Find the Roots (Zeros) f(x)=x^2+14x+48

$f (x) = x^{2} + 14 x + 48$

Step 1

Set $x^{2} + 14 x + 48$ equal to $0$ .

$x^{2} + 14 x + 48 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} + 14 x + 48$ 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 $48$ and whose sum is $14$ .

$6, 8$

Write the factored form using these integers.

$(x + 6) (x + 8) = 0$

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

$x + 6 = 0$

$x + 8 = 0$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $(x + 6) (x + 8) = 0$ true.

$x = - 6, - 8$

Step 3

Find the Roots (Zeros) f(x)=x^2+14x+48

Related Questions

Support LCMGCF.com by disabling your adblocker.