Question

What are all the roots of the following polynomial f(x)=x^4-13x^2+36

Answer 1

$\bf f(x)=x^4-13x^2+36\implies \stackrel{f(x)}{0}=x^4-13x^2+36 \\\\\\ 0=(x^2)^2-13x^2+36~~ \begin{cases} -4-9=\stackrel{middle~term}{-13}\\ -4\cdot -9=\stackrel{constant}{36} \end{cases} \\\\\\ 0=(x^2-4)(x^2-9)\\\\ -------------------------------\\\\ 0=x^2-4\implies 4=x^2\implies \pm\sqrt{4}=x\implies \pm 2 =x\\\\ -------------------------------\\\\ 0=x^2-9\implies 9=x^2\implies \pm\sqrt{9}=x\implies \pm 3= x$