This is modular arithmetic

Question

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9) Evaluate $(3x+6)(3x-2)$ in mod 9

Expand the product.
$(3x+6)(3x-2) = (3x)(3x) + (3x)(-2) + (6)(3x) + (6)(-2)$ $= 9x^2 - 6x + 18x - 12$ $= 9x^2 + 12x - 12$
Reduce the coefficients modulo 9.
$9x^2 \equiv 0x^2 \equiv 0 \pmod{9}$ $12x \equiv 3x \pmod{9} \quad (\text{since } 12 = 1 \times 9 + 3)$ $-12 \equiv 6 \pmod{9} \quad (\text{since } -12 = -2 \times 9 + 6 \text{ or } -12 + 18 = 6)$
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
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.
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 $\boxed{x+1}$ .

Step-by-step solution