Factorise 15a (2p – 3q) – 10b (2p – 3q)

(i) 15a (2p – 3q) – 10b (2p – 3q)

(ii) 3a(x² + y²) + 6b (x² + y²)

Sol: (i) 15a (2p – 3q) – 10b (2p – 3q)

Take out common in both terms,

Then, 5(2p – 3q) [3a – 2b]

Therefore, HCF of 15a (2p – 3q) and 10b (2p – 3q) is 5(2p – 3q).

(ii) 3a(x² + y²) + 6b (x² + y²)

3a(x² + y²) + 6b (x² + y²)

Take out common in all terms,

Then, 3(x² + y²) (a + 2b)

Therefore, HCF of 3a(x² + y²) and 6b (x² + y²) is 3(x² + y²).

More Solutions:

• Factorise a2b – ab2 + 3a – 3b
• Factorise 6xy2 – 3xy – 10y + 5
• Factorise 1 – a – b + ab
• Factorise x2 + xy (1 + y) + y3
• Factorise ab2 + (a – 1)b – 1
• Factorise 5ph – 10qk + 2rph – 4qrk
• Factorise ab(x2 + y2) – xy(a2 + b2)
• Factorise a3 + ab(1 – 2a) – 2b2
• Factorise a2b + ab2 –abc – b2c + axy + bxy
• Factorise ax2 – bx2 + ay2 – by2 + az2 – bz2
• Factorise x – 1 – (x – 1)2 + ax – a