Chapter 11 - Section 11.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 790: 79

Answers may vary. One possible equation is $x^3-39x+70=0$

$x=2$, $x=5$, and $x=-7$ are solutions $(x-2)(x-5)(x+7)$ $(x*x+x*(-5)+(-2)*x+(-2)(-5))*(x+7)$ $(x^2-5x-2x+10)*(x+7)$ $(x^2-7x+10)*(x+7)$ $x^2*x+x^2*7+(-7x)*x+(-7x)*7+10*x+10*7$ $x^3+7x^2-7x^2-49x+10x+70$ $x^3-39x+70$