PARCC Algebra II Arithmetic with Polynomials and Rational Expressions. Practice it free.

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

Since $(x-2)$ is a factor, dividing by $(x-2)$ gives $p(x)=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1)$. The zeros come from the linear factors, so set each equal to zero: $(x-2)=0 \Rightarrow x=2$; $(x-3)=0 \Rightarrow x=3$; $(x+1)=0 \Rightarrow x=-1$. The zeros are $2$, $3$, and $-1$.

Add exponents on matching variables: $a^{3+(-1)} b^{-4+2} = a^2 b^{-2}\textsf{.}$

Distribute the minus sign: $(5x^2 + 3x + 2) - 2x^2 + 5\textsf{.}$ Combine $x^2$ terms: $5x^2 - 2x^2 = 3x^2\textsf{.}$ The $x$ term stays $3x\textsf{.}$ Combine constants: $2 + 5 = 7\textsf{.}$ Result: $3x^2 + 3x + 7\textsf{.}$

Expand: $(x - 8)(x + 1) = x^2 + x - 8x - 8 = x^2 - 7x - 8$. Then add $7$: $x^2 - 7x - 8 + 7 = x^2 - 7x - 1$.

Factor by grouping: $\dfrac{(x^{2}-1)(x+2)}{(x+2)(x-1)}=\dfrac{(x-1)(x+1)(x+2)}{(x+2)(x-1)}=x+1$ for $x \neq -2$ and $x \neq 1\textsf{.}$

$(2x + 1)^{2} = (2x)^{2} + 2(2x)(1) + 1^{2} = 4x^{2} + 4x + 1$.

$C(6,3) = \dfrac{6!}{3!\,(6-3)!} = 20$.