Solve by Factoring (y^2+6y)^2+14(y^2+6y)+45=0

${(y^{2} + 6 y)}^{2} + 14 (y^{2} + 6 y) + 45 = 0$

Step 1

Let $u = y^{2} + 6 y$ . Substitute $u$ for all occurrences of $y^{2} + 6 y$ .

$u^{2} + 14 u + 45 = 0$

Step 2

Factor $u^{2} + 14 u + 45$ 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 $45$ and whose sum is $14$ .

$5, 9$

Write the factored form using these integers.

$(u + 5) (u + 9) = 0$

Step 3

Replace all occurrences of $u$ with $y^{2} + 6 y$ .

$(y^{2} + 6 y + 5) (y^{2} + 6 y + 9) = 0$

Step 4

Factor $y^{2} + 6 y + 5$ 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 $5$ and whose sum is $6$ .

$1, 5$

Write the factored form using these integers.

$(y + 1) (y + 5) (y^{2} + 6 y + 9) = 0$

Step 5

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$(y + 1) (y + 5) (y^{2} + 6 y + 3^{2}) = 0$

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

$6 y = 2 \cdot y \cdot 3$

Rewrite the polynomial.

$(y + 1) (y + 5) (y^{2} + 2 \cdot y \cdot 3 + 3^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = y$ and $b = 3$ .

$(y + 1) (y + 5) {(y + 3)}^{2} = 0$

Step 6

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

$y + 1 = 0$

$y + 5 = 0$

${(y + 3)}^{2} = 0$

Step 7

Set $y + 1$ equal to $0$ and solve for $y$ .

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 8

Set $y + 5$ equal to $0$ and solve for $y$ .

Set $y + 5$ equal to $0$ .

$y + 5 = 0$

Subtract $5$ from both sides of the equation.

$y = - 5$

Step 9

Set ${(y + 3)}^{2}$ equal to $0$ and solve for $y$ .

Set ${(y + 3)}^{2}$ equal to $0$ .

${(y + 3)}^{2} = 0$

Solve ${(y + 3)}^{2} = 0$ for $y$ .

Set the $y + 3$ equal to $0$ .

$y + 3 = 0$

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 10

The final solution is all the values that make $(y + 1) (y + 5) {(y + 3)}^{2} = 0$ true.

$y = - 1, - 5, - 3$

Solve by Factoring (y^2+6y)^2+14(y^2+6y)+45=0

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