Remember the following algebraic rules:

$$
begin{equation*} dfrac{ab}{ac}=dfrac{cancel{a}b}{cancel{a}c}=dfrac{b}{c}end{equation*}
$$

$$
begin{equation*} dfrac{x^a}{x^b}=x^{a-b} end{equation*}
$$

If there are variables in the denominators, the value that will make the denominator zero will be the restrictions.

$$
begin{align*}
&text{a);;;} frac{6-4t}{2}=frac{2(3-2t)}{2}=3-2t \ \
&text{b);;;} frac{9x^2}{6x^3}=frac{3(3)x^{2-3}}{2(3)}=frac{3x^{-1}}{2}=frac{3}{2x};; ;;;xneq 0\\
&text{c);;;} frac{7a^2b^3}{21a^4b}=frac{7a^{2-4}b^{3-1}}{7times 3}=frac{a^{-2}b^2}{3}=frac{b^2}{3a^2};; ;;; aneq 0\\
end{align*}
$$

a) $3-2t$

b) $dfrac{3}{2x}$ , $xneq 0$

c) $dfrac{b^2}{3a^2}$ , $aneq 0$

Remember the following algebraic rules:

$$
begin{equation*} dfrac{ab}{ac}=dfrac{cancel{a}b}{cancel{a}c}=dfrac{b}{c}end{equation*}
$$

$$
begin{equation*} dfrac{x^a}{x^b}=x^{a-b} end{equation*}
$$

If there are variables in the denominators, the value that will make the denominator zero will be the restrictions.

a) $dfrac{5(x+3)}{(x+3)(x-3)}=dfrac{5cancel{(x+3)}}{cancel{(x+3)}(x-3)}=dfrac{5}{x-3}$ ; $xneq -3,;3$

b) $dfrac{6x-9}{2x-3}=dfrac{3(2x-3)}{2x-3}=dfrac{3cancel{(2x-3)}}{cancel{(2x-3)}}=3$ ; $xneq dfrac{3}{2}$

c) $dfrac{4a^2b-2ab^2}{(2a-b)^2}=dfrac{2ab(2a-b)}{(2a-b)(2a-b)}=dfrac{2abcancel{(2a-b)}}{cancel{(2a-b)}(2a-b)}=dfrac{2ab}{2a-b}$ ; $bneq 2a$

a) $dfrac{5}{x-3}$ , $xneq -3,;3$

b) $3, xneq dfrac{3}{2}$

c) $dfrac{2ab}{2a-b}$ , $bneq 2a$

Remember the following algebraic rules:

$$
begin{equation*} dfrac{ab}{ac}=dfrac{cancel{a}b}{cancel{a}c}=dfrac{b}{c}end{equation*}
$$

$$
begin{equation*} dfrac{x^a}{x^b}=x^{a-b} end{equation*}
$$

If there are variables in the denominators, the value that will make the denominator zero will be the restrictions.

a) $dfrac{(x-1)(x-3)}{(x+2)(x-1)}=dfrac{cancel{(x-1)}(x-3)}{(x+2)cancel{(x-1)}}=dfrac{x-3}{x+2}$ ; $xneq -2, ;1$

b) $dfrac{5x^2+x-4}{25x^2-40x+16}=dfrac{(5x-4)(x+1)}{(5x-4)(5x-4)}=dfrac{x+1}{5x-4}$ ; $xneq dfrac{4}{5}$

c) $dfrac{x^2-7xy+10y^2}{x^2+xy-6y^2}=dfrac{(x-5y)(x-2y)}{(x+3y)(x-2y)}=dfrac{x-5y}{x+3y}$ ; $xneq -3y,; 2y$

a) $dfrac{x-3}{x+2}$ ; $xneq -2,;1$

b) $dfrac{x+1}{5x-4}$ ; $xneq dfrac{4}{5}$

c) $dfrac{x-5y}{x+3y}$ ; $xneq -3y,;2y$

a) The factored form of the given expression, $dfrac{14x^3-7x^2+21x}{7x}
,$ is

$$
begin{align*}
&
dfrac{7x(2x^2-x+3)}{7x}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{cancel{7x}(2x^2-x+3)}{cancel{7x}}
\\&=
2x^2-x+3
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $2x^2-x+3
text{, }xne0
.$

