Combine similar terms and then simplify. In subtracting expressions, change the sign of the subtrahend then add.

$$
begin{equation*} a-(b-c+d)=a+(-b+c-d)end{equation*}
$$

$$
begin{align*} bold{a);;;} &(7x^2-2x+1)+(9x^2-4x+5)-(4x^2+6x-7) \\&=(7x^2-2x+1)+(9x^2-4x+5)+(-4x^2-6x+7)
\\
&=(7x^2+9x^2-4x^2)+(-2x-4x-6x)+(1+5+7)\\
&=12x^2-14x+13\\\
bold{b);;;} &(7a^2-4ab+9b^2)-(-a^2+2ab+6b^2)\\
&=(7a^2-4ab+9b^2)+(a^2-2ab-6b^2)\\
&=(7a^2+a^2)+(-4ab-2ab)+(9b^2-6b^2)\\
&=8a^2-6ab+3b^2
end{align*}
$$

a) $12x^2-12x+13$

b) $8a^2-6ab+3b^2$

The two non-equivalent functions, $f(x)=x^2$ and $g(x)=x
,$ satisfies the conditions of the problem since

$$
begin{align*}
f(0)&=0^2
\&=
0,
\\
g(0)&=0
;\\\
f(1)&=1^2
\&=
1
,\\
g(1)&=1
.end{align*}
$$

Hence, $f(0)=g(0)$ and $f(1)=g(1).$

$f(x)=x^2,$ $g(x)=x$

Simplifying the given function, $A(n)=-(n+30)+(2n+5)
,$ results to

$$
begin{align*}
A(n)&=-(n+30)+(2n+5)
\&=
-n-30+2n+5
\&=
n-25
.end{align*}
$$

Simplifying the given function, $B(n)=(7-n)-(32-2n)
,$ results to

$$
begin{align*}
B(n)&=(7-n)-(32-2n)
\&=
7-n-32+2n
\&=
n-25
.end{align*}
$$

Simplifying the given function, $C(n)=(n-26)-(n+4)+(n-3)
,$ results to

$$
begin{align*}
C(n)&=(n-26)-(n+4)+(n-3)
\&=
n-26-n-4+n-3
\&=
n-33
.end{align*}
$$

a) Since $A(n), B(n),$ and $C(n)$ are not equivalent functions, then the daughters are not triplets.

b) Astrid and Bernice are twins since the functions associated with their ages, $A(n)$ and $B(n),$ are equivalent.

c) Since $C(n)=n-33,$ which means that Cassandra’s age is $33$ years less than Ms. Flanagan, Then Ms. Flanagan was $33 text{ years old}$ when Cassandra was born.

a) NO triplets

b) Astrid and Bernice are twins

c) $33 text{ years old}$

a) Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $-3(7x-5)(4x-7)
,$ is equivalent to

$$
begin{align*}
&
-3[(7x(4x)+7x(-7)-5(4x)-5(-7)]
\&=
-3(28x^2-49x-20x+35)
\&=
-3(28x^2-69x+35)
.end{align*}
$$

Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to

$$
begin{align*}
&
-3(28x^2)-3(-69x)-3(35)
\&=
-84x^2+207x-105
.end{align*}
$$

b) Using Distributive Property, the given expression, $-(y^2-4y+7)(3y^2-5y-3)
,$ is equivalent to

$$
begin{align*}
&
-[y^2(3y^2)+y^2(-5y)+y^2(-3)-4y(3y^2)-4y(-5y)-4y(-3)+7(3y^2)+7(-5y)+7(-3)]
\&=
-[3y^4-5y^3-3y^2-12y^3+20y^2+12y+21y^2-35y-21]
\&=
-[3y^4+(-5y^3-12y^3)+(-3y^2+20y^2+21y^2)+(12y-35y)-21]
\&=
-[3y^4-17y^3+38y^2-23y-21]
\&=
-3y^4+17y^3-38y^2+23y+21
.end{align*}
$$

c) Using the cube of a binomial which is given by

$$
begin{align*}
(a+b)^3&=a^3+3a^2b+3ab^2+b^3
\&text{or}\
(a-b)^3&=a^3-3a^2b+3ab^2-b^3
,end{align*}
$$
the given expression, $2(a+b)^3
,$ is equivalent to

$$
begin{align*}
&
2[(a)^3+3(a)^2(b)+3(a)(b)^2+(b)^3]
\&=
2[a^3+3(a^2)(b)+3(a)(b^2)+b^3]
\&=
2[a^3+3a^2b+3ab^2+b^3]
.end{align*}
$$

Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to

$$
begin{align*}
&
2(a^3)+2(3a^2b)+2(3ab^2)+2(b^3)
\&=
2a^3+6a^2b+6ab^2+2b^3
.end{align*}
$$

d) Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the given expression, $3(x^2-2)^2(2x-3)^2
,$ is equivalent to

$$
begin{align*}
&
3[(x^2)^2-2(x^2)(2)+(2)^2][(2x)^2-2(2x)(3)+(3)^2]
\&=
3[x^4-4x^2+4][4x^2-12x+9]
.end{align*}
$$

Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to

$$
begin{align*}
&
3[x^4(4x^2)+x^4(-12x)+x^4(9)-4x^2(4x^2)-4x^2(-12x)-4x^2(9)+4(4x^2)+4(-12x)+4(9)]
\&=
3[4x^6-12x^5+9x^4-16x^4+48x^3-36x^2+16x^2-48x+36]
\&=
3[4x^6-12x^5+(9x^4-16x^4)+48x^3+(-36x^2+16x^2)-48x+36]
\&=
3[4x^6-12x^5-7x^4+48x^3-20x^2-48x+36]
\&=
12x^6-36x^5-21x^4+114x^3-60x^2-144x+108
.end{align*}
$$

a) $-84x^2+207x-105$

b) $-3y^4+17y^3-38y^2+23y+21$

c) $2a^3+6a^2b+6ab^2+2b^3$

d) $12x^6-36x^5-21x^4+114x^3-60x^2-144x+108$

We are given the formula for the volume $V$ of a cone with radius $r$ and height $h$ as

$$
begin{equation*} V=frac{1}{3}pi r^2 hend{equation*}
$$

If the radius is increased by $x$ $(r to r+x)$ and height is increased by 2x $(h to h+2x)$, the new volume $V_2$ would be

$$
begin{align*} V_2&=dfrac{1}{3}pi(r+x)^2(h+2x)\
&=dfrac{1}{3}pi(r^2+2rx+x^2)(h+2x)\
&=dfrac{1}{3}pi[r^2(h+2x)+2rx(h+2x)+x^2(h+2x)]\
&=dfrac{1}{3}pi[(r^2h+2r^2x)+(2rhx+4rx^2)+(x^2h+2x^3)]\
V_2&=dfrac{1}{3}pi(r^2h+2r^2x+2rhx+4rx^2+x^2h+2x^3)\
end{align*}
$$

$V=dfrac{1}{3}pi(r+x)^2(h+2x)$

$V=dfrac{1}{3}pi(r^2h+2r^2x+2rhx+4rx^2+x^2h+2x^3)$

a) Removing the grouping symbols and then combining the like terms, the given expression, $(2x^4-3x^2-6)+(6x^4-x^3+4x^2+5)
,$ is equivalent to

$$
begin{align*}
&
2x^4-3x^2-6+6x^4-x^3+4x^2+5
\&=
(2x^4+6x^4)-x^3+(-3x^2+4x^2)+(-6+5)
\&=
8x^4-x^3+x^2-1
.end{align*}
$$

b) Using the Distributive Property, the given expression, $(x^2-4)(2x^2+5x-2)
,$ is equivalent to

$$
begin{align*}
&
x^2(2x^2)+x^2(5x)+x^2(-2)-4(2x^2)-4(5x)-4(-2)
\&=
2x^4+5x^3-2x^2-8x^2-20x+8
\&=
2x^4+5x^3-10x^2-20x+8
.end{align*}
$$

c) Using the Distributive Property, the given expression, $-7x(x^2+x-1)-3x(2x^2-5x+6)
,$ is equivalent to

$$
begin{align*}
&
-7x(x^2)-7x(x)-7x(-1)-3x(2x^2)-3x(-5x)-3x(6)
\&=
-7x^3-7x^2+7x-6x^3+15x^2-18x
\&=
(-7x^3-6x^3)+(-7x^2+15x^2)+(7x-18x)
\&=
-13x^3+8x^2-11x
.end{align*}
$$

d) Using the Distributive Property, the given expression, $-2x^2(3x^3-7x+2)-x^3(5x^3+2x-8)
,$ is equivalent to

$$
begin{align*}
&
-2x^2(3x^3)-2x^2(-7x)-2x^2(2)-x^3(5x^3)-x^3(2x)-x^3(-8)
\&=
-6x^5+14x^3-4x^2-5x^6-2x^4+8x^3
.end{align*}
$$

Combining like terms, the expression above is equivalent to

$$
begin{align*}
&
-5x^6-6x^5-2x^4+(14x^3+8x^3)-4x^2
\&=
-5x^6-6x^5-2x^4+22x^3-4x^2
.end{align*}
$$

e) Removing the grouping symbols and then combining like terms, the given expression, $-2x[5x-(2x-7)]+6x[3x-(1+2x)]
,$ is equivalent to

$$
begin{align*}
&
-2x[5x-2x+7]+6x[3x-1-2x]
\&=
-2x[3x+7]+6x[x-1]
.end{align*}
$$

Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to

$$
begin{align*}
&
-2x(3x)-2x(7)+6x(x)+6x(-1)
\&=
-6x^2-14x+6x^2-6x
.end{align*}
$$

