Georgia Milestones Algebra I Arithmetic with Polynomials and Rational Expressions. Practice it free.

$3! = 3 \times 2 \times 1 = 6$.

Since $(x-1)$ is a factor, dividing by $(x-1)$ gives $p(x)=(x-1)(x^2+x-6)=(x-1)(x+3)(x-2)$. The zeros are the roots of the linear factors, so set each equal to zero: $(x-1)=0 \Rightarrow x=1$; $(x+3)=0 \Rightarrow x=-3$; $(x-2)=0 \Rightarrow x=2$. The zeros are $1$, $2$, and $-3$.

Add exponents on matching variables: $p^{5+1} q^{2+4} r^{-3+2} = p^6 q^6 r^{-1}\textsf{.}$

Combine $x^2$ terms: $x^2 + 4x^2 = 5x^2\textsf{.}$ The constants cancel: $6 - 6 = 0\textsf{.}$ Result: $5x^2\textsf{.}$

Expand: $(x - 5)(x + 2) = x^2 + 2x - 5x - 10 = x^2 - 3x - 10$. Then add $10$: $x^2 - 3x - 10 + 10 = x^2 - 3x$.

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

$(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$; with $a = 2x$ and $b = 1$, this gives $8x^{3} + 12x^{2} + 6x + 1$.

$C(5,3) = \dfrac{5!}{3!\,(5-3)!} = 10$.