Algebra Examples

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

Step 1

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

$2 x - 7 = 0$

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

Step 2

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

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Step 2.1

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

$2 x - 7 = 0$

Step 2.2

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

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Step 2.2.1

Add $7$ to both sides of the equation.

$2 x = 7$

Step 2.2.2

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

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Step 2.2.2.1

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

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

Step 2.2.2.2

Simplify the left side.

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Step 2.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.2.2.1.1

Cancel the common factor.

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

Step 2.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

Factor using the perfect square rule.

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Step 3.2.1.1

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

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

Step 3.2.1.2

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

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

Step 3.2.1.3

Rewrite the polynomial.

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

Step 3.2.1.4

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

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

Step 3.2.2

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

$x - 2 = 0$

Step 3.2.3

Add $2$ to both sides of the equation.

$x = 2$

Step 4

The final solution is all the values that make $(2 x - 7) (x^{2} - 4 x + 4) > 0$ true.

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

Step 5

The solution consists of all of the true intervals.

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

Step 6

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 7