Combining like terms, the expression above is equivalent to

$$
begin{align*}
&
(-6x^2+6x^2)+(-14x-6x)
\&=
-20x
.end{align*}
$$

f) Since $a^2b^2=(ab)^2,$ the given expression, $(x+2)^2(x-1)^2-(x-4)^2(x+4)^2
,$ is equivalent to

$$
begin{align*}
&
[(x+2)(x-1)]^2-[(x-4)(x+4)]^2
.end{align*}
$$

Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to

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

Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent

$$
begin{align*}
&
[x^2+x-2]^2-[(x)^2-(4)^2]^2
\&=
[x^2+x-2]^2-[x^2-16]^2
.end{align*}
$$

Using the square of a multinomial which is given by $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2b
,$ the expression above is equivalent to

$$
begin{align*}
&
[(x^2)^2+(x)^2+(-2)^2+2(x^2)(x)+2(x^2)(-2)+2(x)(-2)]-[x^2-16]^2
\&=
[x^4+x^2+4+2x^3-4x^2-4x]-[x^2-16]^2
\&=
[x^4+2x^3+(x^2-4x^2)-4x+4]-[x^2-16]^2
\&=
[x^4+2x^3-3x^2-4x+4]-[x^2-16]^2
.end{align*}
$$

Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to

$$
begin{align*}
&
[x^4+2x^3-3x^2-4x+4]-[(x^2)^2-2(x^2)(16)+(16)^2]
\&=
[x^4+2x^3-3x^2-4x+4]-[x^4-32x^2+256]
.end{align*}
$$

Removing the grouping symbols and combining like terms, the expression above is equivalent to

$$
begin{align*}
&
x^4+2x^3-3x^2-4x+4-x^4+32x^2-256
\&=
(x^4-x^4)+2x^3+(-3x^2+32x^2)-4x+(4-256)
\&=
2x^3+29x^2-4x-252
.end{align*}
$$

g) Using the square of a multinomial which is given by $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2b
,$ the given expression, $(x^2+5x-3)^2
,$ is equivalent to

$$
begin{align*}
&
(x^2)^2+(5x)^2+(-3)^2+2(x^2)(5x)+2(x^2)(-3)+2(5x)(-3)
\&=
x^4+25x^2+9+10x^3-6x^2-30x
\&=
x^4+10x^3+(25x^2-6x^2)-30x+9
\&=
x^4+10x^3+19x^2-30x+9
.end{align*}
$$

a) $8x^4-x^3+x^2-1$

b) $2x^4+5x^3-10x^2-20x+8$

c) $-13x^3+8x^2-11x$

d) $-5x^6-6x^5-2x^4+22x^3-4x^2$

e) $-20x$

f) $2x^3+29x^2-4x-252$

g) $x^4+10x^3+19x^2-30x+9$

a) Factoring the $GCF=
6m^2n^2
,$ the given expression, $12m^2n^3+18m^3n^2
,$ is equivalent to

$$
begin{align*}
&
6m^2n^2(2n+3m)
.end{align*}
$$

b) Using the factoring of trinomials in the form $x^2+bx+c,$ the
expression

$$
begin{align*}
x^2-9x+20
end{align*}
$$
has $c=
20$ and $b=
-9
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-4,-5
right}.$ Using these two numbers, the
expression
above is equivalent to

$$
begin{align*}
(x-4)(x-5)
.end{align*}
$$

c) Factoring the $GCF=
3
,$ the given $text{
expression,
}
3x^2+24x+45
,$ is equivalent to

$$
begin{align*}
3(x^2+8x+15)
.end{align*}
$$

Using the factoring of trinomials in the form $x^2+bx+c,$ the expression

$$
begin{align*}
x^2+8x+15
end{align*}
$$
has $c=
15$ and $b=
8
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
3,5
right}.$ Using these two numbers, the
expression
above is equivalent to

$$
begin{align*}
3(x+3)(x+5)
.end{align*}
$$

d) Factoring the $GCF=
2
,$ the given
expression,
$50x^2-72
,$ is equivalent to

$$
begin{align*}
2(25x^2-36)
.end{align*}
$$

Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the
expression
above is equivalent to

$$
begin{align*}
&
2[(5x)^2-(6)^2]
\&=
2(5x+6)(5x-6)
.end{align*}
$$

e) Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression

$$
begin{align*}
9x^2-6x+1
end{align*}
$$
has $ac=
9(1)=9$ and $b=
-6
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-3,-3
right}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to

$$
begin{align*}
9x^2-3x-3x+1
.end{align*}
$$

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to

$$
begin{align*}
(9x^2-3x)-(3x-1)
.end{align*}
$$

Factoring the $GCF$ in each group results to

$$
begin{align*}
3x(3x-1)-(3x-1)
.end{align*}
$$

Factoring the $GCF=
(3x-1)$ of the entire expression above results to

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

f) Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression

$$
begin{align*}
10a^2+a-3
end{align*}
$$
has $ac=
10(-3)=-30$ and $b=
1
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
6,-5
right}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to

