Factorise the expressions. a) \( 2 a^{2}+2 a d+2 a c-a b-b d-b c \) b) \( x^{3}-2 x^{2}-x y+x z-2 z+2 y \) c) \( 2 x^{2}+2 x y+x z+x z^{2}+y z+y z^{2} \) d) \( p^{2} x-2 q^{2} y-2 p q x+p q y \)
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To factorise the expressions, let's start with the first one: a) \( 2a^2 + 2ad + 2ac - ab - bd - bc \) First, we can group the terms: \( 2a^2 + 2ad + 2ac - ab - bd - bc = 2a(a + d + c) - b(a + d + c) = (2a - b)(a + d + c) \). b) \( x^3 - 2x^2 - xy + xz - 2z + 2y \) Grouping the terms gives us: \( x^3 - 2x^2 + x(z - y) - 2(z - y) = (x^3 - 2x^2) + (x - 2)(z - y) \). Now factoring out \( x^2 \) from the first group: \( x^2(x - 2) + (x - 2)(z - y) = (x - 2)(x^2 + z - y) \). c) \( 2x^2 + 2xy + xz + xz^2 + yz + yz^2 \) We can factor out common terms: Grouping gives us \( 2x^2 + 2xy + xz(1 + z) + y(z + z^2) \). This allows us to factor out \( 2 \) and \( y \): \( 2x(x + y) + z(x(1 + z) + y(1 + z)) = (2x + z)(x + y) \). d) \( p^2x - 2q^2y - 2pqx + pqy \) We can arrange terms slightly: Grouping gives us \( p^2x - 2pqx + pqy - 2q^2y = (p^2 - 2pq)x + (pq - 2q^2)y \). Further factoring gives us \( p(px - 2q) + q(y - 2q) \). Thus it becomes: \( (p + q)(px - 2q) \).