b) The factored form of the given expression, $dfrac{-5x^3y^2}{10xy^3}
,$ is

$$
begin{align*}
&
dfrac{5xy^2(-x^2)}{5xy^2(2y)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{cancel{5xy^2}(-x^2)}{cancel{5xy^2}(2y)}
\\&=
dfrac{-x^2}{2y}
\\&=
-dfrac{x^2}{2y}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $-dfrac{x^2}{2y}
text{, }xne0text{, }yne0
.$

c) The given expression, $dfrac{2t(5-t)}{5t^2(t-5)}
,$ is equivalent to

$$
begin{align*}
&
dfrac{2t(-1)(-5+t)}{5t^2(t-5)}
\\&=
dfrac{-2t(t-5)}{5t^2(t-5)}
\\&=
dfrac{-2cdot t(t-5)}{5tcdot t(t-5)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{-2cdot cancel{t(t-5)}}{5tcdot cancel{t(t-5)}}
\\&=
dfrac{-2}{5t}
\\&=
-dfrac{2}{5t}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $-dfrac{2}{5t}
text{, }tne{ 0, 5 }
.$

d) The factored form of the given expression, $dfrac{5ab}{15a^4b-10a^2b^2}
,$ is

$$
begin{align*}
&
dfrac{5ab}{5ab(3a^3-2ab)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{cancel{5ab}}{cancel{5ab}(3a^3-2ab)}
\\&=
dfrac{1}{3a^3-2ab}
\\&=
dfrac{1}{a(3a^2-2b)}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{1}{3a^3-2ab}
text{, }aneleft{0, pmsqrt{dfrac{2b}{3}} right}text{, }bne0text{, }
.$

e) The factored form of the given expression, $dfrac{2x^2+10x}{-3x-15}
,$ is

$$
begin{align*}
&
dfrac{2x(x+5)}{-3(x+5)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{2x(cancel{x+5})}{-3(cancel{x+5})}
\\&=
dfrac{2x}{-3}
\\&=
-dfrac{2x}{3}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $-dfrac{2x}{3}
text{, }xne-5
.$

f) The factored form of the given expression, $dfrac{2ab-6a}{9a-3ab}
,$ is

$$
begin{align*}
&
dfrac{2a(b-3)}{3a(3-b)}
\\&=
dfrac{2a(-1)(-b+3)}{3a(3-b)}
\\&=
dfrac{-2a(3-b)}{3a(3-b)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{-2cancel{a(3-b)}}{3cancel{a(3-b)}}
\\&=
dfrac{-2}{3}
\\&=
-dfrac{2}{3}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $-dfrac{2}{3}
text{, }ane0text{, }bne3
.$

a) $2x^2-x+3
text{, }xne0$

b) $-dfrac{x^2}{2y}
text{, }xne0text{, }yne0$

c) $-dfrac{2}{5t}
text{, }tne{ 0, 5 }$

d) $dfrac{1}{3a^3-2ab}
text{, }aneleft{0, pmsqrt{dfrac{2b}{3}} right}text{, }bne0text{, }$

e) $-dfrac{2x}{3}
text{, }xne-5$

f) $-dfrac{2}{3}
text{, }ane0text{, }bne3$

a) The factored form of the given expression, $dfrac{a+4}{a^2+3a-4}
,$ is

$$
begin{align*}
&
dfrac{a+4}{(a+4)(a-1)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{cancel{a+4}}{(cancel{a+4})(a-1)}
\\&=
dfrac{1}{a-1}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{1}{a-1}
text{, }ane{ -4,1 }
.$

b) The factored form of the given expression, $dfrac{x^2-9}{15-5x}
,$ is

$$
begin{align*}
&
dfrac{(x+3)(x-3)}{5(3-x)}
\\&=
dfrac{(x+3)(-1)(-x+3)}{5(3-x)}
\\&=
dfrac{-(x+3)(3-x)}{5(3-x)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{-(x+3)(cancel{3-x})}{5(cancel{3-x})}
\\&=
dfrac{-x-3}{5}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{-x-3}{5}
text{, }xne3
.$