$$
begin{align*}
10a^2+6a-5a-3
.end{align*}
$$

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to

$$
begin{align*}
(10a^2+6a)-(5a+3)
.end{align*}
$$

Factoring the $GCF$ in each group results to

$$
begin{align*}
2a(5a+3)-(5a+3)
.end{align*}
$$

Factoring the $GCF=
(5a+3)$ of the entire expression above results to

$$
begin{align*}
&
(5a+3)(2a-1)
.end{align*}
$$

a) $6m^2n^2(2n+3m)$

b) $(x-4)(x-5)$

c) $3(x+3)(x+5)$

d) $2(5x+6)(5x-6)$

e) $(3x-1)^2$

f) $(5a+3)(2a-1)$

a) Factoring the $GCF=
2x^2y
,$ the given $text{
expression,
}
2x^2y^4-6x^5y^3+8x^3y
,$ is equivalent to

$$
begin{align*}
2x^2y(y^3-3x^3y^2+4x)
.end{align*}
$$

b) Factoring the $GCF=
(x+4)
,$ the given $text{
expression,
}
2x(x+4)+3(x+4)
,$ is equivalent to

$$
begin{align*}
(x+4)(2x+3)
.end{align*}
$$

c) Using the factoring of trinomials in the form $x^2+bx+c,$ the
expression

$$
begin{align*}
x^2-3x-10
end{align*}
$$
has $c=
-10$ and $b=
-3
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-5,2
right}.$ Using these two numbers, the
expression
above is equivalent to

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

d) Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression

$$
begin{align*}
15x^2-53x+42
end{align*}
$$
has $ac=
15(42)=630$ and $b=
-53
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-18,-35
right}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to

$$
begin{align*}
15x^2-18x-35x+42
.end{align*}
$$

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to

$$
begin{align*}
(15x^2-18x)-(35x-42)
.end{align*}
$$

Factoring the $GCF$ in each group results to

$$
begin{align*}
3x(5x-6)-7(5x-6)
.end{align*}
$$

Factoring the $GCF=
(5x-6)$ of the entire expression above results to

$$
begin{align*}
(5x-6)(3x-7)
.end{align*}
$$

e) Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the given
expression,
$a^4-16
,$ is equivalent to

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

f) Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the given
expression,
$(m-n)^2-(2m+3n)^2
,$ is equivalent to

$$
begin{align*}
&
[(m-n)+(2m+3n)][(m-n)-(2m+3n)]
\&=
(m-n+2m+3n)(m-n-2m-3n)
.end{align*}
$$

a) $2x^2y(y^3-3x^3y^2+4x)$

b) $(x+4)(2x+3)$

c) $(x-5)(x+2)$

d) $(5x-6)(3x-7)$

e) $(a^2+4)(a+2)(a-2)$

f) $(m-n+2m+3n)(m-n-2m-3n)$

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

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

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

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

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $-2a^2-3c^2
text{, }bne0
.$

b) The factored form of the given expression, $dfrac{30x^2y^3-20x^2z^2+50x^2}{10x^2}
,$ is

$$
begin{align*}
&
dfrac{10x^2(3y^3-2z^2+5)}{10x^2}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{10x^2}(3y^3-2z^2+5)}{cancel{10x^2}}
\\&=
3y^3-2z^2+5
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $3y^3-2z^2+5
text{, }xne0
.$

c) The factored form of the given expression, $dfrac{xy-xyz}{xy}
,$ is

$$
begin{align*}
&
dfrac{xy(1-z)}{xy}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{xy}(1-z)}{cancel{xy}}
\\&=
1-z
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $1-z
text{, }xne0text{, }yne0
.$

d) The factored form of the given expression, $dfrac{16mnr-24mnp+40kmn}{8mn}
,$ is

$$
begin{align*}
&
dfrac{8mn(2r-3p+5k)}{8mn}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{8mn}(2r-3p+5k)}{cancel{8mn}}
\\&=
2r-3p+5k
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $2r-3p+5k
text{, }mne0text{, }nne0
.$

a) $-2a^2-3c^2
text{, }bne0$

b) $3y^3-2z^2+5
text{, }xne0$

c) $1-z
text{, }xne0text{, }yne0$

d) $2r-3p+5k
text{, }mne0text{, }nne0$

a) Cancelling the common factors between the numerator and the denominator, the given expression, $8xy^2+12x^2y-dfrac{6x^3}{2xy}
,$ is equivalent to

$$
begin{align*}
&
8xy^2+12x^2y-dfrac{2xcdot3x^2}{2xcdot y}
\\&=
8xy^2+12x^2y-dfrac{cancel{2x}cdot3x^2}{cancel{2x}cdot y}
\\&=
8xy^2+12x^2y-dfrac{3x^2}{y}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $8xy^2+12x^2y-dfrac{3x^2}{y}
text{, }xne0text{, }yne0
.$

b) The factored form of the given expression, $dfrac{7a-14b}{2(a-2b)}
,$ is

$$
begin{align*}
&
dfrac{7(a-2b)}{2(a-2b)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{7(cancel{a-2b})}{2(cancel{a-2b})}
\\&=
dfrac{7}{2}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{7}{2}
text{, }ane2b
.$

