Solve by Factoring b^2+3b-4=0

$b^{2} + 3 b - 4 = 0$

Step 1

Factor $b^{2} + 3 b - 4$ 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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

$(b - 1) (b + 4) = 0$

Step 2

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

$b - 1 = 0$

$b + 4 = 0$

Step 3

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

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

$b - 1 = 0$

Add $1$ to both sides of the equation.

$b = 1$

Step 4

Set $b + 4$ equal to $0$ and solve for $b$ .

Set $b + 4$ equal to $0$ .

$b + 4 = 0$

Subtract $4$ from both sides of the equation.

$b = - 4$

Step 5

The final solution is all the values that make $(b - 1) (b + 4) = 0$ true.

$b = 1, - 4$

Solve by Factoring b^2+3b-4=0

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