c) The factored form of the given expression, $dfrac{x^2-5x+6}{x^2+3x-10}
,$ is

$$
begin{align*}
&
dfrac{(x-3)(x-2)}{(x+5)(x-2)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{(x-3)(cancel{x-2})}{(x+5)(cancel{x-2})}
\\&=
dfrac{x-3}{x+5}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{x-3}{x+5}
text{, }xne{-5,2}
.$

d) The factored form of the given expression, $dfrac{10+3p-p^2}{25-p^2}
,$ is

$$
begin{align*}
&
dfrac{(5-p)(2+p)}{(5-p)(5+p)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{(cancel{5-p})(2+p)}{(cancel{5-p})(5+p)}
\\&=
dfrac{2+p}{5+p}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{2+p}{5+p}
text{, }pne{-5,5}
.$

e) The factored form of the given expression, $dfrac{t^2-7t+12}{t^3-6t^2+9t}
,$ is

$$
begin{align*}
&
dfrac{(t-4)(t-3)}{t(t^2-6t+9)}
\\&=
dfrac{(t-4)(t-3)}{t(t-3)(t-3)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{(t-4)(cancel{t-3})}{t(t-3)(cancel{t-3})}
\\&=
dfrac{t-4}{t(t-3)}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{t-4}{t-3}
text{, }tne{ 0,3 }
.$

f) The factored form of the given expression, $dfrac{6t^2-t-2}{2t^2-t-1}
,$ is

$$
begin{align*}
&
dfrac{(3t-2)(2t+1)}{(2t+1)(t-1)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{(3t-2)(cancel{2t+1})}{(cancel{2t+1})(t-1)}
\\&=
dfrac{3t-2}{t-1}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{3t-2}{t-1}
text{, }tneleft{ -dfrac{1}{2}, 1 right}
.$

a) $dfrac{1}{a-1}
text{, }ane{ -4,1 }$

b) $dfrac{-x-3}{5}
text{, }xne3$

c) $dfrac{x-3}{x+5}
text{, }xne{-5,2}$

d) $dfrac{2+p}{5+p}
text{, }pne{-5,5}$

e) $dfrac{t-4}{t-3}
text{, }tne{ 0,3 }$

f) $dfrac{3t-2}{t-1}
text{, }tneleft{ -dfrac{1}{2}, 1 right}$

a) The denominator of the given expression, $f(x)=dfrac{2+x}{x}
,$ cannot be equal to zero. Hence,

$$
begin{align*}
xne0
.end{align*}
$$

The domain of the given function is $left{x|
xne0
right}
.$

b) The denominator of the given expression, $g(x)=dfrac{3}{x(x-2)}
,$ cannot be equal to zero. Hence,

$$
begin{align*}
begin{array}{rcl}
xne0 &text{ OR }& x-2ne0
\
&& xne2
end{array}
.end{align*}
$$

The domain of the given function is $left{x|
xne0text{ or }xne2
right}
.$

c) The denominator of the given expression, $h(x)=dfrac{-3}{(x+5)(x-5)}
,$ cannot be equal to zero. Hence,

$$
begin{align*}
begin{array}{rcl}
x+5ne0 &text{ OR }& x-5ne0
\
xne-5 && xne5
end{array}
.end{align*}
$$

The domain of the given function is $left{x|
xne-5text{ or }xne5
right}
.$

d) The denominator of the given expression, $f(x)=dfrac{1}{x^2-1}
,$ cannot be equal to zero. Hence,

$$
begin{align*}
x^2-1&ne0
\
x^2&ne1
\
x&nepmsqrt{1}
\
x&nepm1
.end{align*}
$$

The domain of the given function is $left{x|
xne-1text{ or }xne1
right}
.$

e) The denominator of the given expression, $g(x)=dfrac{1}{x^2+1}
,$ cannot be equal to zero. Since for any $x,$ the denominator does not become zero, then there are no restrictions for the values of $x.$ Hence, the domain is
the set of all real numbers.

