Expanding two binomials

(a) $x^2 + 8x + 15$

1. Multiply each term in the first bracket by each term in the second.
   $$x^2 + 5x + 3x + 15$$
2. Collect the $x$ terms.
   $$x^2 + 8x + 15$$

• M1 At least three of the four terms correct: $x^2$, $5x$, $3x$, $15$.
• A1 $x^2 + 8x + 15$.

(b) $x^2 - 2x - 8$

1. $$x^2 + 2x - 4x - 8 = x^2 - 2x - 8$$

• M1 At least three of the four terms correct: $x^2$, $2x$, $-4x$, $-8$.
• A1 The correct answer, $x^2 - 2x - 8$.

(a) $(x^2 + 6x + 8)\ \text{cm}^2$

1. $$(x + 4)(x + 2) = x^2 + 2x + 4x + 8 = x^2 + 6x + 8$$

• M1 Writing $(x + 4)(x + 2)$ and expanding with at least three correct terms.
• A1 The correct answer, $x^2 + 6x + 8$.

(b) $35\ \text{cm}^2$

1. $(3 + 4)(3 + 2) = 7 \times 5 = 35$, or $9 + 18 + 8 = 35$.

• B1 The correct answer, $35$.

(a) $x^{2}-2x-15$

1. Multiply each term in the first bracket by each term in the second.
   $$x^{2}-5x+3x-15$$
2. Therefore $x^{2}-2x-15$.

• M1 Multiply each term in the first bracket by each term in the second.
• A1 Correct answer: $x^{2}-2x-15$

(b) $20$ cm²

1. Substitute into the original lengths to check the expanded result.
   $$(7+3)(7-5)$$
2. Therefore $20$ cm².

• M1 Substitute into the original lengths to check the expanded result.
• A1 Correct answer: $20$ cm²

(a) $x^{2}+8x+16$

1. Multiply every term in one bracket by every term in the other.
   $$(x+4)(x+4)$$
2. Therefore $x^{2}+8x+16$.

• M1 Multiply every term in one bracket by every term in the other.
• A1 Correct answer: $x^{2}+8x+16$

(b) $2x+9$

1. Subtract the original area from the new area.
   $$(x+5)^{2}-(x+4)^{2}$$
2. Therefore $2x+9$.

• M1 Subtract the original area from the new area.
• A1 Correct answer: $2x+9$

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

1. Write all four products before collecting like terms.
   $$2x^{2}+8x-3x-12$$
2. Therefore $2x^{2}+5x-12$.

• M1 Write all four products before collecting like terms.
• A1 Correct answer: $2x^{2}+5x-12$

(b) $9-x^{2}$

1. Therefore $9-x^{2}$.

• B1 Correct answer: $9-x^{2}$

(a) $x^{2}+10x+25$

1. Write the square as a product of two brackets.
   $$(x+5)(x+5)$$
2. Therefore $x^{2}+10x+25$.

• M1 Write the square as a product of two brackets.
• A1 Correct answer: $x^{2}+10x+25$

(b) Only x = 0: the two sides differ by 10x, which is zero only when x = 0.

1. Only x = 0: the two sides differ by 10x, which is zero only when x = 0.

• C1 Correct conclusion with supporting reasoning: Only x = 0: the two sides differ by 10x, which is zero only when x = 0.

(a) $x^2 + 12x + 36$

1. Write it as two brackets.
   $$(x + 6)(x + 6) = x^2 + 6x + 6x + 36 = x^2 + 12x + 36$$

• M1 Writing $(x + 6)(x + 6)$ and expanding with at least three correct terms.
• A1 $x^2 + 12x + 36$.

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

1. Expand the first product.
   $$(2x + 1)(x - 3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$$
2. Expand the square.
   $$(x - 1)^2 = x^2 - 2x + 1$$
3. Subtract the whole of the second expression.
   $$2x^2 - 5x - 3 - x^2 + 2x - 1 = x^2 - 3x - 4$$

• M1 Expanding $(2x + 1)(x - 3)$ to $2x^2 - 5x - 3$.
• M1 Expanding $(x - 1)^2$ to $x^2 - 2x + 1$.
• A1 Subtracting with every sign changed to reach $x^2 - 3x - 4$.

(a) $(x^2 + 6x + 9) - (x^2 + 6x - 7) = 16$

1. Square:
   $$(x + 3)^2 = x^2 + 6x + 9$$
2. Rectangle:
   $$(x + 7)(x - 1) = x^2 - x + 7x - 7 = x^2 + 6x - 7$$
3. Difference:
   $$(x^2 + 6x + 9) - (x^2 + 6x - 7) = 9 + 7 = 16$$
4. The $x$ terms cancel, so the difference is 16 for every value of $x$.

• M1 Expanding the square's area to $x^2 + 6x + 9$.
• M1 Expanding the rectangle's area to $x^2 + 6x - 7$.
• C1 Subtracting correctly to get 16 and stating that it does not depend on $x$.

(b) $6$ cm

1. Form an equation.
   $$x^2 + 6x - 7 = 20 \Rightarrow x^2 + 6x - 27 = 0$$
2. Factorise.
   $$(x + 9)(x - 3) = 0 \Rightarrow x = 3 \text{ or } x = -9$$
3. Since $x > 1$, $x = 3$, so the square's side is $3 + 3 = 6$ cm.

• P1 Forming $x^2 + 6x - 27 = 0$.
• P1 Solving to get $x = 3$ (rejecting $-9$).
• A1 The correct answer, $6$ cm.