Answer
$f(x)=16x^{4}-80x^{3}+32x^{2}+128x$
Work Step by Step
If $r$ is a real zero of a polynomial function $f$ .
then $(x-r)$ is a factor of $f.$
$f(x)=a(x-(-1))(x-0)(x-2)(x-4)$
$f(x)=ax(x+1)(x-2)(x-4)$
We find a by substituting $f(\displaystyle \frac{1}{2})=63$
$63=a(\displaystyle \frac{1}{2})(\frac{1}{2}+1)(\frac{1}{2}-2)(\frac{1}{2}-4)$
$63=a(\displaystyle \frac{1}{2})(\frac{3}{2})(\frac{-3}{2})(\frac{-7}{2})$
$63=\displaystyle \frac{63}{16}a$
$a=16\quad \Rightarrow\quad f(x)=16x(x+1)(x-2)(x-4)$
$f(x)=16x(x^{2}-x-2)(x-4)$
$=16x(x^{3}-x^{2}-2x-4x^{2}+4x+8)$
$=16x(x^{3}-5x^{2}+2x+8)$
$=16x^{4}-80x^{3}+32x^{2}+128x$