c) The factored form of the given expression, $dfrac{m+3}{m^2+10m+21}
,$ is

$$
begin{align*}
&
dfrac{m+3}{(m+7)(m+3)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{m+3}}{(m+7)(cancel{m+3})}
\\&=
dfrac{1}{m+7}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{1}{m+7}
text{, }mne{ -7,-3 }
.$

d) The factored form of the given expression, $dfrac{4x^2-4x-3}{4x^2-9}
,$ is

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

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

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

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{2x+1}{2x+3}
text{, }xneleft{ -dfrac{3}{2},dfrac{3}{2} right}
.$

e) The factored form of the given expression, $dfrac{3x^2-21x}{7x^2-28x+21}
,$ is

$$
begin{align*}
&
dfrac{3x(x-7)}{7(x^2-4x+3)}
\\&=
dfrac{3x(x-7)}{7(x-3)(x-1)}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{3x(x-7)}{7(x-3)(x-1)}
text{, }xneleft{ 1,3 right}
.$

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

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

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

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

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{x-y}{x+y}
text{, }xneleft{ -dfrac{y}{3},-y right}
.$

a) $8xy^2+12x^2y-dfrac{3x^2}{y}
text{, }xne0text{, }yne0$

b) $dfrac{7}{2}
text{, }ane2b$

c) $dfrac{1}{m+7}
text{, }mne{ -7,-3 }$

d) $dfrac{2x+1}{2x+3}
text{, }xneleft{ -dfrac{3}{2},dfrac{3}{2} right}$

e) $dfrac{3x(x-7)}{7(x-3)(x-1)}
text{, }xneleft{ 1,3 right}$

f) $dfrac{x-y}{x+y}
text{, }xneleft{ -dfrac{y}{3},-y right}$

No. The two rational functions, $f(x)=dfrac{1}{x}$ and $g(x)=dfrac{2}{x},$ have the same restrictions, but $f(x)ne g(x).$

a) The factored form of the given expression, $dfrac{6x}{8y}timesdfrac{2y^2}{3x}
,$ is

$$
begin{align*}
&
dfrac{3xcdot2}{2ycdot2cdot2}timesdfrac{2ycdot y}{3x}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{3x}cdotcancel2}{cancel{2y}cdotcancel2cdot2}timesdfrac{cancel{2y}cdot y}{cancel{3x}}
\\&=
dfrac{y}{2}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{y}{2}
text{, }xne0text{, }yne0
.$

b) The factored form of the given expression, $dfrac{10m^2}{3n}timesdfrac{6mn}{20m^2}
,$ is

$$
begin{align*}
&
dfrac{10m^2}{3n}timesdfrac{3ncdot2cdot m}{10m^2cdot2}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{10m^2}}{cancel{3n}}timesdfrac{cancel{3n}cdotcancel{2}cdot m}{cancel{10m^2}cdotcancel{2}}
\\&=
m
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $m
text{, }mne0text{, }nne0
.$

c) Multiplying by the reciprocal of the divisor, the given expression, $dfrac{2ab}{5bc}divdfrac{6ac}{10b}
,$ is equivalent to

$$
begin{align*}
&
dfrac{2ab}{5bc}cdotdfrac{10b}{6ac}
.end{align*}
$$

The factored form of expression above is

$$
begin{align*}
&
dfrac{2acdot b}{5bcdot c}cdotdfrac{5bcdot2}{2acdot3c}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{2a}cdot b}{cancel{5b}cdot c}cdotdfrac{cancel{5b}cdot2}{cancel{2a}cdot3c}
\\&=
dfrac{2b}{3c^2}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{2b}{3c^2}
text{, }ane0text{, }bne0text{, }cne0
.$

d) Multiplying by the reciprocal of the divisor, the given expression, $dfrac{5p}{8pq}divdfrac{3p}{12q}
,$ is equivalent to

$$
begin{align*}
&
dfrac{5p}{8pq}cdotdfrac{12q}{3p}
.end{align*}
$$

The factored form of expression above is

$$
begin{align*}
&
dfrac{5cdot p}{4qcdot2p}cdotdfrac{4qcdot3}{3cdot p}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{5cdot cancel{p}}{cancel{4q}cdot2p}cdotdfrac{cancel{4q}cdotcancel3}{cancel3cdot cancel{p}}
\\&=
dfrac{5}{2p}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{5}{2p}
text{, }pne0text{, }qne0text{, }cne0
.$

a) $dfrac{y}{2}
text{, }xne0text{, }yne0$

b) $m
text{, }mne0text{, }nne0$

c) $dfrac{2b}{3c^2}
text{, }ane0text{, }bne0text{, }cne0$

d) $dfrac{5}{2p}
text{, }pne0text{, }qne0text{, }cne0$

a) Multiplying by the reciprocal of the divisor, the given expression, $dfrac{x^2}{2xy}timesdfrac{x}{2y^2}divdfrac{(3x)^2}{xy^2}
,$ is equivalent to

