Find the Roots (Zeros) (x^2+4x+3)(x^2-4)

$(x^{2} + 4 x + 3) (x^{2} - 4)$

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

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

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 $0$ .

$x^{2} + 4 x + 3 = 0$

$x^{2} - 4 = 0$

Set $x^{2} + 4 x + 3$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4 x + 3$ equal to $0$ .

$x^{2} + 4 x + 3 = 0$

Solve $x^{2} + 4 x + 3 = 0$ for $x$ .

Factor $x^{2} + 4 x + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

$(x + 1) (x + 3) = 0$

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

$x + 1 = 0$

$x + 3 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

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 $(x^{2} + 4 x + 3) (x^{2} - 4) = 0$ true.

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

Find the Roots (Zeros) (x^2+4x+3)(x^2-4)

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