f) The denominator of the given expression, $h(x)=dfrac{x-1}{x^2-1}
,$ cannot be equal to zero. Hence,

$$
begin{align*}
x^2-1&ne0
\
x^2&ne1
\
x&nepmsqrt{1}
\
x&nepm1
.end{align*}
$$

The domain of the given function is $left{x|
xne-1text{ or }xne1
right}
.$

a) $left{x|
xne0
right}$

b) $left{x|
xne0text{ or }xne2
right}$

c) $left{x|
xne-5text{ or }xne5
right}$

d) $left{x|
xne-1text{ or }xne1
right}$

e) the set of all real numbers

f) $left{x|
xne-1text{ or }xne1
right}$

a) Cancelling the common factors between the numerator and the denominator, the given expression, $g(x)=dfrac{6x^2+3x-21}{3}
,$ is equivalent to

$$
begin{align*}
g(x)&=dfrac{3(2x^2+x-7)}{3}
\\&=
dfrac{cancel3(2x^2+x-7)}{cancel3}
\\&=
2x^2+x-7
.end{align*}
$$

This is equivalent to the given function, $f(x)=2x^2+x-7
.$ Hence, $f(x)$ and $g(x)$ are
equivalent.

b) The given functions, $h(x)=3x^2+5x+1$ and $j(x)=dfrac{3x^3+5x^2+x}{x},xne0
,$ are
NOT equivalent
since $j(x)$ contains a restriction while $h(x)$ does not have a restriction.

a) Using $P=a+b+c$ or the formula for the perimeter of a triangle, with two sides having lengths $(9x+3)$ and the perimeter equal to $(30x+10),$ then

$$
begin{align*}
30x+10&=(9x+3)+(9x+3)+c
\
30x+10&=18x+6+c
\
30x-18x+10-6&=c
\
c&=12x+4
.end{align*}
$$

Hence, the ratio of the base, $c,$ and the perimeter is

$$
begin{align*}
&
dfrac{12x+4}{30x+10}
\\&=
dfrac{4(3x+1)}{10(3x+1)}
\\&=
dfrac{2cdot2(3x+1)}{5cdot2(3x+1)}
\\&=
dfrac{2cdotcancel{2(3x+1)}}{5cdotcancel{2(3x+1)}}
\\&=
dfrac{2}{5}
.end{align*}
$$

With the ratio given by the expression, $dfrac{12x+4}{30x+10}
,$ then $xne-dfrac{1}{3}$.

b) Without the restriction, $x$ can be any number including $-dfrac{1}{3}.$ This value of $x,$ however, will make the denominator of the ratio equal to zero and hence, will given an undefined expression. Therefore, the restriction on $x$ is necessary.

a) $dfrac{2}{5}$; $xne-dfrac{1}{3}$

b) see explanation

Using $V=dfrac{1}{3}pi r^2h
,$ with a radius of $5$ and a height of $2,$ then

$$
begin{align*}
V_1&=dfrac{1}{3}pi (5)^2(2)
\
V_1&=dfrac{1}{3}pi (25)(2)
\
V_1&=dfrac{50pi}{3}
.end{align*}
$$

Using $V=dfrac{1}{3}pi r^2h
,$ with a radius of $4$ and a height of $3,$ then

$$
begin{align*}
V_2&=dfrac{1}{3}pi (4)^2(3)
\
V_2&=dfrac{1}{3}pi (16)(3)
\
V_2&=16pi
.end{align*}
$$

Hence, the ratio of the two cones is

$$
begin{align*}
&
dfrac{V_1}{V_2}
\\&=
dfrac{dfrac{50pi}{3}}{16pi}
\\&=
dfrac{50pi}{3}div16pi
\\&=
dfrac{50pi}{3}cdotdfrac{1}{16pi}
\\&=
dfrac{50pi(1)}{3(16pi)}
\\&=
dfrac{(50pidiv2pi)(1)}{3(16pidiv2pi)}
&qty(text{Cancel }2pi)
\\&=
dfrac{(25)(1)}{3(8)}
\\&=
dfrac{25}{24}
.end{align*}
$$

$$
dfrac{25}{24}
$$

a) The factored form of the given expression, $dfrac{20t^3+15t^2-5t}{5t}
,$ is

$$
begin{align*}
&
dfrac{5t(4t^2+3t-1)}{5t}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{cancel{5t}(4t^2+3t-1)}{cancel{5t}}
\\&=
4t^2+3t-1
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $4t^2+3t-1
text{, }tne0
.$

