-12$

1. Multiply all four term pairs and collect the middle terms.
   $$2x^{2}-8x+3x-12$$
2. Therefore $2x^{2}-5x-12$.

• M1 Multiply all four term pairs and collect the middle terms.
• A1 Correct answer: $2x^{2}-5x-12$

(b) $2x^{3}-3x^{2}-17x-12$

1. Distribute both x and 1 across all three terms.
   $$(2x^{2}-5x-12)(x+1)$$
2. Therefore $2x^{3}-3x^{2}-17x-12$.

• M1 Distribute both x and 1 across all three terms.
• A1 Correct answer: $2x^{3}-3x^{2}-17x-12$

(a) $(5x+1)(2x-3)$

1. Find two numbers with product 10 × (−3) = −30 and sum −13.
   $$-15\times 2=-30$$
2. Therefore $(5x+1)(2x-3)$.

• M1 Find two numbers with product 10 × (−3) = −30 and sum −13.
• A1 Correct answer: $(5x+1)(2x-3)$

(b) $-1/5, 3/2$

1. Set each factor equal to zero.
   $$5x+1=0$$
2. Therefore $-1/5, 3/2$.

• M1 Set each factor equal to zero.
• A1 Correct answer: $-1/5, 3/2$

(a) $3x-4$

1. Factorise the area with (2x + 5) as one factor.
   $$(2x+5)(3x-4)$$
2. Therefore $3x-4$.

• M1 Factorise the area with (2x + 5) as one factor.
• A1 Correct answer: $3x-4$

(b) $\frac{18}{5}$

1. Form an equation for the perimeter.
   $$2\times ((2x+5)+(3x-4))=38$$
2. Therefore $\frac{18}{5}$.

• P1 Form an equation for the perimeter.
• A1 Correct answer: $\frac{18}{5}$

(a) $2(x+4)(3x+7)$

1. Split the middle term: 38x = 14x + 24x, since 14 × 24 = 6 × 56.
   $$6x^{2}+14x+24x+56$$
2. Therefore $2(x+4)(3x+7)$.

• M1 Split the middle term: 38x = 14x + 24x, since 14 × 24 = 6 × 56.
• A1 Correct answer: $2(x+4)(3x+7)$

(b) $-4, -7/3$

1. Set each non-monic factor equal to zero.
   $$2x+8=0$$
2. Therefore $-4, -7/3$.

• M1 Set each non-monic factor equal to zero.
• A1 Correct answer: $-4, -7/3$