## Precalculus (6th Edition) Blitzer

The complete factorization is $\underline{f\left( x \right)=4\left( x+2 \right)\left( x+\frac{1}{2} \right)\left( x-\frac{1}{2} \right)\left( x-3 \right)}$
It is observed that the function crosses the $x-$ axis at $x=-2,-\frac{1}{2},-\frac{1}{2},3$ Therefore $\left( x+2 \right),\left( x+\frac{1}{2} \right),\left( x-\frac{1}{2} \right),\left( x-3 \right)$ are the four factors of the polynomial $f\left( x \right)$; hence the polynomial $f\left( x \right)$ can be considered as $f\left( x \right)=\left( x+2 \right)\left( x+\frac{1}{2} \right)\left( x-\frac{1}{2} \right)\left( x-3 \right)g\left( x \right)$ Where $g\left( x \right)$ is a polynomial of degree $\le 4$. As the degree of the polynomial $f\left( x \right)$ is 4, hence the degree of $g\left( x \right)$ is zero, which is a constant and therefore $f\left( x \right)=C\left( x+2 \right)\left( x+\frac{1}{2} \right)\left( x-\frac{1}{2} \right)\left( x-3 \right)$ Equating the coefficients of ${{x}^{4}}$ on both sides $4=C$ Therefore $f\left( x \right)=4\left( x+2 \right)\left( x+\frac{1}{2} \right)\left( x-\frac{1}{2} \right)\left( x-3 \right)$ The complete factorization is, $f\left( x \right)=4\left( x+2 \right)\left( x+\frac{1}{2} \right)\left( x-\frac{1}{2} \right)\left( x-3 \right)$.