Solve by Factoring y^4-25y^2+144=0

$y^{4} - 25 y^{2} + 144 = 0$

Step 1

Rewrite $y^{4}$ as ${(y^{2})}^{2}$ .

${(y^{2})}^{2} - 25 y^{2} + 144 = 0$

Step 2

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

$u^{2} - 25 u + 144 = 0$

Step 3

Factor $u^{2} - 25 u + 144$ 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 $144$ and whose sum is $- 25$ .

$- 16, - 9$

Write the factored form using these integers.

$(u - 16) (u - 9) = 0$

Step 4

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

$(y^{2} - 16) (y^{2} - 9) = 0$

Step 5

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

$(y^{2} - 4^{2}) (y^{2} - 9) = 0$

Step 6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 4$ .

$(y + 4) (y - 4) (y^{2} - 9) = 0$

Step 7

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

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

Step 8

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 3$ .

$(y + 4) (y - 4) ((y + 3) (y - 3)) = 0$

Remove unnecessary parentheses.

$(y + 4) (y - 4) (y + 3) (y - 3) = 0$

Step 9

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

$y + 4 = 0$

$y - 4 = 0$

$y + 3 = 0$

$y - 3 = 0$

Step 10

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.

$y = - 4$

Step 11

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

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

$y - 4 = 0$

Add $4$ to both sides of the equation.

$y = 4$

Step 12

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

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

$y + 3 = 0$

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 13

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

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

$y - 3 = 0$

Add $3$ to both sides of the equation.

$y = 3$

Step 14

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

$y = - 4, 4, - 3, 3$

Solve by Factoring y^4-25y^2+144=0

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