Find the Roots (Zeros) (x^2-1)(9x^2+16x+21)

$(x^{2} - 1) (9 x^{2} + 16 x + 21)$

Set $(x^{2} - 1) (9 x^{2} + 16 x + 21)$ equal to $0$ .

$(x^{2} - 1) (9 x^{2} + 16 x + 21) = 0$

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 $0$ .

$x^{2} - 1 = 0$

$9 x^{2} + 16 x + 21 = 0$

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 $9 x^{2} + 16 x + 21$ equal to $0$ and solve for $x$ .

Set $9 x^{2} + 16 x + 21$ equal to $0$ .

$9 x^{2} + 16 x + 21 = 0$

Solve $9 x^{2} + 16 x + 21 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 9$ , $b = 16$ , and $c = 21$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (9 \cdot 21)}}{2 \cdot 9}$

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 9 \cdot 21}}{2 \cdot 9}$

Multiply $- 4$ by $9$ .

$x = \frac{- 16 \pm \sqrt{256 - 36 \cdot 21}}{2 \cdot 9}$

Multiply $- 36$ by $21$ .

$x = \frac{- 16 \pm \sqrt{256 - 756}}{2 \cdot 9}$

Subtract $756$ from $256$ .

$x = \frac{- 16 \pm \sqrt{- 500}}{2 \cdot 9}$

Rewrite $- 500$ as $- 1 (500)$ .

$x = \frac{- 16 \pm \sqrt{- 1 \cdot 500}}{2 \cdot 9}$

Rewrite $\sqrt{- 1 (500)}$ as $\sqrt{- 1} \cdot \sqrt{500}$ .

$x = \frac{- 16 \pm \sqrt{- 1} \cdot \sqrt{500}}{2 \cdot 9}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 16 \pm i \cdot \sqrt{500}}{2 \cdot 9}$

Rewrite $500$ as $10^{2} \cdot 5$ .

Factor $100$ out of $500$ .

$x = \frac{- 16 \pm i \cdot \sqrt{100 (5)}}{2 \cdot 9}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{- 16 \pm i \cdot \sqrt{10^{2} \cdot 5}}{2 \cdot 9}$

Pull terms out from under the radical.

$x = \frac{- 16 \pm i \cdot (10 \sqrt{5})}{2 \cdot 9}$

Move $10$ to the left of $i$ .

$x = \frac{- 16 \pm 10 i \sqrt{5}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{- 16 \pm 10 i \sqrt{5}}{18}$

Simplify $\frac{- 16 \pm 10 i \sqrt{5}}{18}$ .

$x = \frac{- 8 \pm 5 i \sqrt{5}}{9}$

The final answer is the combination of both solutions.

$x = - \frac{8 - 5 i \sqrt{5}}{9}, - \frac{8 + 5 i \sqrt{5}}{9}$

The final solution is all the values that make $(x^{2} - 1) (9 x^{2} + 16 x + 21) = 0$ true.

$x = 1, - 1, - \frac{8 - 5 i \sqrt{5}}{9}, - \frac{8 + 5 i \sqrt{5}}{9}$

Find the Roots (Zeros) (x^2-1)(9x^2+16x+21)

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