Answer to Question #251157 in Calculus for Shawty

Question

If a third degree polynomial has a line x-intercept at x=a, discuss what this implies about the linear and quadratic factors of that polynomial

Answer 1

Taking into account that a third degree polynomial $f$ has alone $x$-intercept at $x=a$, it has a unique root $x=a.$ If the multiplicity of this root is 3, then $f(x)=b(x-a)^3,$ and in this case the polynomial has three linear factors. If the root $x=a$ has multiplicity 2, then it has linear factor $x-c$ different from $x-a$ , and polinomial has the root $x=c\ne a,$ which is impossible according to uniqueness of a root. Thererfore, $x=a$ can not be of multiplicity 2. If the multiplicity of this root is 1, then $f(x)=(x-a)(mx^2+nx+k),$ and the polynomial $h(x)=mx^2+nx+k$ has no roots. It follows that the polynomial $g$ is irreducible, and hence it is the quadratic factor of the polynomial $f$.