Solve for $x$ for the following expression.

$$
3x^3-3x^2-7x+5=x^3-2x^2-1
$$

Isolate the variables on the left side as follows:

$$
2x^3-x^2-7x+6=0
$$

Substitution $f(1)$ to know it factored or not as follows:

$$
f(1)=2-1-7+6=0 (x-1) text{is the factor for the expression}
$$

$$
1 | 2 -1 -7 6
$$

$$
0 2 1 -6
$$

$$
2 1 -6 0
$$

$$
2x^3-x^2-7x+6=(x-1)(2x^2+x-6)=0
$$

$$
2x^3-x^2-7x+6=(x-1)(2x-3)(x+2)=0
$$

Use zero property as follows:

$$
x-1=0 x=1
$$

$$
2x-3=0 x=dfrac{3}{2}
$$

$$
x+2=0 x=-2
$$

$$
text{color{Brown} $x=1, x=dfrac{3}{2}, x=-2$}
$$

Depending on the graph on the textbook, answer for the following question.

$$
text{color{#4257b2}(a) Determine where $f(x)$ is positive, negative and zero}
$$

$f(x)$ is positive in the interval of $x$ at $0le xle2$

$f(x)$ is negative in the interval of $x$ at $-2le xle0$

$f(x)$ is equal zero at $x=0$

$$
text{color{#4257b2}(b) Determine where instantaneous rate of change of $f(x)$ is positive, negative and zero. Then determine the rate of change from $x=1$ to $x=2$ }
$$

Rate of change is positive at $0le x<2$

Rate of change is negative at $0le x<-2$

Rate of change is zero at $x=0$

Rate of change from $x=1, x=2$

$$
dfrac{0-1}{2-1}=dfrac{-1}{1}=-1
$$

Rate of change from $x=1$ to $x=2$ is $-1$

$$
text{color{Brown}(a)
\ \
$f(x)$ is positive in the interval of $x$ at $0le xle2$
\ \
$f(x)$ is negative in the interval of $x$ at $-2le xle0$
\ \
$f(x)$ is equal zero at $x=0$
\ \
(b) \ \
Rate of change is positive at $0le x<2$
\ \
Rate of change is negative at $0le x<-2$
\ \
Rate of change is zero at $x=0$
\ \
Rate of change from $x=1$ to $ x=2$ is $-1$}
$$

A pizza company advertised a special card with $50$ dollar but this card allowance for owner to purchases pizza for $5$ dollar for each one fully year. Pizza normally is $12$ dollars.

$$
text{color{#4257b2}(a) Write the expression that represent for $n$ of pizza with card and without the card}
$$

With card $a_{n}=5 n$ where, $n$ the number of pizza.

Without card $a_{n}=12 n$ where, $n$ the number of pizza.

$$
text{color{#4257b2}(b) How many pizza that make the card worthwhile?}
$$

$$
a_{n}=5 n 50=5 n
$$

$$
n=dfrac{50}{5} n=10
$$

Number of pizza that make card worthwhile is $10$ pizzas.

$$
text{color{Brown}(a) $a_{n}=5 n$ $a_{n}=12 n $
\ \
(b) Number of pizza that make card worthwhile is $10$ pizzas.}
$$

Solve the following inequalities.

$$
color{#4257b2}text{(a)} 4x-5<-2(x+1)
$$

Use distributive property as follows:

$$
4x-5<-2x-2
$$

Isolate the variables on the left side as follows:

$$
4x+2x<-2+5 6x<3
$$

Divide both of sides by $(6)$ as follows:

$$
x<dfrac{1}{2}
$$

$therefore$ The solution set is $left(-infty, dfrac{1}{2} right)$

$$
color{#4257b2}text{(b)} -4le-(3x+1)le5
$$

Use distributive property as follows:

$$
-4le-3x-1le5
$$

Add both of sides by $(1)$ as follows:

$$
-3le-3xle6
$$

Divide both of sides by $(-3)$ as follows:

$$
1ge xge-2
$$

$therefore$ The solution set is $left[-2, 1 right]$

$$
color{#4257b2}text{(c)} (x+1)(x-5)(x+2)>0
$$

$$
x=-2, x=-1, x=5
$$

Interval of $(x)$ are:

$$
x<-2, -2<x<-1, -1<x5
$$

Determine the value of polynomial according each interval as follows:

