ATLAS Algebra I Algebra. Practice it free.

Let $u=x^2$. Then $u^2-10u+9=(u-1)(u-9)$, so $p(x)=(x^2-1)(x^2-9)=(x-1)(x+1)(x-3)(x+3)$. Set each linear factor equal to zero: $(x-1)=0 \Rightarrow x=1$; $(x+1)=0 \Rightarrow x=-1$; $(x-3)=0 \Rightarrow x=3$; $(x+3)=0 \Rightarrow x=-3$. The zeros are $-3$, $-1$, $1$, and $3$.

In slope-intercept form $y = mx + b$, the coefficient of $x$ is the slope. Here $m = -5$.

The slope is $m = \dfrac{1 - (-3)}{6 - 2} = 1$. Substitute $(2, -3)$: $-3 = 1(2) + b$, so $b = -5$. The equation is $y = x - 5$.

Substitute $y = 2$. From $2x - y \ge 4$, $2x - 2 \ge 4$, so $x \ge 3$. From $x + y \le 7$, $x \le 5$. The value $x = 4$ satisfies $3 \le 4 \le 5$.

Subtract the third-year salary from the first: $\text{\char36}36{,}400 - \text{\char36}32{,}500 = \text{\char36}3{,}900$. Over three years there are two equal raises, so divide by $3 - 1 = 2$ to get $x = \text{\char36}1{,}950$. Add $\text{\char36}1{,}950$ each year after the first: $\text{\char36}32{,}500$, $\text{\char36}34{,}450$, $\text{\char36}36{,}400$. Sum all three years: $\text{\char36}103{,}350$.

Multiply the coefficients: $3 \cdot 4 = 12\textsf{.}$ Add the exponents on $x$: $x^{2+1} = x^{3}\textsf{.}$ So the product is $12x^{3}\textsf{.}$

Distribute the minus sign: $(5x^2 + 3) - x^2 + 2x\textsf{.}$ Combine $x^2$ terms: $5x^2 - x^2 = 4x^2\textsf{.}$ The $x$ term is $2x\textsf{.}$ The constant stays $3\textsf{.}$ Result: $4x^2 + 2x + 3\textsf{.}$

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