b) The factored form of the given expression, $dfrac{5(4x-2)}{8(2x-1)^2}
,$ is

$$
begin{align*}
&
dfrac{5cdot2(2x-1)}{4cdot2(2x-1)(2x-1)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{5cdotcancel{2(2x-1)}}{4cdotcancel{2(2x-1)}(2x-1)}
\\&=
dfrac{5}{4(2x-1)}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{5}{4(2x-1)}
text{, }xnedfrac{1}{2}
.$

c) The factored form of the given expression, $dfrac{x^2-9x+20}{16-x^2}
,$ is

$$
begin{align*}
&
dfrac{(x-5)(x-4)}{(4-x)(4+x)}
\\&=
dfrac{(x-5)(x-4)}{(-1)(-4+x)(4+x)}
\\&=
-dfrac{(x-5)(x-4)}{(x-4)(x+4)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
-dfrac{(x-5)(cancel{x-4})}{(cancel{x-4})(x+4)}
\\&=
-dfrac{x-5}{x+4}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $-dfrac{x-5}{x+4}
text{, }xne{ -4,4 }
.$

d) The factored form of the given expression, $dfrac{2x^2-xy-y^2}{x^2-2xy+y^2}
,$ is

$$
begin{align*}
&
dfrac{(2x+y)(x-y)}{(x-y)(x-y)}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{(2x+y)(cancel{x-y})}{(cancel{x-y})(x-y)}
\\&=
dfrac{2x+y}{x-y}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{2x+y}{x-y}
text{, }xne y
.$

a) $4t^2+3t-1
text{, }tne0$

b) $dfrac{5}{4(2x-1)}
text{, }xnedfrac{1}{2}$

c) $-dfrac{x-5}{x+4}
text{, }xne{ -4,4 }$

d) $dfrac{2x+y}{x-y}
text{, }xne y$

The area $A$ and perimeter $P$ of a rectangle with width $w$ and length $l$ are given by

$$
begin{equation*}A=lwend{equation*}
$$

$$
begin{equation*}P=2(l+w)end{equation*}
$$

Since length is six times its width, $l=6w$, the ratio of its area to its perimeter is

$$
begin{equation*}dfrac{A}{P}=dfrac{lw}{2(l+w)}=dfrac{(6w)(w)}{2(6w+w)}=dfrac{6w^2}{2(7w)}=dfrac{2(3)w^{2-1}}{2(7)}=dfrac{3w}{7}end{equation*}
$$

$$
dfrac{3w}{7}
$$

An example of a polynomial with a quotient of $3x-2$ and a restriction of $xne4$ is

$$
begin{align*}
&
3x-2cdotdfrac{x-4}{x-4}
\\&=
dfrac{(3x-2)(x-4)}{x-4}
\\&=
dfrac{3x(x)+3x(-4)-2)(x)-2(-4)}{x-4}
\\&=
dfrac{3x^2-12x-2x+8}{x-4}
\\&=
dfrac{3x^2-14x+8}{x-4}
.end{align*}
$$

$$
dfrac{3x^2-14x+8}{x-4}
$$

An example of a polynomial with $3$ restrictions that are consecutive numbers is

$$
begin{align*}
&
dfrac{1}{x(x-1)(x-2)}
\\&=
dfrac{1}{x[(x(x)+x(-2)-1(x)-1(-2)}
\\&=
dfrac{1}{x(x^2-2x-x+2)}
\\&=
dfrac{1}{x(x^2-3x+2)}
\\&=
dfrac{1}{x(x^2)+x(-3x)+x(2)}
\\&=
dfrac{1}{x^3-3x^2+2x}
.end{align*}
$$

$$
dfrac{1}{x^3-3x^2+2x}
$$

a.i) An example of a polynomial that simplifies to $dfrac{2x+1}{x-4}
,$ with restrictions $xne-1,4,$ is

$$
begin{align*}
&
dfrac{2x+1}{x-4}cdotdfrac{x+1}{x+1}
\\&=
dfrac{(2x+1)(x+1)}{(x-4)(x+1)}
\\&=
dfrac{2x(x)+2x(1)+1(x)+1(1)}{x(x)+x(1)-4(x)-4(1)}
\\&=
dfrac{2x^2+2x+x+1}{x^2+x-4x-4}
\\&=
dfrac{2x^2+3x+1}{x^2-3x-4}
.end{align*}
$$