For $x<-2 f(-3)=-16$ For $-2<x<-1 f(-1.5)=4.875$

For $-1<x5 f(6)=63$

According the inequality $f(x)<0$, the true intervals which make the inequality true are:

$$
-2<x5
$$

$therefore$ The solution set is $left[-2, -1 right] cup left(5, infty right)$

$$
color{#4257b2}text{(d)} (2x-4)^2(x+3)ge0
$$

$$
x=-3, x=2
$$

Interval of $(x)$ are:

$$
xle-3, -3le xle2, xge2
$$

Determine the value of polynomial according each interval as follows:

For $xle-3 f(-4)=-144$

For $-3le xle2 f(0)=48$

For $xge2 f(3)=24$

According the inequality $f(x)<0$, the true intervals which make the inequality true are:

$$
-3le xle2 xge2
$$

$therefore$ The solution set is $left[-3, 2 right] cup left[2, infty right)$

$$
text{color{Brown}(a) $left(-infty, dfrac{1}{2} right)$
\ \ \
(b) $left[-2, 1 right]$
\ \
(c) $left[-2, -1 right] cup left(5, infty right)$
\ \
(d) $left[-3, 2 right] cup left[2, infty right)$}
$$

The following expression represent the height in meter for projectile. answer for the following:

$$
h(t)=-5t^2+20t+15
$$

$$
text{color{#4257b2}(a) Find the height at the moment of launch.}
$$

At the moment of launch $t=0$ $h(t)=15$ meter

$$
text{color{#4257b2}(b) Find the time when the projectile in the ground.}
$$

At the moment of launch $h(t)=0$

$$
h(t)=-5t^2+20t+15=0
$$

Divide the function by $(-5)$ as follows:

$$
t^2-4t-3=0
$$

Use completing square property as follows:

$$
t^2-4t+4=3+4 (t-2)^2=7
$$

Use zero property as follows:

$$
t-2=7 t=9 text{second}
$$

The time when the projectile in the ground is $9$ seconds

$$
text{color{Brown}(a) $h(t)=15$ meters (b) $t=9$ second}
$$

Consider the following expression, then answer for the following terms.

$$
x^3+x^2+1
$$

$$
text{color{#4257b2}(a) Estimate the slope of the tangent line at $x=1$}
$$

From the graph, the tangent line cross with curve at point $(y=3)$

Slope $=dfrac{3-1}{1-0}$ Slope $=dfrac{2}{1}=2$

$$
text{color{#4257b2}(b) What are the coordinates of point of tangent?}
$$

$$
(1, 3)
$$

$$
text{color{#4257b2}(c) Determine the equation of tangent line}
$$

Note that, slope $=2$ and the point on the curve is $(1, 3)$, so:

According the rule of linear equation as follows:

$$
y=mx+b 3=2+b
$$

$$
b=3-2 b=1
$$

The equation of tangent line is $y=2x+1$

$$
text{color{Brown}(a) Slope $=2$ (b) $(1, 3)$
\ \
(c) The equation of tangent line is $y=2x+1$}
$$

Explain why polynomial function of $f(x)=4x^{2008}+2008x^4+4$ has no zeros.

Because when we solve this polynomial, we use the logarithmic and exponential laws and general rules in this partition, not quadratic formula equations rules, so we not need zeros to solve it.

$$
text{color{Brown}Because when we solve this polynomial, we use the logarithmic and exponential laws and general rules in this partition, not quadratic formula equations rules, so we not need zeros to solve it.}
$$

According the number line in the textbook, answer for the following terms.

$$
text{color{#4257b2}(a) Write the solution using set notation.}
$$

$$
(-2, 7)
$$

$$
text{color{#4257b2}(b) Create an inequality that represent the number line.}
$$

$$
2x+2<3x+4<2x+11
$$

$$
text{color{Brown}(a) $(-2, 7)$ (b) $2x+2<3x+4<2x+11$}
$$