Monic quadratics and difference of squares

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

1. Find two numbers that multiply to 12 and add to 7: $3$ and $4$.
2. $$x^2 + 7x + 12 = (x + 3)(x + 4)$$

• M1 Brackets of the form $(x \pm a)(x \pm b)$ with $ab = 12$ or $a + b = 7$.
• A1 $(x + 3)(x + 4)$.

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

1. The numbers multiply to $+6$ and add to $-5$, so both are negative: $-2$ and $-3$.
2. $$x^2 - 5x + 6 = (x - 2)(x - 3)$$

• M1 Brackets with $ab = 6$ or $a + b = -5$.
• A1 $(x - 2)(x - 3)$.

(a) $(y + 3)(y + 7)$

1. $3 \times 7 = 21$ and $3 + 7 = 10$.

• M1 Brackets with $ab = 21$ or $a + b = 10$.
• A1 $(y + 3)(y + 7)$.

(b) $(p + 3)(p - 3)$

1. $p^2 - 3^2 = (p + 3)(p - 3)$.

• B1 $(p + 3)(p - 3)$.

(a) $(x+3)(x+5)$

1. Find two integers whose sum is the x coefficient and whose product is the constant.
   $$3+5=8$$
2. Therefore $(x+3)(x+5)$.

• M1 Find two integers whose sum is the x coefficient and whose product is the constant.
• A1 Correct answer: $(x+3)(x+5)$

(b) $-3, -5$

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

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

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

1. Recognise a difference of two squares.
   $$x^{2}-4^{2}$$
2. Therefore $(x-4)(x+4)$.

• M1 Recognise a difference of two squares.
• A1 Correct answer: $(x-4)(x+4)$

(b) $20$

1. Substitute and multiply the two simple factors.
   $$(6-4)(6+4)$$
2. Therefore $20$.

• M1 Substitute and multiply the two simple factors.
• A1 Correct answer: $20$

(a) $(x-6)(x+4)$

1. Find two numbers with product −24 and sum −2.
   $$-6\times 4=-24$$
2. Therefore $(x-6)(x+4)$.

• M1 Find two numbers with product −24 and sum −2.
• A1 Correct answer: $(x-6)(x+4)$

(b) $6, -4$

1. Set each factor equal to zero.
   $$x-6=0$$
2. Therefore $6, -4$.

• M1 Set each factor equal to zero.
• A1 Correct answer: $6, -4$

(a) $x+5$

1. Factorise the area expression.
   $$(x+4)(x+5)$$
2. Therefore $x+5$.

• M1 Factorise the area expression.
• A1 Correct answer: $x+5$

(b) $5$

1. Form an equation for the perimeter.
   $$2\times ((x+4)+(x+5))=38$$
2. Therefore $5$.

• P1 Form an equation for the perimeter.
• A1 Correct answer: $5$

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

1. Numbers that multiply to $-15$ and add to $+2$: $+5$ and $-3$.

• M1 Brackets with $ab = -15$ or $a + b = 2$.
• A1 $(x + 5)(x - 3)$.

(b) $(x + 7)(x - 7)$

1. This is a difference of two squares: $x^2 - 7^2 = (x + 7)(x - 7)$.

• B1 $(x + 7)(x - 7)$.

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

1. Take out the common factor 3 first.
   $$3x^2 - 12 = 3(x^2 - 4)$$
2. Then use the difference of two squares.
   $$3(x + 2)(x - 2)$$

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

(a) $12\,000$

1. Use the difference of two squares.
   $$1003^2 - 997^2 = (1003 + 997)(1003 - 997)$$
2. $$= 2000 \times 6 = 12\,000$$

• M1 Writing $(1003 + 997)(1003 - 997)$.
• A1 The correct answer, $12\,000$.

(b) $n^2 + 5n + 6 = (n + 2)(n + 3)$, a product of two whole numbers greater than 1.

1. Factorise.
   $$n^2 + 5n + 6 = (n + 2)(n + 3)$$
2. When $n \ge 1$, both $n + 2 \ge 3$ and $n + 3 \ge 4$, so the number has factors other than 1 and itself.
3. So it is never prime.

• M1 Factorising to $(n + 2)(n + 3)$.
• C1 Stating that both factors are greater than 1 for positive $n$, so the number is not prime.