Let p(x) be a polynomial of the form p(x)=2x^3+ax^2+38x+c, such that 4 and 6 are roots of p(x). What is the third root?

Let p(x) be a polynomial of the form p(x)=2x^3+ax^2+38x+c, such that 4 and 6 are roots of p(x).

(a) What is the third root? 

let the roots be 4,6 and L

\( 4L+6L+4*6=\frac{38}{2}\\ 10L+24=19\\ 10L=-5\\ L=-0.5 \)

The third root is  -0.5

(b) Compute ac.

\(-0.5*4*6=\frac{-c}{2}\\ -24=-c\\ c=24\\ 4+6+-0.5=\frac{-a}{2}\\ 19=-a\\ a=-19\\~\\ ac=-19*24=-456\\~\\ \)

I have not checked my answers Debae, I will leave that task to you.  :)