1°) Sendo f(X)=x²+7x+10=0, caucule:
a) f(0)= b) f(1)=
c) f(2)= d) f(-1)=
e) f(-3)= f) f(-5)=
a) f(0)= b) f(1)=
c) f(2)= d) f(-1)=
e) f(-3)= f) f(-5)=
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f(0)=10 f(1)=1+7+10=18
c) f(2)=2^2+7.2+10 d) f(-1)= (-1)^2+7.(-1)+10
f(2)=4+14+10=28 f(-1)=1-7+10=4
e) f(-3)=(-3)^2+7.(-3)+10 f) f(-5)= (-5)^2+7.(-5)+10
f(-3)=9-21+10=-2 f(-5)=25-35+10=0