Factor \( b^{2}-2 b^{2}+3 b-6 \)

Let's factor the expression: \[ b^{2} - 2b^{2} + 3b - 6 \] First, simplify the expression by combining like terms: \[ b^{2} - 2b^{2} + 3b - 6 = -b^{2} + 3b - 6 \] To factor this quadratic expression, it's often helpful to factor out the leading coefficient (in this case, \(-1\)): \[ -b^{2} + 3b - 6 = -\left(b^{2} - 3b + 6\right) \] Now, let's try to factor the quadratic inside the parentheses: \[ b^{2} - 3b + 6 \] To factor \(b^{2} - 3b + 6\), we look for two numbers that multiply to \(6\) and add to \(-3\). However, there are no real numbers that satisfy these conditions because the discriminant (\(b^2 - 4ac\)) is negative: \[ \text{Discriminant} = (-3)^2 - 4(1)(6) = 9 - 24 = -15 \] Since the discriminant is negative, \(b^{2} - 3b + 6\) does not factor over the real numbers. **Conclusion:** The expression cannot be factored further using real numbers. However, if you're working in the context of real numbers, you can express it as: \[ -b^{2} + 3b - 6 = -\left(b^{2} - 3b + 6\right) \] If you need to factor it over the complex numbers, it would be: \[ -b^{2} + 3b - 6 = -\left(b - \frac{3}{2} + \frac{\sqrt{15}}{2}i\right)\left(b - \frac{3}{2} - \frac{\sqrt{15}}{2}i\right) \] But typically, unless specified, factoring is done over the real numbers. **Final Answer:** The expression factors as –( b² – 3b + 6 ), which cannot be factored further using real numbers.