Simplify 3<3+y^2+8+6y

$3 < 3 + y^{2} + 8 + 6 y$

Rewrite so $y$ is on the left side of the inequality.

$3 + y^{2} + 8 + 6 y > 3$

Add $3$ and $8$ .

$y^{2} + 11 + 6 y > 3$

Convert the inequality to an equation.

$y^{2} + 11 + 6 y = 3$

Subtract $3$ from both sides of the equation.

$y^{2} + 11 + 6 y - 3 = 0$

Subtract $3$ from $11$ .

$y^{2} + 6 y + 8 = 0$

Factor $y^{2} + 6 y + 8$ 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 $8$ and whose sum is $6$ .

$2, 4$

Write the factored form using these integers.

$(y + 2) (y + 4) = 0$

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

$y + 2 = 0$

$y + 4 = 0$

Set $y + 2$ equal to $0$ and solve for $y$ .

Set $y + 2$ equal to $0$ .

$y + 2 = 0$

Subtract $2$ from both sides of the equation.

$y = - 2$

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

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

$y + 4 = 0$

Subtract $4$ from both sides of the equation.

$y = - 4$

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

$y = - 2, - 4$

Use each root to create test intervals.

$y < - 4$

$- 4 < y < - 2$

$y > - 2$

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 $y < - 4$ to see if it makes the inequality true.

Choose a value on the interval $y < - 4$ and see if this value makes the original inequality true.

$y = - 6$

Replace $y$ with $- 6$ in the original inequality.

$3 < 3 + {(- 6)}^{2} + 8 + 6 (- 6)$

The left side $3$ is less than the right side $11$ , which means that the given statement is always true.

True

Test a value on the interval $- 4 < y < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < y < - 2$ and see if this value makes the original inequality true.

$y = - 3$

Replace $y$ with $- 3$ in the original inequality.

$3 < 3 + {(- 3)}^{2} + 8 + 6 (- 3)$

The left side $3$ is not less than the right side $2$ , which means that the given statement is false.

False

Test a value on the interval $y > - 2$ to see if it makes the inequality true.

Choose a value on the interval $y > - 2$ and see if this value makes the original inequality true.

$y = 0$

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

$3 < 3 + {(0)}^{2} + 8 + 6 (0)$

The left side $3$ is less than the right side $11$ , which means that the given statement is always true.

True

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

$y < - 4$ True

$- 4 < y < - 2$ False

$y > - 2$ True

$y < - 4$ True

$- 4 < y < - 2$ False

$y > - 2$ True

The solution consists of all of the true intervals.

$y < - 4$ or $y > - 2$

The result can be shown in multiple forms.

Inequality Form:

$y < - 4 or y > - 2$

Interval Notation:

$(- \infty, - 4) \cup (- 2, \infty)$

Simplify 3<3+y^2+8+6y

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