Find the Roots (Zeros) f(x)=2x^2-6x-8

$f (x) = 2 x^{2} - 6 x - 8$

Step 1

Set $2 x^{2} - 6 x - 8$ equal to $0$ .

$2 x^{2} - 6 x - 8 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 6 x - 8$ .

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

$2 (x^{2}) - 6 x - 8 = 0$

Factor $2$ out of $- 6 x$ .

$2 (x^{2}) + 2 (- 3 x) - 8 = 0$

Factor $2$ out of $- 8$ .

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

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

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

Factor $2$ out of $2 (x^{2} - 3 x) + 2 \cdot - 4$ .

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

Factor.

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

$- 4, 1$

Write the factored form using these integers.

$2 ((x - 4) (x + 1)) = 0$

Remove unnecessary parentheses.

$2 (x - 4) (x + 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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.

$x = - 1$

The final solution is all the values that make $2 (x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Step 3

Find the Roots (Zeros) f(x)=2x^2-6x-8

Related Questions

Support LCMGCF.com by disabling your adblocker.