Prove that roots of cubic equation are real.

Question

The equation is $(x+4)(x+2)(x-3)+(x+3)(x+1)(x-5)=0$.

How do I prove that the all the roots are real and distinct?

Answer 1

Let $f(x)=(x+4)(x+2)(x-3)+(x+3)(x+1)(x-5)$. Then

• since $-27=f(-4)<0<6=f(-3)$, there is a root in $(-4,-3)$;
• since $7=f(-2)>0>-12=f(-1)$, there is a root in $(-2,-1)$;
• since $-48=f(3)<0<13=f(4)$, there is a root in $(3,4)$.

Since it is a cubic equation, it can have no more roots. Therefore, all roots are real.