Solve by Factoring b^2+5b-35=3b

$b^{2} + 5 b - 35 = 3 b$

Step 1

Subtract $3 b$ from both sides of the equation.

$b^{2} + 5 b - 35 - 3 b = 0$

Step 2

Subtract $3 b$ from $5 b$ .

$b^{2} + 2 b - 35 = 0$

Step 3

Factor $b^{2} + 2 b - 35$ 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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Write the factored form using these integers.

$(b - 5) (b + 7) = 0$

Step 4

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

$b - 5 = 0$

$b + 7 = 0$

Step 5

Set $b - 5$ equal to $0$ and solve for $b$ .

Set $b - 5$ equal to $0$ .

$b - 5 = 0$

Add $5$ to both sides of the equation.

$b = 5$

Step 6

Set $b + 7$ equal to $0$ and solve for $b$ .

Set $b + 7$ equal to $0$ .

$b + 7 = 0$

Subtract $7$ from both sides of the equation.

$b = - 7$

Step 7

The final solution is all the values that make $(b - 5) (b + 7) = 0$ true.

$b = 5, - 7$

Solve by Factoring b^2+5b-35=3b

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