Simplify 6-1>=5+8(2+y)+y^2

$6 - 1 \geq 5 + 8 (2 + y) + y^{2}$

Step 1

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

$5 + 8 (2 + y) + y^{2} \leq 6 - 1$

Step 2

Simplify $5 + 8 (2 + y) + y^{2}$ .

Simplify each term.

Apply the distributive property.

$5 + 8 \cdot 2 + 8 y + y^{2} \leq 6 - 1$

Multiply $8$ by $2$ .

$5 + 16 + 8 y + y^{2} \leq 6 - 1$

Add $5$ and $16$ .

$21 + 8 y + y^{2} \leq 6 - 1$

Step 3

Subtract $1$ from $6$ .

$21 + 8 y + y^{2} \leq 5$

Step 4

Convert the inequality to an equation.

$21 + 8 y + y^{2} = 5$

Step 5

Subtract $5$ from both sides of the equation.

$21 + 8 y + y^{2} - 5 = 0$

Step 6

Subtract $5$ from $21$ .

$8 y + y^{2} + 16 = 0$

Step 7

Factor the left side of the equation.

Let $u = y$ . Substitute $u$ for all occurrences of $y$ .

$8 u + u^{2} + 16 = 0$

Factor using the perfect square rule.

Rearrange terms.

$u^{2} + 8 u + 16 = 0$

Rewrite $16$ as $4^{2}$ .

$u^{2} + 8 u + 4^{2} = 0$

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

$8 u = 2 \cdot u \cdot 4$

Rewrite the polynomial.

$u^{2} + 2 \cdot u \cdot 4 + 4^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 4$ .

${(u + 4)}^{2} = 0$

Replace all occurrences of $u$ with $y$ .

${(y + 4)}^{2} = 0$

Step 8

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

$y + 4 = 0$

Step 9

Subtract $4$ from both sides of the equation.

$y = - 4$

Step 10

Simplify 6-1>=5+8(2+y)+y^2

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