Question

Help solving for the formula of p(x) for pre cal

Answer 1

`\begin{cases} x = 4\implies &x-4=0\\\\ x=0\implies &x=0\\\\ x=-4\implies &x+4=0 \end{cases} \\\\\\ a\stackrel{multiplicities}{\stackrel{two}{(x-4)^2} \stackrel{one}{(x)^1} \stackrel{one}{(x+4)^1}}~~ = ~~\stackrel{y}{0}\implies a(x-4)^2(x)(x+4)=y \\\\\\ \textit{we also know that} \begin{cases} x=-2\\ y = 144 \end{cases}\implies a(-2-4)^2 (-2) (-2+4)=144`

`a(36)(-2)(2)=144\implies -144a=144\implies a=\cfrac{144}{-144}\implies \boxed{a=-1} \\\\[-0.35em] ~\dotfill\\\\ -(x-4)^2(x)(x+4)=y\implies {\Large \begin{array}{llll} -x^4+4x^3+16x^2-64x=y \end{array}}`