$$
begin{align*}
&
dfrac{x^2}{2xy}timesdfrac{x}{2y^2}cdotdfrac{xy^2}{(3x)^2}
\\&=
dfrac{x^2}{2xy}timesdfrac{x}{2y^2}cdotdfrac{xy^2}{9x^2}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{x^2}}{2cancel{x}y}timesdfrac{cancel{x}}{2cancel{y^2}}cdotdfrac{xcancel{y^2}}{9cancel{x^2}}
\\&=
dfrac{x}{36y}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{x}{36y}
text{, }xne0text{, }yne0
.$

b) Multiplying by the reciprocal of the divisor, the given expression, $dfrac{x^2-5x+6}{x^2-1}timesdfrac{x^2-4x-5}{x^2-4}divdfrac{x-5}{x^2+3x+2}
,$ is equivalent to

$$
begin{align*}
&
dfrac{x^2-5x+6}{x^2-1}timesdfrac{x^2-4x-5}{x^2-4}timesdfrac{x^2+3x+2}{x-5}
.end{align*}
$$

The factored form of expression above is

$$
begin{align*}
&
dfrac{(x-2)(x-3)}{(x+1)(x-1)}timesdfrac{(x-5)(x+1)}{(x+2)(x-2)}timesdfrac{(x+2)(x+1)}{x-5}
.end{align*}
$$

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

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

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{(x-3)(x+1)}{x-1}
text{, }xneleft{ -2,-1,1,2,5 right}
.$

c) Multiplying by the reciprocal of the divisor, the given expression, $dfrac{1-x^2}{1+y}timesdfrac{1-y^2}{x+x^2}divdfrac{y^3-y}{x^2}
,$ is equivalent to

$$
begin{align*}
&
dfrac{1-x^2}{1+y}timesdfrac{1-y^2}{x+x^2}timesdfrac{x^2}{y^3-y}
.end{align*}
$$

The factored form of expression above is

$$
begin{align*}
&
dfrac{(1+x)(1-x)}{1+y}timesdfrac{-(y^2-1)}{x(1+x)}timesdfrac{x^2}{y(y^2-1)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{(cancel{1+x})(1-x)}{1+y}timesdfrac{-(cancel{y^2-1})}{cancel{x}(cancel{1+x})}timesdfrac{cancel{x}cdot x}{y(cancel{y^2-1})}
\\&=
dfrac{-x(1-x)}{y(1+y)}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{-x(1-x)}{y(1+y)}
text{, }xneleft{ -1,0 right}text{, }yneleft{ -1,0,1 right}
.$

d) Multiplying by the reciprocal of the divisor, the given expression, $dfrac{x^2-y^2}{4x^2-y^2}timesdfrac{4x^2+8xy+3y^2}{x+y}divdfrac{2x+3y}{2x-y}
,$ is equivalent to

$$
begin{align*}
&
dfrac{x^2-y^2}{4x^2-y^2}timesdfrac{4x^2+8xy+3y^2}{x+y}timesdfrac{2x-y}{2x+3y}
.end{align*}
$$

The factored form of expression above is

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

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

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

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $x-y
text{, }xneleft{ -dfrac{3y}{2},-y,-dfrac{y}{2},dfrac{y}{2} right}
.$

a) $dfrac{x}{36y}
text{, }xne0text{, }yne0$

b) $dfrac{(x-3)(x+1)}{x-1}
text{, }xneleft{ -2,-1,1,2,5 right}$

c) $dfrac{-x(1-x)}{y(1+y)}
text{, }xneleft{ -1,0 right}text{, }yneleft{ -1,0,1 right}$

d) $x-y
text{, }xneleft{ -dfrac{3y}{2},-y,-dfrac{y}{2},dfrac{y}{2} right}$

a) Multiplying the fractions by an expression equal to $1$ to find an equivalent fraction that uses the $LCD=
15x
,$ the given expression, $dfrac{4}{5x}-dfrac{2}{3x}
,$ is equivalent to

$$
begin{align*}
&
dfrac{4}{5x}cdotdfrac{3}{3}-dfrac{2}{3x}cdotdfrac{5}{5}
\\&=
dfrac{12}{15x}-dfrac{10}{15x}
.end{align*}
$$

Operating on the numerators and copying the common denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{12-10}{15x}
\\&=
dfrac{2}{15x}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{2}{15x}
text{, }xne0
.$

b) Multiplying the fractions by an expression equal to $1$ to find an equivalent fraction that uses the $LCD=
(x+1)(x-1)
,$ the given expression, $dfrac{5}{x+1}-dfrac{2}{x-1}
,$ is equivalent to

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

Operating on the numerators and copying the common denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{5x-5-2x-2}{(x+1)(x-1)}
\\&=
dfrac{3x-7}{(x+1)(x-1)}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{3x-7}{(x+1)(x-1)}
text{, }xneleft{ -1,1 right}
.$

c) The factored form of the given expression, $dfrac{1}{x^2+3x-4}+dfrac{1}{x^2+x-12}
,$ is

$$
begin{align*}
&
dfrac{1}{(x+4)(x-1)}+dfrac{1}{(x+4)(x-3)}
.end{align*}
$$

Multiplying the fractions by an expression equal to $1$ to find an equivalent fraction that uses the $LCD=
(x+4)(x-1)(x-3)
,$ the expression above is equivalent to

