a)
$$
begin{align*}
(4a^{2}-3a+2)-(-2a^{2}-3a+9)&=4a^{2}-3a+2+2a^{2}+3a-9tag{text{remove parentheses}}\
&=6a^{2}-7tag{text{combine like terms}}\
end{align*}
$$

b)
$$
begin{align*}
(2x^{2}-4xy+y^{2})-(4x^{2}+7xy-2y^{2})+(3x^{2}+6y^{2})&=\
&=2x^{2}-4xy+y^{2}-4x^{2}-7xy+2y^{2}+3x^{2}+6y^{2}
tag{text{remove parentheses}}\
&=2x^{2}-4x^{2}+3x^{2}-4xy-7xy+y^{2}+2y^{2}+6y^{2}tag{text{change places of the terms}}\
&=x^{2}-11xy+9y^{2}tag{text{combine like terms}}\
end{align*}
$$

c)
$$
begin{align*}
-(3d^{2}-2cd+d)+d(2c-5d)-3c(2c+d)&=\
&=-3d^{2}+2cd-d+2cd-5d^{2}-6c^{2}-3cd&
tag{text{remove parentheses use distributive property}}\
&=-8d^{2}+cd-d-6c^{2}tag{text{combine like terms}}\
end{align*}
$$

d)
$$
begin{align*}
3x(2x+y)-4x[5-(3x+2)]&=3x(2x-y)-4x[5-3x-2] tag{text{remove second parentheses in the middle}}\
&=3x(2x+y)-4x(3-3x)\
&=6x^{2}+3xy-12x+12x^{2} tag{text{remove parentheses use distributive property}}\
&=18x^{2}+3xy-12x tag{text{combine like terms}}\
end{align*}
$$

e)
$$
begin{align*}
2a(3a-5b+4)-6(3-2a-b)&=6a^{2}-10ab+8a-18+12a+6b tag{text{remove parentheses use distributive property}}\
&=6a^{2}-10ab+20a+6b-18tag{text{combine like terms}}\
end{align*}
$$

f)
$$
begin{align*}
7x(2x^{2}+3y-3)-3x(9-2x+4y)&=14x^{3}+21xy-21x-27x+6x^{2}-12xy tag{text{remove parentheses use distributive property}}\
&=14x^{3}+6x^{2}+9xy-48xtag{text{combine like terms}}\
end{align*}
$$

a) $6a^{2}-7$

b)$x^{2}-11xy+9y^{2}$

c) $-8d^{2}+cd-d-6c^{2}$

d) $18x^{2}+3xy-12x$

e) $6a^{2}-10ab+20a+6b-18$

f) $14x^{3}+6x^{2}+9xy-48x$

a) Given functions are:

$g(t)=(t-2)^{5}$ and $h(t)=(2-t)^{5}$

If $t=3$ then:

$g(3)=(3-2)^{5}=1^{5}=1$ and

$h(3)=(2-3)^{5}=(-1)^{5}=-1$

$g(3)ne h(3)$, so functions are not eqivalent.

Answer is: $text{textcolor{#4257b2}{No}}$

b) Given functions are: $f(x)$ and $g(x)$.

The first, we need simplify algebraically these functions and then compare them.

$$
begin{align*}
f(x)&=(x^{2}-6x)-(x^{2}+x-4)+(2x^{2}+1)\
&=x^{2}-6x-x^{2}-x+4+2x^{2}+1 tag{text{remove parentheses}}\
&=2x^{2}-7x-5 tag{text{combine like terms}}\
end{align*}
$$

$$
begin{align*}
g(x)&=(4x^{2}-7x-3)-(2x^{2}-8)\
&=4x^{2}-7x-3-2x^{2}+8 tag{text{remove parentheses}}\
&=2x^{2}-7x+5tag{text{combine like terms}}\
end{align*}
$$

$f(x)=x^{2}-7x+5$ and $g(x)=x^{2}-7x+5$, so we can conclude $f(x)=g(x)$ for each $xin R$

Answer is: $text{textcolor{#4257b2}{Yes}}$

c) Given functions are: $h(x)$ and $d(x)$.

