Respuesta :

gmany

Answer:

[tex]\large\boxed{x\in\{-5,\ -1,\ 0,\ 4,\ 5\}}[/tex]

Step-by-step explanation:

The zeros of g(x):

[tex]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[/tex]

MrRoyal

The zeros of the function are x = 0, x = 5, x = -5, x = 4 and x = -1

The equation of the function is given as:

[tex]g(x)=x(x^2-25)(x^2-3x-4)[/tex]

Factorize the equation

[tex]g(x)=x(x-5)(x+ 5)(x^2-3x-4)[/tex]

Expand the equation

[tex]g(x)=x(x-5)(x+ 5)(x^2+x-4x-4)[/tex]

Factorize

[tex]g(x)=x(x-5)(x+ 5)(x(x+1)-4(x+1))[/tex]

Factor out x + 1

[tex]g(x)=x(x-5)(x+ 5)(x-4)(x+1)[/tex]

Equate to 0

[tex]x(x-5)(x+ 5)(x-4)(x+1) = 0[/tex]

Solve for x

x = 0, x = 5, x = -5, x = 4 and x = -1

Hence, the zeros of the function are x = 0, x = 5, x = -5, x = 4 and x = -1

Read more about zeros of the functions at:

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