Answer
$\color{blue}{f(x)=5x^2-30x+40}$
Work Step by Step
RECALL:
The zeros of the quadratic function $f(x) =c(x-a)(x-b)=0$ are $x=a$ and $x=b$.
The given quadratic function has the zeros $x=4$ and $x=2$.
Using the rule mentioned in the recall part above, then the function is:
$f(x)=c(x-4)(x-2)
\\f(x)=c[x(x)+x(-2) - 4(x) - 4(-2)]
\\f(x)=c(x^2-2x-4x+8)
\\f(x)=c(x^2-6x+8)$
With $f(0) = 40$, substitute $40$ to $f(0)$ and $0$ to $x$ to obtain:
$f(0) = c(0^2-6\cdot 0+8)
\\40 = c(0-0+8)
\\40=c(8)
\\\frac{40}{8}=c
\\5=c$
Thus, the function is:
$f(x) = 5(x^2-6x+8)
\\\color{blue}{f(x)=5x^2-30x+40}$
