The problem asks us to divide the polynomial $9b^4 - 12b^3 - b^2$ by the monomial $3b^2$.AlgebraPolynomial DivisionMonomial DivisionAlgebraic Manipulation2025/3/221. Problem DescriptionThe problem asks us to divide the polynomial 9b4−12b3−b29b^4 - 12b^3 - b^29b4−12b3−b2 by the monomial 3b23b^23b2.2. Solution StepsWe need to divide each term of the polynomial by 3b23b^23b2.9b4−12b3−b23b2=9b43b2−12b33b2−b23b2\frac{9b^4 - 12b^3 - b^2}{3b^2} = \frac{9b^4}{3b^2} - \frac{12b^3}{3b^2} - \frac{b^2}{3b^2}3b29b4−12b3−b2=3b29b4−3b212b3−3b2b2Now we simplify each term:9b43b2=93⋅b4b2=3b4−2=3b2\frac{9b^4}{3b^2} = \frac{9}{3} \cdot \frac{b^4}{b^2} = 3b^{4-2} = 3b^23b29b4=39⋅b2b4=3b4−2=3b212b33b2=123⋅b3b2=4b3−2=4b\frac{12b^3}{3b^2} = \frac{12}{3} \cdot \frac{b^3}{b^2} = 4b^{3-2} = 4b3b212b3=312⋅b2b3=4b3−2=4bb23b2=13⋅b2b2=13⋅1=13\frac{b^2}{3b^2} = \frac{1}{3} \cdot \frac{b^2}{b^2} = \frac{1}{3} \cdot 1 = \frac{1}{3}3b2b2=31⋅b2b2=31⋅1=31Putting these together, we get:3b2−4b−133b^2 - 4b - \frac{1}{3}3b2−4b−313. Final Answer3b2−4b−133b^2 - 4b - \frac{1}{3}3b2−4b−31
