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

Let's factor the expression \( b^{2} - 2b^{2} + 3b - 6 \). ### Step 1: Simplify the Expression Combine like terms: \[ b^{2} - 2b^{2} + 3b - 6 = -b^{2} + 3b - 6 \] ### Step 2: Factor Out the Common Factor Notice that each term has a common factor of \(-1\): \[ - b^{2} + 3b - 6 = - (b^{2} - 3b + 6) \] ### Step 3: Check for Further Factorization Now, consider the quadratic inside the parentheses: \( b^{2} - 3b + 6 \). To factor this quadratic, we look for two numbers that multiply to \(6\) (the constant term) and add up to \(-3\) (the coefficient of \(b\)). However, there are no such integer pairs. If you attempt to factor it over the real numbers, you'll find that the discriminant is negative: \[ \Delta = (-3)^2 - 4 \times 1 \times 6 = 9 - 24 = -15 \] Since the discriminant is negative, the quadratic does not factor over the real numbers. ### Final Factored Form Thus, the expression cannot be factored further using real numbers. The fully factored form is: \[ - (b^{2} - 3b + 6) \] **Answer:** After simplifying, the factored form is – (b² – 3 b + 6): \[ - (b^{2} - 3b + 6) \]