Find the Roots (Zeros) f(x)=4x^3-4x^2-16x+16

$f (x) = 4 x^{3} - 4 x^{2} - 16 x + 16$

Step 1

Set $4 x^{3} - 4 x^{2} - 16 x + 16$ equal to $0$ .

$4 x^{3} - 4 x^{2} - 16 x + 16 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $4$ out of $4 x^{3} - 4 x^{2} - 16 x + 16$ .

Factor $4$ out of $4 x^{3}$ .

$4 (x^{3}) - 4 x^{2} - 16 x + 16 = 0$

Factor $4$ out of $- 4 x^{2}$ .

$4 (x^{3}) + 4 (- x^{2}) - 16 x + 16 = 0$

Factor $4$ out of $- 16 x$ .

$4 (x^{3}) + 4 (- x^{2}) + 4 (- 4 x) + 16 = 0$

Factor $4$ out of $16$ .

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

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

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

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

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

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

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$4 (x^{2} (x - 1) - 4 (x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$4 ((x - 1) (x^{2} - 4)) = 0$

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

$4 ((x - 1) (x^{2} - 2^{2})) = 0$

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

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

$x - 1 = 0$

$x + 2 = 0$

$x - 2 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 1, - 2, 2$

Step 3

Find the Roots (Zeros) f(x)=4x^3-4x^2-16x+16

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