Okay, I will continue with the modular arithmetic problems, ensuring all math is in LaTeX and following the specified format. 9) Evaluate $(3x+6)(3x-2)$ in mod 9 Step 1: Expand the product. $$(3x+6)(3x-2) = (3x)(3x) + (3x)(-2) + (6)(3x) + (6)(-2)$$ $$= 9x^2 - 6x + 18x - 12$$ $$= 9x^2 + 12x - 12$$ Step 2: Reduce the coefficients modulo 9. $$9x^2 \equiv 0x^2 \equiv 0 \pmod{9}$$ $$12x \equiv 3x \pmod{9} \quad (\text{since } 12 = 1 \times 9 + 3)$$ $$-12 \equiv 6 \pmod{9} \quad (\text{since } -12 = -2 \times 9 + 6 \text{ or } -12 + 18 = 6)$$ Step 3: Substitute the reduced coefficients back into the expression. $$9x^2 + 12x - 12 \equiv 0 + 3x + 6 \pmod{9}$$ The expression in mod 9 is $\boxed{3x + 6}$. 10) Evaluate $(x^2+2x+1) \div (x+1)$ in mod 5 Step 1: Reduce the coefficients modulo 5. The coefficients in $x^2+2x+1$ and $x+1$ are already in their smallest non-negative residues modulo 5. Step 2: Perform polynomial division modulo 5. We can recognize that the numerator is a perfect square: $x^2+2x+1 = (x+1)^2$. So, the division becomes: $$\frac{(x+1)^2}{x+1} = x+1$$ Alternatively, using polynomial long division modulo 5: ` x + 1 _________ x + 1 | x^2 + 2x + 1 -(x^2 + x) ___________ x + 1 -(x + 1) _________ 0 ` The quotient is $x+1$ with a remainder of $0$. The evaluated expression in mod 5 is $\boxed{x+1}$.