Question

Factorise :


a²bx + 2a_cx - 2aby - 4acy + bz + 2cz

(ii) a + b + c + ab + bc + ac + 1 + abc


2.
+
+2 - 2x
(iii) x2
x²
Х​

Answer 1

Answer:

a²bx + 2a²cx - 2aby - 4acy + bz + 2cz

= a²x (b + 2c) - 2ay (b + 2c) + z (b + 2c)

= (b + 2c) (a²x - 2ay + z),

Answer 2

Step-by-step explanation:

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