1. **Problem statement:** Given the polynomial $p(x) = 6x^3 + ax^2 + bx + c$ (assuming the polynomial form since the problem states $p(x)=6+a$ is incomplete, we interpret it as a cubic polynomial with coefficients involving $a$ and $b$), and that $x-2$ is a factor of $p(x)$. Given: $p(1) = 2p(0)$. Find the values of $a$ and $b$. 2. **Using the factor theorem:** Since $x-2$ is a factor, $p(2) = 0$. 3. **Express $p(x)$:** Since the problem is ambiguous, we assume $p(x) = 6x^3 + ax^2 + bx + c$ with $c$ possibly 0 or unknown. But since $p(x) = 6 + a$ is given, we interpret $p(x) = 6x^3 + ax^2 + bx + 6 + a$ (assuming constant term $6+a$). 4. **Apply $p(1) = 2p(0)$:** Calculate $p(1) = 6(1)^3 + a(1)^2 + b(1) + 6 + a = 6 + a + b + 6 + a = 12 + 2a + b$. Calculate $p(0) = 6 + a$. Given $p(1) = 2p(0)$, so: $$12 + 2a + b = 2(6 + a) = 12 + 2a$$ Simplify: $$12 + 2a + b = 12 + 2a$$ Subtract $12 + 2a$ from both sides: $$b = 0$$ 5. **Apply factor theorem $p(2) = 0$:** Calculate $p(2) = 6(8) + a(4) + b(2) + 6 + a = 48 + 4a + 0 + 6 + a = 54 + 5a$. Set equal to zero: $$54 + 5a = 0$$ Solve for $a$: $$5a = -54$$ $$a = -\frac{54}{5}$$ Since $a$ and $b$ are integers, this contradicts the assumption. So the initial assumption about $p(x)$ is likely incorrect. **Re-examining the problem:** The problem states "The factor $p(x) = 6 + a$, where $a$ and $b$ are integers has a factor of $x-2$" which is ambiguous. Assuming $p(x) = 6x^2 + ax + b$ (a quadratic polynomial), and $x-2$ is a factor. 6. **Apply factor theorem $p(2) = 0$:** $$p(2) = 6(2)^2 + a(2) + b = 24 + 2a + b = 0$$ 7. **Apply $p(1) = 2p(0)$:** Calculate $p(1) = 6(1)^2 + a(1) + b = 6 + a + b$. Calculate $p(0) = b$. Given $p(1) = 2p(0)$: $$6 + a + b = 2b$$ Simplify: $$6 + a = b$$ 8. **Substitute $b = 6 + a$ into $p(2) = 0$:** $$24 + 2a + (6 + a) = 0$$ $$24 + 2a + 6 + a = 0$$ $$30 + 3a = 0$$ $$3a = -30$$ $$a = -10$$ 9. **Find $b$:** $$b = 6 + a = 6 - 10 = -4$$ 10. **Find remainder when $p(x)$ is divided by $2x - 1$:** The remainder when dividing by $2x - 1$ is $p(\frac{1}{2})$. Calculate: $$p\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 + a\left(\frac{1}{2}\right) + b = 6\times \frac{1}{4} + (-10)\times \frac{1}{2} + (-4) = \frac{6}{4} - 5 - 4 = 1.5 - 5 - 4 = -7.5$$ 11. **Factorize $p(x)$:** $$p(x) = 6x^2 - 10x - 4$$ Factor out 2: $$2(3x^2 - 5x - 2)$$ Factor quadratic: Find two numbers that multiply to $3 \times (-2) = -6$ and add to $-5$: these are $-6$ and $1$. Rewrite: $$2(3x^2 - 6x + x - 2) = 2[(3x^2 - 6x) + (x - 2)] = 2[3x(x - 2) + 1(x - 2)] = 2(x - 2)(3x + 1)$$ **Final answers:** $$a = -10, \quad b = -4$$ Remainder when divided by $2x - 1$ is $-\frac{15}{2}$. Factorization: $$p(x) = 2(x - 2)(3x + 1)$$