Find the Roots (Zeros) p(x)=(x^4-1)(x-1)

$p (x) = (x^{4} - 1) (x - 1)$

Step 1

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

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

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 $0$ .

$x^{4} - 1 = 0$

$x - 1 = 0$

Set $x^{4} - 1$ equal to $0$ and solve for $x$ .

Set $x^{4} - 1$ equal to $0$ .

$x^{4} - 1 = 0$

Solve $x^{4} - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$x^{4} = 1$

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

$x = \pm \sqrt[4]{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

Set $x - 1$ equal to $0$ and solve for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 1, - 1$

Step 3