Precalculus Examples

${(y^{2} + 4 y)}^{2} + 7 (y^{2} + 4 y) + 12 = 0$

Step 1

Factor the left side of the equation.

Tap for more steps...

Step 1.1

Let $u = y^{2} + 4 y$ . Substitute $u$ for all occurrences of $y^{2} + 4 y$ .

$u^{2} + 7 u + 12 = 0$

Step 1.2

Factor $u^{2} + 7 u + 12$ using the AC method.

Tap for more steps...

Step 1.2.1

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 $12$ and whose sum is $7$ .

$3, 4$

Step 1.2.2

Write the factored form using these integers.

$(u + 3) (u + 4) = 0$

Step 1.3

Replace all occurrences of $u$ with $y^{2} + 4 y$ .

$(y^{2} + 4 y + 3) (y^{2} + 4 y + 4) = 0$

Step 2

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

$y^{2} + 4 y + 3 = 0$

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

Step 3

Set $y^{2} + 4 y + 3$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 3.1

Set $y^{2} + 4 y + 3$ equal to $0$ .

$y^{2} + 4 y + 3 = 0$

Step 3.2

Solve $y^{2} + 4 y + 3 = 0$ for $y$ .

Tap for more steps...

Step 3.2.1

Factor $y^{2} + 4 y + 3$ using the AC method.

Tap for more steps...

Step 3.2.1.1

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 $3$ and whose sum is $4$ .

$1, 3$

Step 3.2.1.2

Write the factored form using these integers.

$(y + 1) (y + 3) = 0$

Step 3.2.2

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

$y + 1 = 0$

$y + 3 = 0$

Step 3.2.3

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

Tap for more steps...

Step 3.2.3.1

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

$y + 1 = 0$

Step 3.2.3.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 3.2.4

Set $y + 3$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 3.2.4.1

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

$y + 3 = 0$

Step 3.2.4.2

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 3.2.5

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

$y = - 1, - 3$

Step 4

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

Tap for more steps...

Step 4.1

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

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

Step 4.2

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

Tap for more steps...

Step 4.2.1

Factor using the perfect square rule.

Tap for more steps...

Step 4.2.1.1

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

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

Step 4.2.1.2

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

$4 y = 2 \cdot y \cdot 2$

Step 4.2.1.3

Rewrite the polynomial.

$y^{2} + 2 \cdot y \cdot 2 + 2^{2} = 0$

Step 4.2.1.4

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

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

Step 4.2.2

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

$y + 2 = 0$

Step 4.2.3

Subtract $2$ from both sides of the equation.

$y = - 2$

Step 5

The final solution is all the values that make $(y^{2} + 4 y + 3) (y^{2} + 4 y + 4) = 0$ true.

$y = - 1, - 3, - 2$