Solve for x (2x-7)(x^2-4x+4)>0

$(2 x - 7) (x^{2} - 4 x + 4) > 0$

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

$2 x - 7 = 0$

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

Set $2 x - 7$ equal to $0$ and solve for $x$ .

Set $2 x - 7$ equal to $0$ .

$2 x - 7 = 0$

Solve $2 x - 7 = 0$ for $x$ .

Add $7$ to both sides of the equation.

$2 x = 7$

Divide each term in $2 x = 7$ by $2$ and simplify.

Divide each term in $2 x = 7$ by $2$ .

$\frac{2 x}{2} = \frac{7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Divide $x$ by $1$ .

$x = \frac{7}{2}$

Set $x^{2} - 4 x + 4$ equal to $0$ and solve for $x$ .

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

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

Solve $x^{2} - 4 x + 4 = 0$ for $x$ .

Factor using the perfect square rule.

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 2 + 2^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

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

Set the $x - 2$ 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 $(2 x - 7) (x^{2} - 4 x + 4) > 0$ true.

$x = \frac{7}{2}, 2$

The solution consists of all of the true intervals.

$x > \frac{7}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x > \frac{7}{2}$

Interval Notation:

$(\frac{7}{2}, \infty)$