Polynomial Division 0Ce06C

1. **State the problem:** We need to perform polynomial division for two expressions: a) $$\frac{-15y^2 - 11y + 2y^5 + 5}{2y - 1}$$ b) $$\frac{-15y^3 + 12y^2 + 12y^4 - 11y + 5}{3y^2 + 2y - 1}$$ 2. **Arrange polynomials in descending powers of $y$:** a) Dividend: $$2y^5 - 15y^2 - 11y + 5$$ Divisor: $$2y - 1$$ b) Dividend: $$12y^4 - 15y^3 + 12y^2 - 11y + 5$$ Divisor: $$3y^2 + 2y - 1$$ 3. **Polynomial division formula:** $$\text{Dividend} = (\text{Divisor}) \times (\text{Quotient}) + \text{Remainder}$$ 4. **Divide (a):** $$\frac{2y^5 - 15y^2 - 11y + 5}{2y - 1}$$ - First term: $$\frac{2y^5}{2y} = y^4$$ - Multiply divisor by $$y^4$$: $$y^4(2y - 1) = 2y^5 - y^4$$ - Subtract: $$(2y^5 - 15y^2 - 11y + 5) - (2y^5 - y^4) = y^4 - 15y^2 - 11y + 5$$ - Next term: $$\frac{y^4}{2y} = \frac{1}{2}y^3$$ - Multiply divisor by $$\frac{1}{2}y^3$$: $$\frac{1}{2}y^3(2y - 1) = y^4 - \frac{1}{2}y^3$$ - Subtract: $$(y^4 - 15y^2 - 11y + 5) - (y^4 - \frac{1}{2}y^3) = \frac{1}{2}y^3 - 15y^2 - 11y + 5$$ - Next term: $$\frac{\frac{1}{2}y^3}{2y} = \frac{1}{4}y^2$$ - Multiply divisor by $$\frac{1}{4}y^2$$: $$\frac{1}{4}y^2(2y - 1) = \frac{1}{2}y^3 - \frac{1}{4}y^2$$ - Subtract: $$(\frac{1}{2}y^3 - 15y^2 - 11y + 5) - (\frac{1}{2}y^3 - \frac{1}{4}y^2) = -\frac{59}{4}y^2 - 11y + 5$$ - Next term: $$\frac{-\frac{59}{4}y^2}{2y} = -\frac{59}{8}y$$ - Multiply divisor by $$-\frac{59}{8}y$$: $$-\frac{59}{8}y(2y - 1) = -\frac{59}{4}y^2 + \frac{59}{8}y$$ - Subtract: $$(-\frac{59}{4}y^2 - 11y + 5) - (-\frac{59}{4}y^2 + \frac{59}{8}y) = -\frac{147}{8}y + 5$$ - Next term: $$\frac{-\frac{147}{8}y}{2y} = -\frac{147}{16}$$ - Multiply divisor by $$-\frac{147}{16}$$: $$-\frac{147}{16}(2y - 1) = -\frac{147}{8}y + \frac{147}{16}$$ - Subtract: $$(-\frac{147}{8}y + 5) - (-\frac{147}{8}y + \frac{147}{16}) = \frac{80}{16} - \frac{147}{16} = -\frac{67}{16}$$ - Remainder: $$-\frac{67}{16}$$ - Quotient: $$y^4 + \frac{1}{2}y^3 + \frac{1}{4}y^2 - \frac{59}{8}y - \frac{147}{16}$$ 5. **Divide (b):** $$\frac{12y^4 - 15y^3 + 12y^2 - 11y + 5}{3y^2 + 2y - 1}$$ - First term: $$\frac{12y^4}{3y^2} = 4y^2$$ - Multiply divisor by $$4y^2$$: $$4y^2(3y^2 + 2y - 1) = 12y^4 + 8y^3 - 4y^2$$ - Subtract: $$(12y^4 - 15y^3 + 12y^2 - 11y + 5) - (12y^4 + 8y^3 - 4y^2) = -23y^3 + 16y^2 - 11y + 5$$ - Next term: $$\frac{-23y^3}{3y^2} = -\frac{23}{3}y$$ - Multiply divisor by $$-\frac{23}{3}y$$: $$-\frac{23}{3}y(3y^2 + 2y - 1) = -23y^3 - \frac{46}{3}y^2 + \frac{23}{3}y$$ - Subtract: $$(-23y^3 + 16y^2 - 11y + 5) - (-23y^3 - \frac{46}{3}y^2 + \frac{23}{3}y) = \frac{94}{3}y^2 - \frac{56}{3}y + 5$$ - Next term: $$\frac{\frac{94}{3}y^2}{3y^2} = \frac{94}{9}$$ - Multiply divisor by $$\frac{94}{9}$$: $$\frac{94}{9}(3y^2 + 2y - 1) = \frac{94}{3}y^2 + \frac{188}{9}y - \frac{94}{9}$$ - Subtract: $$(\frac{94}{3}y^2 - \frac{56}{3}y + 5) - (\frac{94}{3}y^2 + \frac{188}{9}y - \frac{94}{9}) = -\frac{710}{9}y + \frac{139}{9}$$ - Remainder degree less than divisor degree, stop. - Quotient: $$4y^2 - \frac{23}{3}y + \frac{94}{9}$$ - Remainder: $$-\frac{710}{9}y + \frac{139}{9}$$ 6. **Final answers:** a) $$\frac{2y^5 - 15y^2 - 11y + 5}{2y - 1} = y^4 + \frac{1}{2}y^3 + \frac{1}{4}y^2 - \frac{59}{8}y - \frac{147}{16} + \frac{-\frac{67}{16}}{2y - 1}$$ b) $$\frac{12y^4 - 15y^3 + 12y^2 - 11y + 5}{3y^2 + 2y - 1} = 4y^2 - \frac{23}{3}y + \frac{94}{9} + \frac{-\frac{710}{9}y + \frac{139}{9}}{3y^2 + 2y - 1}$$