Find the Roots (Zeros) f(x)=2x^3+24x+16x^2

$f (x) = 2 x^{3} + 24 x + 16 x^{2}$

Step 1

Set $2 x^{3} + 24 x + 16 x^{2}$ equal to $0$ .

$2 x^{3} + 24 x + 16 x^{2} = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2 x$ out of $2 x^{3} + 24 x + 16 x^{2}$ .

Factor $2 x$ out of $2 x^{3}$ .

$2 x (x^{2}) + 24 x + 16 x^{2} = 0$

Factor $2 x$ out of $24 x$ .

$2 x (x^{2}) + 2 x (12) + 16 x^{2} = 0$

Factor $2 x$ out of $16 x^{2}$ .

$2 x (x^{2}) + 2 x (12) + 2 x (8 x) = 0$

Factor $2 x$ out of $2 x (x^{2}) + 2 x (12)$ .

$2 x (x^{2} + 12) + 2 x (8 x) = 0$

Factor $2 x$ out of $2 x (x^{2} + 12) + 2 x (8 x)$ .

$2 x (x^{2} + 12 + 8 x) = 0$

Factor.

Factor $x^{2} + 12 + 8 x$ 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 $12$ and whose sum is $8$ .

$2, 6$

Write the factored form using these integers.

$2 x ((x + 2) (x + 6)) = 0$

Remove unnecessary parentheses.

$2 x (x + 2) (x + 6) = 0$

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

$x = 0$

$x + 2 = 0$

$x + 6 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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.

$x = - 6$

The final solution is all the values that make $2 x (x + 2) (x + 6) = 0$ true.

$x = 0, - 2, - 6$

Step 3

Find the Roots (Zeros) f(x)=2x^3+24x+16x^2

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