Question

Find the H.C.F of a²+7a+10,a²+6a+8anda²+9a+20​

Answer 1

a²+7a+10=0

(a+2)(a+5)=0

∴a+2=0

a= -2

a+5=0

a= -5

a²+6a+8=0

(a+2)(a+4)=0

∴a+2=0

a= -2

a+4=0

a= -4

a²+9a+20

(a+4)(a+5)

∴a+4=0

a= -4

a+5=0

a= -5

Explanation:

Solution on the Math book