1. **Problem 1:** Find the quotient and remainder when $6x^4 + x^3 - x^2 + 5x - 6$ is divided by $2x^2 - x + 1$. 2. **Problem 2:** Find the quotient and remainder when $2x^4 + 1$ is divided by $x^2 - x + 2$. 3. **Problem 3:** Find the quotient and remainder when $2x^4 - 27$ is divided by $x^2 + x + 3$. 4. **Problem 4:** Given $p(x) = 2x^4 + ax^3 + bx - 1$, when divided by $x^2 - x + 1$ the remainder is $3x + 2$. Find $a$ and $b$. --- ### Step-by-step solutions: ### Problem 1: 1. Divide $6x^4 + x^3 - x^2 + 5x - 6$ by $2x^2 - x + 1$ using polynomial long division. 2. Leading term division: $\frac{6x^4}{2x^2} = 3x^2$. 3. Multiply divisor by $3x^2$: $3x^2(2x^2 - x + 1) = 6x^4 - 3x^3 + 3x^2$. 4. Subtract: $$\left(6x^4 + x^3 - x^2 + 5x - 6\right) - \left(6x^4 - 3x^3 + 3x^2\right) = (6x^4 - 6x^4) + (x^3 + 3x^3) + (-x^2 - 3x^2) + 5x - 6 = 4x^3 - 4x^2 + 5x - 6$$ 5. Next term: $\frac{4x^3}{2x^2} = 2x$. 6. Multiply divisor by $2x$: $2x(2x^2 - x + 1) = 4x^3 - 2x^2 + 2x$. 7. Subtract: $$\left(4x^3 - 4x^2 + 5x - 6\right) - \left(4x^3 - 2x^2 + 2x\right) = (4x^3 - 4x^3) + (-4x^2 + 2x^2) + (5x - 2x) - 6 = -2x^2 + 3x - 6$$ 8. Next term: $\frac{-2x^2}{2x^2} = -1$. 9. Multiply divisor by $-1$: $-1(2x^2 - x + 1) = -2x^2 + x - 1$. 10. Subtract: $$\left(-2x^2 + 3x - 6\right) - \left(-2x^2 + x - 1\right) = (-2x^2 + 2x^2) + (3x - x) + (-6 + 1) = 2x - 5$$ 11. The degree of remainder $2x - 5$ is less than divisor degree 2, so stop. **Quotient:** $3x^2 + 2x - 1$ **Remainder:** $2x - 5$ --- ### Problem 2: 1. Divide $2x^4 + 1$ by $x^2 - x + 2$. 2. Leading term division: $\frac{2x^4}{x^2} = 2x^2$. 3. Multiply divisor by $2x^2$: $2x^2(x^2 - x + 2) = 2x^4 - 2x^3 + 4x^2$. 4. Subtract: $$\left(2x^4 + 0x^3 + 0x^2 + 0x + 1\right) - \left(2x^4 - 2x^3 + 4x^2\right) = 0x^4 + 2x^3 - 4x^2 + 0x + 1$$ 5. Next term: $\frac{2x^3}{x^2} = 2x$. 6. Multiply divisor by $2x$: $2x(x^2 - x + 2) = 2x^3 - 2x^2 + 4x$. 7. Subtract: $$\left(2x^3 - 4x^2 + 0x + 1\right) - \left(2x^3 - 2x^2 + 4x\right) = 0x^3 - 2x^2 - 4x + 1$$ 8. Next term: $\frac{-2x^2}{x^2} = -2$. 9. Multiply divisor by $-2$: $-2(x^2 - x + 2) = -2x^2 + 2x - 4$. 10. Subtract: $$\left(-2x^2 - 4x + 1\right) - \left(-2x^2 + 2x - 4\right) = 0x^2 - 6x + 5$$ 11. Degree of remainder $-6x + 5$ is less than divisor degree 2, stop. **Quotient:** $2x^2 + 2x - 2$ **Remainder:** $-6x + 5$ --- ### Problem 3: 1. Divide $2x^4 - 27$ by $x^2 + x + 3$. 2. Leading term division: $\frac{2x^4}{x^2} = 2x^2$. 3. Multiply divisor by $2x^2$: $2x^2(x^2 + x + 3) = 2x^4 + 2x^3 + 6x^2$. 4. Subtract: $$\left(2x^4 + 0x^3 + 0x^2 + 0x - 27\right) - \left(2x^4 + 2x^3 + 6x^2\right) = 0x^4 - 2x^3 - 6x^2 + 0x - 27$$ 5. Next term: $\frac{-2x^3}{x^2} = -2x$. 6. Multiply divisor by $-2x$: $-2x(x^2 + x + 3) = -2x^3 - 2x^2 - 6x$. 7. Subtract: $$\left(-2x^3 - 6x^2 + 0x - 27\right) - \left(-2x^3 - 2x^2 - 6x\right) = 0x^3 - 4x^2 + 6x - 27$$ 8. Next term: $\frac{-4x^2}{x^2} = -4$. 9. Multiply divisor by $-4$: $-4(x^2 + x + 3) = -4x^2 - 4x - 12$. 10. Subtract: $$\left(-4x^2 + 6x - 27\right) - \left(-4x^2 - 4x - 12\right) = 0x^2 + 10x - 15$$ 11. Degree of remainder $10x - 15$ is less than divisor degree 2, stop. **Quotient:** $2x^2 - 2x - 4$ **Remainder:** $10x - 15$ --- ### Problem 4: 1. Given $p(x) = 2x^4 + ax^3 + bx - 1$ and divisor $x^2 - x + 1$, remainder is $3x + 2$. 2. By polynomial division, remainder degree is less than divisor degree 2, so remainder is linear. 3. Write $p(x) = (x^2 - x + 1)Q(x) + 3x + 2$ for some quadratic $Q(x) = cx^2 + dx + e$. 4. Expand: $$p(x) = (x^2 - x + 1)(cx^2 + dx + e) + 3x + 2$$ $$= c x^4 + d x^3 + e x^2 - c x^3 - d x^2 - e x + c x^2 + d x + e + 3x + 2$$ 5. Group terms: $$= c x^4 + (d - c) x^3 + (e - d + c) x^2 + (-e + d + 3) x + (e + 2)$$ 6. Equate coefficients with $p(x) = 2x^4 + a x^3 + b x - 1$: - Coefficient of $x^4$: $c = 2$ - Coefficient of $x^3$: $d - c = a$ so $d - 2 = a$ - Coefficient of $x^2$: $e - d + c = 0$ so $e - d + 2 = 0$ - Coefficient of $x$: $-e + d + 3 = b$ - Constant term: $e + 2 = -1$ so $e = -3$ 7. From $e = -3$, substitute into $e - d + 2 = 0$: $$-3 - d + 2 = 0 \implies -1 - d = 0 \implies d = -1$$ 8. From $d - 2 = a$: $$-1 - 2 = a \implies a = -3$$ 9. From $-e + d + 3 = b$: $$-(-3) + (-1) + 3 = b \implies 3 - 1 + 3 = b \implies b = 5$$ **Final values:** $a = -3$, $b = 5$