a.ii) An example of a polynomial that simplifies to $dfrac{2x+1}{x-4}
,$ with restrictions $xne0,4,$ is

$$
begin{align*}
&
dfrac{2x+1}{x-4}cdotdfrac{x}{x}
\\&=
dfrac{(2x+1)(x)}{(x-4)(x)}
\\&=
dfrac{2x^2+x}{x^2-4x}
.end{align*}
$$

a.iii) An example of a polynomial that simplifies to $dfrac{2x+1}{x-4}
,$ with restrictions $xnedfrac{2}{3},4,$ is

$$
begin{align*}
&
dfrac{2x+1}{x-4}cdotdfrac{3x-2}{3x-2}
\\&=
dfrac{(2x+1)(3x-2)}{(x-4)(3x-2)}
\\&=
dfrac{2x(3x)+2x(-2)+1(3x)+1(-2)}{x(3x)+x(-2)-4(3x)-4(-2)}
\\&=
dfrac{6x^2-4x+3x-2}{3x^2-2x-12x+8}
\\&=
dfrac{6x^2-x-2}{3x^2-14x+8}
.end{align*}
$$

a.iv) An example of a polynomial that simplifies to $dfrac{2x+1}{x-4}
,$ with restrictions $xne-dfrac{1}{2},4,$ is

$$
begin{align*}
&
dfrac{2x+1}{x-4}cdotdfrac{2x+1}{2x+1}
\\&=
dfrac{(2x+1)(2x+1)}{(x-4)(2x+1)}
\\&=
dfrac{2x(2x)+2x(1)+1(2x)+1(1)}{x(2x)+x(1)-4(2x)-4(1)}
\\&=
dfrac{4x^2+2x+2x+1}{2x^2+x-8x-4}
\\&=
dfrac{4x^2+4x+1}{2x^2-7x-4}
.end{align*}
$$

b) The rational expression, $dfrac{2x^2-7x-4}{x^2-8x+16}
,$ satisfies the conditions of the problem since

$$
begin{align*}
&
dfrac{2x^2-7x-4}{x^2-8x+16}
\\&=
dfrac{(2x+1)(x-4)}{(x-4)(x-4)}
\\&=
dfrac{(2x+1)(cancel{x-4})}{(cancel{x-4})(x-4)}
\\&=
dfrac{2x+1}{x-4}text{, }xne4
.end{align*}
$$

a.i) $dfrac{2x^2+3x+1}{x^2-3x-4}$

a.ii) $dfrac{2x^2+x}{x^2-4x}$

a.iii) $dfrac{6x^2-x-2}{3x^2-14x+8}$

a.iv) $dfrac{4x^2+4x+1}{2x^2-7x-4}$

b) Yes, there is.

The rational expressions, $dfrac{x^2+x}{x}$ and $dfrac{2x^2+2x}{2x}
,$ are two different rational expressions that simplify to the same polynomial since

$$
begin{align*}
begin{array}{rcccl}
dfrac{x^2+x}{x} &&&& dfrac{2x^2+2x}{2x}
\\
=dfrac{x(x+1)}{x} &&&& =dfrac{2x(x+1)}{2x}
\\
=dfrac{cancel{x}(x+1)}{cancel{x}} &&&& =dfrac{cancel{2x}(x+1)}{cancel{2x}}
\\
=x+1 &&&& =x+1
end{array}
.end{align*}
$$

We need to evaluate the given function when $x$ approaches to very large positive values $(xto +infty)$ and to very large negative values $(xto-infty)$. To do this, we can arbitrarily use $x=10^{10}$ and $x=-10^{10}$ respectively. You need your calculators to do this.

a) $f(x)=dfrac{50x+73}{x^2-10x-400}$

Evaluate when $x=10^{10}$

$f(10^{10})=dfrac{50(10^{10})+73}{(10^{10})^2-10(10^{10})-400}=5times 10^{-9}approx 0$

