1. **Problem:** Given that $(kx+1)$ is a common factor of the polynomials $2x^2 + 7x + 3$ and $2x^2 - 5x - 3$, find the value of $k$. Then find the remainder when $2x^2 + x^2 - 18x - 9$ is divided by $x + k$. 2. **Step 1: Use the factor condition.** If $(kx+1)$ is a factor, then substituting $x = -\frac{1}{k}$ into each polynomial should give zero. 3. **Step 2: Apply to the first polynomial:** $$2\left(-\frac{1}{k}\right)^2 + 7\left(-\frac{1}{k}\right) + 3 = 0$$ Simplify: $$2\frac{1}{k^2} - \frac{7}{k} + 3 = 0$$ Multiply both sides by $k^2$: $$2 - 7k + 3k^2 = 0$$ Rearranged: $$3k^2 - 7k + 2 = 0$$ 4. **Step 3: Apply to the second polynomial:** $$2\left(-\frac{1}{k}\right)^2 - 5\left(-\frac{1}{k}\right) - 3 = 0$$ Simplify: $$2\frac{1}{k^2} + \frac{5}{k} - 3 = 0$$ Multiply both sides by $k^2$: $$2 + 5k - 3k^2 = 0$$ Rearranged: $$-3k^2 + 5k + 2 = 0$$ Or equivalently: $$3k^2 - 5k - 2 = 0$$ 5. **Step 4: Solve the system:** We have two quadratic equations: $$3k^2 - 7k + 2 = 0$$ $$3k^2 - 5k - 2 = 0$$ Subtract the second from the first: $$(3k^2 - 7k + 2) - (3k^2 - 5k - 2) = 0$$ Simplify: $$-7k + 2 + 5k + 2 = 0 \Rightarrow -2k + 4 = 0$$ Solve for $k$: $$-2k + 4 = 0 \Rightarrow 2k = 4 \Rightarrow k = 2$$ 6. **Step 5: Verify $k=2$ in one equation:** $$3(2)^2 - 7(2) + 2 = 3(4) - 14 + 2 = 12 - 14 + 2 = 0$$ Correct. 7. **Step 6: Find the remainder when dividing $2x^2 + x^2 - 18x - 9$ by $x + k$.** First simplify the polynomial: $$2x^2 + x^2 - 18x - 9 = 3x^2 - 18x - 9$$ Since $k=2$, divisor is $x + 2$. 8. **Step 7: Use the Remainder Theorem:** Remainder is value of polynomial at $x = -2$: $$3(-2)^2 - 18(-2) - 9 = 3(4) + 36 - 9 = 12 + 36 - 9 = 39$$ **Final answers:** - $k = 2$ - Remainder when dividing $3x^2 - 18x - 9$ by $x + 2$ is $39$.