Find the Roots (Zeros) f(x)=-3(x-1)^2(x^2-4)

$f (x) = - 3 {(x - 1)}^{2} (x^{2} - 4)$

Step 1

Set $- 3 {(x - 1)}^{2} (x^{2} - 4)$ equal to $0$ .

$- 3 {(x - 1)}^{2} (x^{2} - 4) = 0$

Step 2

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(x - 1)}^{2} = 0$

$x^{2} - 4 = 0$

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

${(x - 1)}^{2} = 0$

Solve ${(x - 1)}^{2} = 0$ for $x$ .

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x^{2} - 4$ equal to $0$ and solve for $x$ .

Set $x^{2} - 4$ equal to $0$ .

$x^{2} - 4 = 0$

Solve $x^{2} - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

The final solution is all the values that make $- 3 {(x - 1)}^{2} (x^{2} - 4) = 0$ true.

$x = 1, 2, - 2$

Step 3