Solve the Inequality for x x(x+1)>112-5x

$x (x + 1) > 112 - 5 x$

Simplify $x (x + 1)$ .

Rewrite.

$0 + 0 + x (x + 1) > 112 - 5 x$

Simplify by adding zeros.

$x (x + 1) > 112 - 5 x$

Apply the distributive property.

$x \cdot x + x \cdot 1 > 112 - 5 x$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 1 > 112 - 5 x$

Multiply $x$ by $1$ .

$x^{2} + x > 112 - 5 x$

Move all terms containing $x$ to the left side of the inequality.

Add $5 x$ to both sides of the inequality.

$x^{2} + x + 5 x > 112$

Add $x$ and $5 x$ .

$x^{2} + 6 x > 112$

Convert the inequality to an equation.

$x^{2} + 6 x = 112$

Subtract $112$ from both sides of the equation.

$x^{2} + 6 x - 112 = 0$

Factor $x^{2} + 6 x - 112$ 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 $- 112$ and whose sum is $6$ .

$- 8, 14$

Write the factored form using these integers.

$(x - 8) (x + 14) = 0$

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

$x - 8 = 0$

$x + 14 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Set $x + 14$ equal to $0$ and solve for $x$ .

Set $x + 14$ equal to $0$ .

$x + 14 = 0$

Subtract $14$ from both sides of the equation.

$x = - 14$

The final solution is all the values that make $(x - 8) (x + 14) = 0$ true.

$x = 8, - 14$

Use each root to create test intervals.

$x < - 14$

$- 14 < x < 8$

$x > 8$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 14$ to see if it makes the inequality true.

Choose a value on the interval $x < - 14$ and see if this value makes the original inequality true.

$x = - 16$

Replace $x$ with $- 16$ in the original inequality.

$(- 16) ((- 16) + 1) > 112 - 5 \cdot - 16$

The left side $240$ is greater than the right side $192$ , which means that the given statement is always true.

True

Test a value on the interval $- 14 < x < 8$ to see if it makes the inequality true.

Choose a value on the interval $- 14 < x < 8$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$(0) ((0) + 1) > 112 - 5 \cdot 0$

The left side $0$ is not greater than the right side $112$ , which means that the given statement is false.

False

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$(10) ((10) + 1) > 112 - 5 \cdot 10$

The left side $110$ is greater than the right side $62$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 14$ True

$- 14 < x < 8$ False

$x > 8$ True

$x < - 14$ True

$- 14 < x < 8$ False

$x > 8$ True

The solution consists of all of the true intervals.

$x < - 14$ or $x > 8$

The result can be shown in multiple forms.

Inequality Form:

$x < - 14 or x > 8$

Interval Notation:

$(- \infty, - 14) \cup (8, \infty)$

Solve the Inequality for x x(x+1)>112-5x

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