ATLAS Algebra II Algebra. Practice it free.

Factor out $x$: $p(x)=x(x^2+4x+4)=x(x+2)^2$. Set each factor equal to zero: $x=0 \Rightarrow x=0$; $(x+2)^2=0$, so $(x+2)=0 \Rightarrow x=-2$. The zero $-2$ has multiplicity $2$, so the zeros are $0$, $-2$, and $-2$.

Set $y = 0$: $2x = 12$, so $x = 6$.

The slope is $m = \dfrac{5 - (-1)}{2 - (-4)} = 1$. Substitute $(-4, -1)$: $-1 = 1(-4) + b$, so $b = 3$. The equation is $y = x + 3$.

Substitute $x = 4$. From $y \le -x + 10$, $y \le 6$. From $y > x - 1$, $y > 3$. The value $y = 5$ satisfies $3 < 5 \le 6$.

Subtract the third-year base from the first: $\text{\char36}35{,}200 - \text{\char36}31{,}000 = \text{\char36}4{,}200$. Over three years there are two equal increases, so divide by $3 - 1 = 2$ to get $x = \text{\char36}2{,}100$. Add $\text{\char36}2{,}100$ each year after the first: $\text{\char36}31{,}000$, $\text{\char36}33{,}100$, $\text{\char36}35{,}200$. Sum all three years: $\text{\char36}99{,}300$.

Multiply the coefficients: $2 \cdot 5 \cdot 4 = 40\textsf{.}$ Add the exponents on $y$: $y^{1+1} = y^{2}\textsf{.}$ The factor $z$ appears once, so the product is $40y^{2}z\textsf{.}$

Combine $x^2$ terms: $2x^2 - x^2 = x^2\textsf{.}$ Combine $x$ terms: $7x + 2x = 9x\textsf{.}$ Combine constants: $1 - 4 = -3\textsf{.}$ Result: $x^2 + 9x - 3\textsf{.}$

Expand: $(2x + 1)(x - 3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$. Then add $4$: $2x^2 - 5x - 3 + 4 = 2x^2 - 5x + 1$.