Question 7a for my homework said to show that $x=-1$ is a solution of $x^3+px^2+px+1=0$. I just did that by using synthetic division.
Question 7b however says "hence find the range of values of $p$ for the equation to have real roots." I've never encountered a question like this, and dont know what to do. It's a 7 mark question out of 36, so I really need help understanding how to answer it.
Best Answer
it is the same as $$ (x+1) \left(x^2+(p-1) x+1\right)=0 $$ as for quadratic equation we have to require $$ (p-1)^2-4>0 $$ (say because the roots are $\frac{-(p-1)\pm \sqrt{(p-1)^2-4}}{2}$)
which implies $p>3$ or $p<-1$