Polynomial Division De2F81

1. **State the problem:** We want to divide the polynomial $$2x^4 + 4x^3 - 7x^2 + 2$$ by $$2x^2 + 2x - 7$$ using long division. 2. **Recall the long division formula:** When dividing $$f(x)$$ by $$g(x)$$, the result is $$q(x) + \frac{r(x)}{g(x)}$$ where $$q(x)$$ is the quotient and $$r(x)$$ is the remainder with degree less than $$g(x)$$. 3. **Set up the division:** Dividend: $$2x^4 + 4x^3 - 7x^2 + 0x + 2$$ (note the zero coefficient for $$x$$ term) Divisor: $$2x^2 + 2x - 7$$ 4. **Divide the leading terms:** $$\frac{2x^4}{2x^2} = x^2$$ This is the first term of the quotient. 5. **Multiply divisor by $$x^2$$:** $$x^2(2x^2 + 2x - 7) = 2x^4 + 2x^3 - 7x^2$$ 6. **Subtract this from the dividend:** $$\begin{aligned} &(2x^4 + 4x^3 - 7x^2 + 0x + 2) - (2x^4 + 2x^3 - 7x^2) \\ &= (2x^4 - 2x^4) + (4x^3 - 2x^3) + (-7x^2 + 7x^2) + 0x + 2 \\ &= 0 + 2x^3 + 0 + 0x + 2 = 2x^3 + 2 \end{aligned}$$ 7. **Bring down the next term (already included), now divide leading terms:** $$\frac{2x^3}{2x^2} = x$$ Add $$x$$ to the quotient. 8. **Multiply divisor by $$x$$:** $$x(2x^2 + 2x - 7) = 2x^3 + 2x^2 - 7x$$ 9. **Subtract this from the current remainder:** $$\begin{aligned} &(2x^3 + 0x^2 + 0x + 2) - (2x^3 + 2x^2 - 7x) \\ &= (2x^3 - 2x^3) + (0x^2 - 2x^2) + (0x + 7x) + 2 \\ &= 0 - 2x^2 + 7x + 2 = -2x^2 + 7x + 2 \end{aligned}$$ 10. **Divide leading terms:** $$\frac{-2x^2}{2x^2} = -1$$ Add $$-1$$ to the quotient. 11. **Multiply divisor by $$-1$$:** $$-1(2x^2 + 2x - 7) = -2x^2 - 2x + 7$$ 12. **Subtract this from the current remainder:** $$\begin{aligned} &(-2x^2 + 7x + 2) - (-2x^2 - 2x + 7) \\ &= (-2x^2 + 2x^2) + (7x + 2x) + (2 - 7) \\ &= 0 + 9x - 5 = 9x - 5 \end{aligned}$$ 13. **Since the degree of remainder $$9x - 5$$ is less than degree of divisor $$2x^2 + 2x - 7$$, stop here.** 14. **Write the final answer:** $$\boxed{q(x) = x^2 + x - 1, \quad r(x) = 9x - 5}$$ So, $$\frac{2x^4 + 4x^3 - 7x^2 + 2}{2x^2 + 2x - 7} = x^2 + x - 1 + \frac{9x - 5}{2x^2 + 2x - 7}$$