Solve by Factoring z^4-13z^2+36=0

$z^{4} - 13 z^{2} + 36 = 0$

Step 1

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

${(z^{2})}^{2} - 13 z^{2} + 36 = 0$

Step 2

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

$u^{2} - 13 u + 36 = 0$

Step 3

Factor $u^{2} - 13 u + 36$ 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 $36$ and whose sum is $- 13$ .

$- 9, - 4$

Write the factored form using these integers.

$(u - 9) (u - 4) = 0$

Step 4

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

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

Step 5

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

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

Step 6

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

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

Step 7

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

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

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 $a = z$ and $b = 2$ .

$(z + 3) (z - 3) ((z + 2) (z - 2)) = 0$

Remove unnecessary parentheses.

$(z + 3) (z - 3) (z + 2) (z - 2) = 0$

Step 9

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

$z + 3 = 0$

$z - 3 = 0$

$z + 2 = 0$

$z - 2 = 0$

Step 10

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

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

$z + 3 = 0$

Subtract $3$ from both sides of the equation.

$z = - 3$

Step 11

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

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

$z - 3 = 0$

Add $3$ to both sides of the equation.

$z = 3$

Step 12

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

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

$z + 2 = 0$

Subtract $2$ from both sides of the equation.

$z = - 2$

Step 13

Set $z - 2$ equal to $0$ and solve for $z$ .

Set $z - 2$ equal to $0$ .

$z - 2 = 0$

Add $2$ to both sides of the equation.

$z = 2$

Step 14

The final solution is all the values that make $(z + 3) (z - 3) (z + 2) (z - 2) = 0$ true.

$z = - 3, 3, - 2, 2$

Solve by Factoring z^4-13z^2+36=0

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