Evaluate when $x=-10^{10}$

$f(-10^{10})=dfrac{50(-10^{10})}{(-10^{10})^2-10(-10^{10})-400}=-5times 10^{-9}approx 0$

Therefore

$limlimits_{x to +infty }f(x)=0$ and$limlimits_{x to -infty }f(x)=0$

b) $g(x)=dfrac{4x^3-100}{5x^3+87x+28}$

$g(10^{10})=dfrac{4(10^{10})^3-100}{5(10^{10})^3+87(10^{10})+28}=dfrac{4}{5}$

$g(-10^{10})=dfrac{4(-10^{10})^3-100}{5(-10^{10})^3+87(-10^{10})+28}=dfrac{4}{5}$

Therefore

$limlimits_{x to +infty }g(x)=dfrac{4}{5}$ and $limlimits_{x to -infty }g(x)=dfrac{4}{5}$

c) $h(x)=dfrac{-7x^2+3x}{200x+9999}$

$h(10^{10})=dfrac{-7(10^{10})^2+3(10^{10})}{200(10^{10})+9999}=3.5times 10^8 approx +infty$

$h(-10^{10})=dfrac{-7(-10^{10})^2+3(-10^{10})}{200(-10^{10})+9999}=-3.5times 10^8 approx -infty$

Therefore,

$limlimits_{x to +infty }h(x)=+infty$ and $limlimits_{x to -infty }h(x)=-infty$

a) $limlimits_{x to +infty }f(x)=0$ and$limlimits_{x to -infty }f(x)=0$

b) $limlimits_{x to +infty }g(x)=dfrac{4}{5}$ and $limlimits_{x to -infty }g(x)=dfrac{4}{5}$

c) $limlimits_{x to +infty }h(x)=+infty$ and $limlimits_{x to -infty }h(x)=-infty$

a) The factored form of the given function, $a(t)=dfrac{-2(1+t^2)^2+2t(2)(1+t^2)(2t)}{(1+t^2)^4}
,$ is

$$
begin{align*}
a(t)&=dfrac{-2(1+t^2)[(1+t^2)-t(2)(2t)]}{(1+t^2)^4}
\\
a(t)&=dfrac{-2(1+t^2)[(1+t^2)-4t^2]}{(1+t^2)^4}
\\
a(t)&=dfrac{-2(1+t^2)[1+(t^2-4t^2)]}{(1+t^2)^4}
\\
a(t)&=dfrac{-2(1+t^2)(1-3t^2)}{(1+t^2)^4}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
a(t)&=dfrac{-2(cancel{1+t^2})(1-3t^2)}{(1+t^2)^{cancel43}}
\\&=
dfrac{-2(1-3t^2)}{(1+t^2)^3}
.end{align*}
$$

For any value of $x,$ the denominator does not become zero. Hence, there is no restriction.

b) The factored form of the given function, $f(x)=dfrac{2(2x+1)(2)(3x-2)^3-(2x+1)^2(3)(3x-2)^2}{(3x-2)^6}
,$ is

$$
begin{align*}
f(x)&=dfrac{(2x+1)(3x-2)^2[2(2)(3x-2)-(2x+1)(3)}{(3x-2)^6}
\\
f(x)&=dfrac{(2x+1)(3x-2)^2(12x-8-6x-3)}{(3x-2)^6}
\\
f(x)&=dfrac{(2x+1)(3x-2)^2(6x-11)}{(3x-2)^6}
.end{align*}
$$

Cancelling the common factors between the numerator and the denominator, the expression above is equivalent to

$$
begin{align*}
f(x)&=dfrac{(2x+1)cancel{(3x-2)^2}(6x-11)}{(3x-2)^{cancel64}}
\\&=
dfrac{(2x+1)(6x-11)}{(3x-2)^{4}}
.end{align*}
$$

Since the denominator of the original expression cannot be zero, then the simplified form of the given expression with its restriction is $dfrac{(2x+1)(6x-11)}{(3x-2)^{4}}
text{, }xnedfrac{2}{3}
.$

a) $dfrac{-2(1-3t^2)}{(1+t^2)^3}$; no restriction

b) $dfrac{(2x+1)(6x-11)}{(3x-2)^{4}}
text{, }xnedfrac{2}{3}$