**(i) 2a**^{2}x – bx + 2a^{2} – b

^{2}x – bx + 2a

^{2}– b

**(ii) p ^{2} – (a + 2b)p + 2ab**

**Answer :**

**(i) 2a**^{2}x – bx + 2a^{2} – b

^{2}x – bx + 2a

^{2}– b

Rearrange the above terms we get,

2a^{2}x + 2a – bx – b

Take out common in both terms,

2a^{2}(x + 1) – b(x + 1)

(x + 1) (2a^{2} – b)

**(ii) p**^{2} – (a + 2b)p + 2ab

^{2}– (a + 2b)p + 2ab

p^{2} – (a + 2b)p + 2ab

Above terms can be written as,

p^{2} – ap – 2bp + 2ab

Take out common in both terms,

p(p – a) – 2b(p – a)

(p – a) (p – 2b)

**More Solutions:**

- Factorise 21×2 – 59xy + 40y2
- Factorise x2y2 – xy – 72
- Factorise (3a – 2b)2 + 3(3a – 2b) – 10
- Factorise (x2 – 3x) (x2 – 3x + 7) + 10
- Factorise (x2 – x) (4×2 – 4x – 5) – 6
- Factorise x4 + 9x2y2 + 81y4
- Factorise (8/27)x3 – (1/8)y3
- Factorise x6 + 63×3 – 64
- Factorise x3 + x2 – (1/x2) + (1/x3)
- Factorise (x + 1)6 – (x – 1)6
- Find the value of a4 + a2b2 + b4.