Find the Roots (Zeros) p(x)=(x^2-1)(x^2-5x+6)

$p (x) = (x^{2} - 1) (x^{2} - 5 x + 6)$

Set $(x^{2} - 1) (x^{2} - 5 x + 6)$ equal to $0$ .

$(x^{2} - 1) (x^{2} - 5 x + 6) = 0$

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 $0$ .

$x^{2} - 1 = 0$

$x^{2} - 5 x + 6 = 0$

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

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

$x^{2} - 1 = 0$

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

Add $1$ to both sides of the equation.

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

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 $x^{2} - 5 x + 6$ equal to $0$ and solve for $x$ .

Set $x^{2} - 5 x + 6$ equal to $0$ .

$x^{2} - 5 x + 6 = 0$

Solve $x^{2} - 5 x + 6 = 0$ for $x$ .

Factor $x^{2} - 5 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

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

$x - 3 = 0$

$x - 2 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 3, 2$

The final solution is all the values that make $(x^{2} - 1) (x^{2} - 5 x + 6) = 0$ true.

$x = 1, - 1, 3, 2$

Find the Roots (Zeros) p(x)=(x^2-1)(x^2-5x+6)

Related Questions

Support LCMGCF.com by disabling your adblocker.