Question

Find a polynomial P(x) with a degree of three and zeros -2,3,5 and passing through the point (2,24)

Answer 1

$\begin{cases} x=-2\implies &x+2=0\\ x=3\implies &x-3=0\\ x=5\implies &x-5=0 \end{cases}~\hfill a(x+2)(x-3)(x-5)=\stackrel{0}{y} \\\\\\ \stackrel{\textit{we also know that}}{x = 2\qquad y = 24}\qquad \implies a(2+2)(2-3)(2-5)=24 \\\\\\ a(4)(-1)(-3)=24\implies 12a=24\implies a=\cfrac{24}{12}\implies \underline{a=2} \\\\[-0.35em] ~\dotfill$

$2(x+2)(x-3)(x-5)=y\implies 2(x+2)\stackrel{F~O~I~L}{(x^2-8x+15)}=y \\\\\\ 2(x^3-8x^2+15x+2x^2-16x+30)=y\implies 2(x^3-6x^2-x+30)=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill 2x^3-12x^2-2x+60=y~\hfill$

Check the picture below.