Solve by Factoring z^2+4z-5=0

$z^{2} + 4 z - 5 = 0$

Step 1

Factor $z^{2} + 4 z - 5$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$(z - 1) (z + 5) = 0$

Step 2

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

$z - 1 = 0$

$z + 5 = 0$

Step 3

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 4

Set $z + 5$ equal to $0$ and solve for $z$ .

Set $z + 5$ equal to $0$ .

$z + 5 = 0$

Subtract $5$ from both sides of the equation.

$z = - 5$

Step 5

The final solution is all the values that make $(z - 1) (z + 5) = 0$ true.

$z = 1, - 5$

Solve by Factoring z^2+4z-5=0

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