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

To factor the polynomial \( b^{3}-2 b^{2}+3 b-6 \), we can start by trying to find rational roots using the Rational Root Theorem. Testing \( b = 2 \) gives us: \( (2)^3 - 2(2)^2 + 3(2) - 6 = 8 - 8 + 6 - 6 = 0 \) So, \( b - 2 \) is a factor. Now we can perform polynomial long division or synthetic division of \( b^{3}-2 b^{2}+3 b-6 \) by \( b - 2 \): 1. Divide \( b^3 \) by \( b \) to get \( b^2 \). 2. Multiply \( b^2 \) by \( b - 2 \) to get \( b^3 - 2b^2 \). 3. Subtract this from the original polynomial to get \( 5b - 6 \). 4. Divide \( 5b \) by \( b \) to get \( 5 \). 5. Multiply \( 5 \) by \( b - 2 \) to get \( 5b - 10 \). 6. Subtract to find the remainder \( 4 \). Thus, we have: \( b^{3} - 2b^{2} + 3b - 6 = (b - 2)(b^{2} + 5) - 4 \) Now, since \( b^{2} + 5 \) doesn’t factor further over the reals, the final factorization is: \( b^{3} - 2b^{2} + 3b - 6 = (b - 2)(b^{2} + 5) - 4 \). Unfortunately, this polynomial cannot be completely factored into real-number factors. If you move into the complex number field, it can be further analyzed in that regard, as \( b^{2} + 5 \) could be expressed in terms of imaginary numbers.