Solve by Factoring z^4-37z^2+36=0

$z^{4} - 37 z^{2} + 36 = 0$

Step 1

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

${(z^{2})}^{2} - 37 z^{2} + 36 = 0$

Step 2

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

$u^{2} - 37 u + 36 = 0$

Step 3

Factor $u^{2} - 37 u + 36$ 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 $- 37$ .

$- 36, - 1$

Write the factored form using these integers.

$(u - 36) (u - 1) = 0$

Step 4

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

$(z^{2} - 36) (z^{2} - 1) = 0$

Step 5

Rewrite $36$ as $6^{2}$ .

$(z^{2} - 6^{2}) (z^{2} - 1) = 0$

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 $b = 6$ .

$(z + 6) (z - 6) (z^{2} - 1) = 0$

Step 7

Rewrite $1$ as $1^{2}$ .

$(z + 6) (z - 6) (z^{2} - 1^{2}) = 0$

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 $b = 1$ .

$(z + 6) (z - 6) ((z + 1) (z - 1)) = 0$

Remove unnecessary parentheses.

$(z + 6) (z - 6) (z + 1) (z - 1) = 0$

Step 9

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

$z + 6 = 0$

$z - 6 = 0$

$z + 1 = 0$

$z - 1 = 0$

Step 10

Set $z + 6$ equal to $0$ and solve for $z$ .

Set $z + 6$ equal to $0$ .

$z + 6 = 0$

Subtract $6$ from both sides of the equation.

$z = - 6$

Step 11

Set $z - 6$ equal to $0$ and solve for $z$ .

Set $z - 6$ equal to $0$ .

$z - 6 = 0$

Add $6$ to both sides of the equation.

$z = 6$

Step 12

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

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

$z + 1 = 0$

Subtract $1$ from both sides of the equation.

$z = - 1$

Step 13

Set $z - 1$ equal to $0$ and solve for $z$ .

Set $z - 1$ equal to $0$ .

$z - 1 = 0$

Add $1$ to both sides of the equation.

$z = 1$

Step 14

The final solution is all the values that make $(z + 6) (z - 6) (z + 1) (z - 1) = 0$ true.

$z = - 6, 6, - 1, 1$

Solve by Factoring z^4-37z^2+36=0

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