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

Operating on the numerators and copying the common denominator, the expression above is equivalent to

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

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{2(x-2)}{(x+4)(x-1)(x-3)}
text{, }xne{ -4,1,3 }
.$

d) The factored form of the given expression, $dfrac{1}{x^2-5x+6}-dfrac{1}{x^2-9}
,$ is

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

Multiplying the fractions by an expression equal to $1$ to find an equivalent fraction that uses the $LCD=
(x-2)(x-3)(x+3)
,$ the expression above is equivalent to

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

Operating on the numerators and copying the common denominator, the expression above is equivalent to

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

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{5}{(x-2)(x-3)(x+3)}
text{, }xne{ -3,2,3 }
.$

a) $dfrac{2}{15x}
text{, }xne0$

b) $dfrac{3x-7}{(x+1)(x-1)}
text{, }xneleft{ -1,1 right}$

c) $dfrac{2(x-2)}{(x+4)(x-1)(x-3)}
text{, }xne{ -4,1,3 }$

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

a) Multiplying the fractions by an expression equal to $1$ to find an equivalent fraction that uses the $LCD=
6x^3
,$ the given expression, $dfrac{1}{2x}-dfrac{7}{3x^2}+dfrac{4}{x^3}
,$ is equivalent to

$$
begin{align*}
&
dfrac{1}{2x}cdotdfrac{3x^2}{3x^2}-dfrac{7}{3x^2}cdotdfrac{2x}{2x}+dfrac{4}{x^3}cdotdfrac{6}{6}
\\&=
dfrac{3x^2}{6x^3}-dfrac{14x}{6x^3}+dfrac{24}{6x^3}
\\&=
dfrac{3x^2-14x+24}{6x^3}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{3x^2-14x+24}{6x^3}
text{, }xne0
.$

b) Multiplying the fractions by an expression equal to $1$ to find an equivalent fraction that uses the $LCD=
(x+2)(x-6)
,$ the given expression, $dfrac{3x}{x+2}+dfrac{4x}{x-6}
,$ is equivalent to

$$
begin{align*}
&
dfrac{3x}{x+2}cdotdfrac{x-6}{x-6}+dfrac{4x}{x-6}cdotdfrac{x+2}{x+2}
\\&=
dfrac{3x^2-18x}{(x+2)(x-6)}+dfrac{4x^2+8x}{(x+2)(x-6)}
.end{align*}
$$

Operating on the numerators and copying the common denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{3x^2-18x+4x^2+8x}{(x+2)(x-6)}
\\&=
dfrac{7x^2-10x}{(x+2)(x-6)}
\\&=
dfrac{x(7x-10)}{(x+2)(x-6)}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{x(7x-10)}{(x+2)(x-6)}
text{, }xne{ -2,6 }
.$

c) The factored form of the given expression, $dfrac{6x}{x^2-5x+6}-dfrac{3x}{x^2+x-12}
,$ is

$$
begin{align*}
&
dfrac{6x}{(x-2)(x-3)}-dfrac{3x}{(x-3)(x+4)}
.end{align*}
$$

Multiplying the fractions by an expression equal to $1$ to find an equivalent fraction that uses the $LCD=
(x-2)(x-3)(x+4)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{6x}{(x-2)(x-3)}cdotdfrac{x+4}{x+4}-dfrac{3x}{(x-3)(x+4)}cdotdfrac{x-2}{x-2}
\\&=
dfrac{6x^2+24x}{(x-2)(x-3)(x+4)}-dfrac{3x^2-6x}{(x-2)(x-3)(x+4)}
.end{align*}
$$

Operating on the numerators and copying the common denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{6x^2+24x-3x^2+6x}{(x-2)(x-3)(x+4)}
\\&=
dfrac{3x^2+30x}{(x-2)(x-3)(x+4)}
\\&=
dfrac{3x(x+10)}{(x-2)(x-3)(x+4)}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{3x(x+10)}{(x-2)(x-3)(x+4)}
text{, }xne{ -4,2,3 }
.$

d) The factored form of the given expression, $dfrac{2(x-2)^2}{x^2+6x+5}timesdfrac{3x+15}{(2-x)^2}
,$ is

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

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

$$
begin{align*}
&
dfrac{2cancel{(x-2)(x-2)}}{(cancel{x+5})(x+1)}timesdfrac{3(cancel{x+5})}{-cancel{(x-2)(x-2)}}
\\&=
dfrac{6}{-(x+1)}
\\&=
-dfrac{6}{x+1}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $-dfrac{6}{x+1}
text{, }xne{ -5,-1,2 }
.$

e) Multiplying by the reciprocal of the divisor, the given expression, $dfrac{(x-2y)^2}{x^2-y^2}divdfrac{(x-2y)(x+3y)}{(x+y)^2}
,$ is equivalent to

