# Solutions to Problems 1301 - 1310

Q1301 (Suggested by J. Guest, Victoria)

Solve the quartic $$(x+1)(x+5)(x-3)(x-7) = -135$$

ANS: (suggested by David Shaw, Geelong, Victoria)

Rearrange the equation as $$(x^2-2x-3)(x^2-2x-35) = -135$$ By setting $z=x^2-2x$, the above equation becomes $$z^2-38z+240=0$$
which has solutions $z_1=30$ and $z_2=8$. Solving the two equations $x^2-2x=30$ and $x^2-2x=8$ results in 4 solutions to the quartic equation $$x_1 = 1+\sqrt{31}, \quad x_2 = 1-\sqrt{31}, \quad x_3 = 4, \quad\text{and}\quad x_4 = -2$$