1)15a+10b-5ab-30= 5(3a+2b-ab-6)=5( a(3-b) + 2(b-3) ) =5( a(3-b) - 2(3-b) ) =5(a-2)(3-b)
2)-30k+30-10p+10kp= 10(-3(k-1) + p (k-1) ) =10(p-3)(k-1)
1.=(10b-30)-(5ab-15a)=10(b-3)-5ab(b-3)=(b-3)(10-5ab)=5(b-3)(2-ab)
2.=(30-30k)-(10p-10kp)=30(1-k)-10p(1-k)=(1-k)(30-10p)=10(1-k)(3-p)
1)15a+10b-5ab-30= 5(3a+2b-ab-6)=5( a(3-b) + 2(b-3) ) =5( a(3-b) - 2(3-b) ) =5(a-2)(3-b)
2)-30k+30-10p+10kp= 10(-3(k-1) + p (k-1) ) =10(p-3)(k-1)
1.=(10b-30)-(5ab-15a)=10(b-3)-5ab(b-3)=(b-3)(10-5ab)=5(b-3)(2-ab)
2.=(30-30k)-(10p-10kp)=30(1-k)-10p(1-k)=(1-k)(30-10p)=10(1-k)(3-p)