Question

Factorise the expressions:

i) 6ab – b² + 12ac – 2bc

ii) p²q – pr² – pq + r²

iii) x(y + z) - 5(y + z)

Answer 1

i) 6ab – b² + 12ac – 2bc

6ab – b² + 12ac – 2bc

= (6ab – b²) + (12ac – 2bc)

= (6 × a× b – b × b) + (6 × 2 × a × c – 2 × b × c)

= b[6a – b] + 2c[6a – b]

= (6a – b) (b + 2c)

ii) p²q – pr² – pq + r²

p²q – pr² – pq + r²

= (p²q – pr²) – (pq – r²)

= (p × p × q – p × r × r) – (pq – r²)

= P(pq – r²) – (pq – r²) × 1

= (pq – r²)(p – 1)

iii) x (y + z) -5 (y + z)

= x(y + z) – 5(y + z)

= (y + z) (x – 5)