Question

If $f\left(x\right)=x^2+6x+c$, where ${}^{\prime }{c}^{\prime }$ is an integer, then $f\left(0\right)+f(-1)$ is

• A. an even integer
• B. an odd integer always disable by $3$
• C. an odd integer not divisible by $3$
• D. an odd integer may or not be divisible by $3$

Answer 1

Answer: (D)

an odd integer may or not be divisible by $3$

$f\left(0\right)=c$

$f(-1)=c-5$

$f\left(0\right)+f(-1)=2c-5=2(c-3)+1$

As the above is of the form $2k+1$, it is always odd.
For $c=3$, the above is not divisible by 3 but for $c=4$, it is. Therefore, it may or may not be divisible by 3.