1. Factorise each expression: (i) Factorise $1 + 2x + x^2$. This is a perfect square trinomial: $$1 + 2x + x^2 = (1 + x)^2$$ (ii) Factorise $x^2 - 9ax + 18a^2$. Find two numbers that multiply to $18a^2$ and add to $-9a$: these are $-6a$ and $-3a$. $$x^2 - 9ax + 18a^2 = x^2 - 6ax - 3ax + 18a^2$$ Group terms: $$(x^2 - 6ax) - (3ax - 18a^2)$$ Factor each group: $$x(x - 6a) - 3a(x - 6a)$$ Factor out common binomial: $$(x - 3a)(x - 6a)$$ (iii) Factorise $2xy - x + z - 2zy$. Group terms: $$(2xy - 2zy) + (-x + z)$$ Factor each group: $$2y(x - z) - 1(x - z)$$ Factor out common binomial: $$(2y - 1)(x - z)$$ (iv) Factorise $p^4 - 81q^4$. Recognize difference of squares: $$p^4 - (9q^2)^2 = (p^2 - 9q^2)(p^2 + 9q^2)$$ Further factor $p^2 - 9q^2$ as difference of squares: $$(p - 3q)(p + 3q)(p^2 + 9q^2)$$ (v) Factorise $7\sqrt{2}x^2 - 10x - 4\sqrt{2}$. Use trial or grouping: Rewrite as $7\sqrt{2}x^2 - 10x - 4\sqrt{2}$. Try factors of $7\sqrt{2}x^2$ and $-4\sqrt{2}$ that multiply to $-28x^2$ and sum to $-10x$. The factorisation is: $$(7x + 2\sqrt{2})(\sqrt{2}x - 2)$$ (vi) Factorise $x^4 - (2y - 3z)^2$. Difference of squares: $$x^4 - (2y - 3z)^2 = (x^2 - (2y - 3z))(x^2 + (2y - 3z))$$ (vii) Factorise $24\sqrt{3}x^3 - 125y^3$. Recognize difference of cubes: $$24\sqrt{3}x^3 = (2\sqrt[3]{3}x)^3$$ $$125y^3 = (5y)^3$$ So: $$a = 2\sqrt[3]{3}x, b = 5y$$ Use formula: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ Hence: $$(2\sqrt[3]{3}x - 5y)((2\sqrt[3]{3}x)^2 + 2\sqrt[3]{3}x \cdot 5y + (5y)^2)$$ Simplify inside: $$(2\sqrt[3]{3}x - 5y)(4\sqrt[3]{9}x^2 + 10\sqrt[3]{3}xy + 25y^2)$$ (viii) Factorise $a^6x^4 - a^4x^6$. Factor out common terms: $$a^4x^4(a^2 - x^2)$$ Further factor difference of squares: $$a^4x^4(a - x)(a + x)$$ 2. Evaluate using identities: (i) $103 \times 105$. Use identity: $(a + b)(a + c) = a^2 + a(b + c) + bc$ with $a=104$, $b=-1$, $c=1$. Or use $(x - 1)(x + 1) = x^2 - 1$ with $x=104$: $$103 \times 105 = 104^2 - 1 = 10816 - 1 = 10815$$ (ii) $98 \times 99$. Use $(100 - 2)(100 - 1) = 100^2 - 100 - 200 + 2 = 9802$ or $= 100^2 - 3 \times 100 + 2 = 9802$ (iii) $104 \times 95$. Use $(100 + 4)(100 - 5) = 100^2 - 500 + 400 - 20 = 9880$ (iv) $(101)^3$. Use binomial expansion: $$(100 + 1)^3 = 100^3 + 3 \times 100^2 \times 1 + 3 \times 100 \times 1^2 + 1^3 = 1000000 + 30000 + 300 + 1 = 1030301$$ (v) $(399)^3$. Use $(400 - 1)^3 = 400^3 - 3 \times 400^2 + 3 \times 400 - 1$ Calculate: $$400^3 = 64000000$$ $$3 \times 400^2 = 3 \times 160000 = 480000$$ $$3 \times 400 = 1200$$ So: $$399^3 = 64000000 - 480000 + 1200 - 1 = 63501200 - 1 = 63501199$$