Question

What are the zeros of g(x)=x(x^2-25)(x^2-3x-4)

Answer 1

Answer:

$\large\boxed{x\in\{-5,\ -1,\ 0,\ 4,\ 5\}}$

Step-by-step explanation:

The zeros of g(x):

$g(x)=x(x^2-25)(x^2-3x-4)\to g(x)=0\\\\x(x^2-25)(x^2-3x-4)=0\iff x=0\ \vee\ x^2-25=0\ \vee\ x^2-3x-4\\\\x^2-25=0\qquad\text{add 25 to both sides}\\x^2=25\to x=\pm\sqrt{25}\\x=-5\ \vee\ x=5\\\\x^2-3x-4=0\\x^2+x-4x-4=0\\x(x+1)-4(x+1)=0\\(x+1)(x-4)=0\iff x+1=0\ \vee\ x-4=0\\x=-1\ \vee\ x=4$