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

$y = - x^{2} - 2 x + 3$

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

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

Solve for $x$ .

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 2 x + 3$ .

Factor $- 1$ out of $- x^{2}$ .

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

Factor $- 1$ out of $- 2 x$ .

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

Rewrite $3$ as $- 1 (- 3)$ .

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

Factor $- 1$ out of $- (x^{2}) - (2 x)$ .

$- (x^{2} + 2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (- 3)$ .

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

Factor.

Factor $x^{2} + 2 x - 3$ 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 $2$ .

$- 1, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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 $0$ .

$x - 1 = 0$

Add $1$ to 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.

$x = 1, - 3$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.