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10-10-2020
Mathematics

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

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10-10-2020

Answer:

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

Step-by-step explanation:

Solution on the Math book

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