Question

The polynomial equation x(×^2+4)(x^2-x-6)=0 has how many real roots

Answer 1

`x(x^2+4)(x^2-x-6)=0\iff x=0\ \vee\ x^2+4=0\ \vee\ x^2-x-6=0\\\\\boxed{x=0}\\\\x^2+4=0\ \ \ |-4\\x^2=-4 < 0-No\ real\ solution\\\\x^2-x-6=0\\\\x^2+2x-3x-6=0\\\\x(x+2)-3(x+2)=0\\\\(x+2)(x-3)=0\iff \boxed{x=-2}\ \vee\ \boxed{x=3}`

Answer: 3 real roots: -2, 0 and 3.