Solve by Factoring x^3+3x^2-16x-48=0

$x^{3} + 3 x^{2} - 16 x - 48 = 0$

Step 1

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} + 3 x^{2}) - 16 x - 48 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x + 3) - 16 (x + 3) = 0$

Step 2

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

$(x + 3) (x^{2} - 16) = 0$

Step 3

Rewrite $16$ as $4^{2}$ .

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

Step 4

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$(x + 3) ((x + 4) (x - 4)) = 0$

Remove unnecessary parentheses.

$(x + 3) (x + 4) (x - 4) = 0$

Step 5

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

$x + 3 = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 6

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.

$x = - 3$

Step 7

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8

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.

$x = 4$

Step 9

The final solution is all the values that make $(x + 3) (x + 4) (x - 4) = 0$ true.

$x = - 3, - 4, 4$