Question
b^2+5b-35=3b solve factoring
Answers
GPT-4o mini
To solve the equation \( b^2 + 5b - 35 = 3b \) by factoring, first, we want to set the equation to zero. We can do this by subtracting \( 3b \) from both sides:
\[
b^2 + 5b - 3b - 35 = 0
\]
This simplifies to:
\[
b^2 + 2b - 35 = 0
\]
Next, we will factor the quadratic expression \( b^2 + 2b - 35 \). We are looking for two numbers that multiply to \(-35\) (the constant term) and add to \(2\) (the coefficient of \(b\)). The numbers \(7\) and \(-5\) work, since:
\[
7 \times (-5) = -35 \quad \text{and} \quad 7 + (-5) = 2
\]
So we can factor the quadratic as:
\[
(b + 7)(b - 5) = 0
\]
Now, we can set each factor to zero:
1. \( b + 7 = 0 \)
\[
b = -7
\]
2. \( b - 5 = 0 \)
\[
b = 5
\]
Thus, the solutions to the equation \( b^2 + 5b - 35 = 3b \) are:
\[
b = -7 \quad \text{and} \quad b = 5
\]
\[
b^2 + 5b - 3b - 35 = 0
\]
This simplifies to:
\[
b^2 + 2b - 35 = 0
\]
Next, we will factor the quadratic expression \( b^2 + 2b - 35 \). We are looking for two numbers that multiply to \(-35\) (the constant term) and add to \(2\) (the coefficient of \(b\)). The numbers \(7\) and \(-5\) work, since:
\[
7 \times (-5) = -35 \quad \text{and} \quad 7 + (-5) = 2
\]
So we can factor the quadratic as:
\[
(b + 7)(b - 5) = 0
\]
Now, we can set each factor to zero:
1. \( b + 7 = 0 \)
\[
b = -7
\]
2. \( b - 5 = 0 \)
\[
b = 5
\]
Thus, the solutions to the equation \( b^2 + 5b - 35 = 3b \) are:
\[
b = -7 \quad \text{and} \quad b = 5
\]