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

The factored form of the expression above is

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

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

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

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{(x-2y)(x+y)}{(x-y)(x+3y)}
text{, }xneleft{ -3y,-y,y, 2y right}
.$

f) The factored form of the given expression, $dfrac{2b-5}{b^2-2b-15}+dfrac{3b}{b^2+b-30}timesdfrac{b^2+8b+12}{b+3}
,$ is

$$
begin{align*}
&
dfrac{2b-5}{(b-5)(b+3)}+dfrac{3b}{(b+6)(b-5)}timesdfrac{(b+6)(b+2)}{b+3}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{2b-5}{(b-5)(b+3)}+dfrac{3b}{(cancel{b+6})(b-5)}timesdfrac{(cancel{b+6})(b+2)}{b+3}
\\&=
dfrac{2b-5}{(b-5)(b+3)}+dfrac{3b(b+2)}{(b-5)(b+3)}
\\&=
dfrac{2b-5}{(b-5)(b+3)}+dfrac{3b^2+6b}{(b-5)(b+3)}
.end{align*}
$$

Operating on the numerators and copying the common denominator, the expression above is equivalent to

$$
begin{align*}
&
dfrac{2b-5+3b^2+6b}{(b-5)(b+3)}
\\&=
dfrac{3b^2+8b-5}{(b-5)(b+3)}
.end{align*}
$$

Since the denominator of the original expression cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{3b^2+8b-5}{(b-5)(b+3)}
text{, }bne{ -6,-3,5 }
.$

a) $dfrac{3x^2-14x+24}{6x^3}
text{, }xne0$

b) $dfrac{x(7x-10)}{(x+2)(x-6)}
text{, }xne{ -2,6 }$

c) $dfrac{3x(x+10)}{(x-2)(x-3)(x+4)}
text{, }xne{ -4,2,3 }$

d) $-dfrac{6}{x+1}
text{, }xne{ -5,-1,2 }$

e) $dfrac{(x-2y)(x+y)}{(x-y)(x+3y)}
text{, }xneleft{ -3y,-y,y, 2y right}$

f) $dfrac{3b^2+8b-5}{(b-5)(b+3)}
text{, }bne{ -6,-3,5 }$

Fred’s final mark is the weighted sum of the mid-term exam $(25%)$ and final exam $(75%)$ marks .

mid-term exam: $dfrac{40}{x}$

final exam: $dfrac{60}{2x}$

The expression for the final mark based on his exam results is

$0.25left(dfrac{40}{x}right)+0.75left(dfrac{60}{2x}right)=0.5$

Solve for $x$

$dfrac{10}{x}+dfrac{45}{2x}=0.5implies dfrac{20}{2x}+dfrac{45}{2x}=0.5$

$dfrac{20+45}{2x}=0.5implies dfrac{65}{2x}=0.5$

$2xleft(dfrac{65}{2x}right)=0.5(2x)implies 65=x$

Therefore, the value of $x$ is 65

a) Remember that $dfrac{a}{b}div dfrac{c}{d}=dfrac{a}{b}timesdfrac{d}{c}$. It is recommended to factor expressions first before performing operations.

$$
begin{align*}
P(n)&=dfrac{24}{n^3-3n^2+2n}div dfrac{3}{n}\\&=dfrac{8(3)}{n(n^2-3n+2)}div dfrac{3}{n}\\
&=dfrac{8(3)}{n(n-2)(n-1)}div dfrac{3}{n}\\&=dfrac{8(3)}{n(n-2)(n-1)}timesdfrac{n}{3}\\
&=dfrac{8(cancel{3})}{cancel{n}(n-2)(n-1)}timesdfrac{cancel{n}}{cancel{3}}\\
&=dfrac{8}{(n-2)(n-1)};; ;;; nneq 0,;1,;2
end{align*}
$$

Evaluate the simplified expression for the given value of $n$

b) $P(5)=dfrac{8}{(5-2)(5-1)}=dfrac{8}{3(4)}=dfrac{8}{12}=dfrac{2(4)}{3(4)}=dfrac{2}{3}$

$P(4)=dfrac{8}{(4-2)(4-1)}=dfrac{8}{2(3)}=dfrac{8}{6}=dfrac{4(2)}{3(2)}=dfrac{4}{3}$

Notice that the probability exceeds 1 which is absurd.

Note: It is likely a $bold{typo; in; the; question}$ and $P(n)$ should have been

$P(n)=dfrac{18}{n^3-3n^2+2n}div dfrac{3}{n}=dfrac{6}{(n-2)(n-1)} ; ; ; nneq 0,;1,;2$

In this case,

$P(5)=dfrac{6}{(5-2)(5-1)}=dfrac{6}{3(4)}=dfrac{1}{2}$

$$
P(4)=dfrac{6}{(4-2)(4-1)}=dfrac{6}{(2)(3)}=1
$$

a) $dfrac{6}{(n-2)(n-1)}$ ; $nneq 0,;1,;2$

b) $P(5)=dfrac{1}{2}$ , $P(4)=1$

Note: There’s a typo in the question.