Consider the function f(x)=x2+bx−16, where b is a constant.
Part A
If the function has an axis of symmetry at x=5, what is the value of b?
Part B
If b=−6, what are the zero(s) of the function?

F(x)=x^2+bx-16
x=5
F(5)=5^2+b(5)-16
=25+5b-16=0
=5b+9=0
=5b=-9
=b= -9/5
By remainder theorem

21 is the answer the answer is 21

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