Solve by Factoring z^2+32=12z

$z^{2} + 32 = 12 z$

Step 1

Subtract $12 z$ from both sides of the equation.

$z^{2} + 32 - 12 z = 0$

Step 2

Factor $z^{2} + 32 - 12 z$ 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 $32$ and whose sum is $- 12$ .

$- 8, - 4$

Write the factored form using these integers.

$(z - 8) (z - 4) = 0$

Step 3

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

$z - 8 = 0$

$z - 4 = 0$

Step 4

Set $z - 8$ equal to $0$ and solve for $z$ .

Set $z - 8$ equal to $0$ .

$z - 8 = 0$

Add $8$ to both sides of the equation.

$z = 8$

Step 5

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

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

$z - 4 = 0$

Add $4$ to both sides of the equation.

$z = 4$

Step 6

The final solution is all the values that make $(z - 8) (z - 4) = 0$ true.

$z = 8, 4$

Solve by Factoring z^2+32=12z

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