The first, we need simplify algebraically these functions and then compare them.

$$
begin{align*}
h(x)&=(x-4)(x+7)(x+4)\
&=(x+7)(x-4)(x+4) tag{text{change places of the parentheses}}\
&=(x+7)(x^{2}-16)tag{text{use difference of squares}}\
end{align*}
$$

$$
begin{align*}
d(x)&=(x+7)(x^{2}-16)
end{align*}
$$

$h(x)=(x+7)(x^{2}-16)$ and $d(x)=(x+7)(x^{2}-16)$, so we can conclude $h(x)=d(x)$ for each $xin R$

Answer is: $text{textcolor{#4257b2}{Yes}}$

b) Given functions are: $b(t)$ and $c(t)$.

The first, we need simplify algebraically these functions and then compare them.

$$
begin{align*}
b(t)&=(3t+1)^{3}\
&=(3t)^{3}+3cdot(3t)^{2}cdot1+3cdot3tcdot1^{2}+1^{3}tag{text{use cube of a binomial}}\
&=27t^{3}+277t^{2}+9t+1\
end{align*}
$$

$$
begin{align*}
c(t)&=27t^{3}+277t^{2}+9t-1
end{align*}
$$

$b(t)=27t^{3}+277t^{2}+9t+1$ and $c(t)=27t^{3}+277t^{2}+9t-1$, so we can conclude $b(t)ne c(t)$.

Answer is: $text{textcolor{#4257b2}{No}}$

A quadratic polynomial as a degree of 2 while a linear has a degree of 1. Multiplying the two together results in:

$x(x^2)=x^3$

This is a third degree polynomial.

The sum of their ages is equal $5x-99$

There $x$ is Dion’s age, and Pam is five years younger than Dion so, Pams’s age can write like $x-5$

The sum of Dion and Pam’s ages is equal $x+(x-5)=2x-5$.

Now, we can determine ages of their children as below:

$$
begin{align*}
(5x-99)-(2x-5)&=5x-99-2x+5tag{text{remove parentheses}}\
&=3x-94tag{text{combine like terms}}\
end{align*}
$$

Answer is: $text{textcolor{#4257b2}{ sum of the ages of children is $3x-94$}}$

$$
3x-94
$$

a) Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $2(x-5)(3x-4)
,$ is equivalent to

$$
begin{align*}
&
2[x(3x)+x(-4)-5(3x)-5(-4)]
\&=
2[3x^2-4x-15x+20]
\&=
2(3x^2-19x+20)
.end{align*}
$$

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

$$
begin{align*}
&
2(3x^2)+2(-19x)+2(20)
\&=
6x^2-38x+40
.end{align*}
$$

b) 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, $(3x-1)^3
,$ is equivalent to

$$
begin{align*}
&
(3x)^3-3(3x)^2(1)+3(3x)(1)^2-(1)^3
\&=
(27x^3)-3(9x^2)(1)+3(3x)(1)-(1)
\&=
27x^3-27x^2+9x-1
.end{align*}
$$

c) The given expression, $2(x^2-3x+4)(-x^2+3x-4)
,$ is equivalent to

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

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

$$
begin{align*}
&
-2[(x^2)^2+(-3x)^2+(4)^2+2(x^2)(-3x)+2(x^2)(4)+2(-3x)(4)]
\&=
-2[x^4+9x^2+16-6x^3+8x^2-24x]
\&=
-2[x^4-6x^3+(9x^2+8x^2)-24x+16]
\&=
-2[x^4-6x^3+17x^2-24x+16]
.end{align*}
$$

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

$$
begin{align*}
&
-2(x^4)-2(-6x^3)-2(17x^2)-2(-24x)-2(16)
\&=
-2x^4+12x^3-34x^2+48x-32
.end{align*}
$$

d) Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $(5x-4)(3x-5)-(2x-3)^2
,$ is equivalent to

$$
begin{align*}
&
5x(3x)+5x(-5)-4(3x)-4(-5)-(2x-3)^2
\&=
15x^2-25x-12x+20-(2x-3)^2
\&=
15x^2-37x+20-(2x-3)^2
.end{align*}
$$

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

$$
begin{align*}
&
15x^2-37x+20-[(2x)^2-2(2x)(3)+(3)^2]
\&=
15x^2-37x+20-[4x^2-12x+9]
\&=
15x^2-37x+20-4x^2+12x-9
\&=
(15x^2-4x^2)+(-37x+12x)+(20-9)
\&=
11x^2-25x+11
.end{align*}
$$

e) Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the given expression, $3(2x-5)-9(4x-5)
,$ is equivalent to

$$
begin{align*}
&
3(2x)+3(-5)-9(4x)-9(-5)
\&=
6x-15-36x+45
\&=
-30x+30
.end{align*}
$$

f) 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, $-(x-y)^3
,$ is equivalent to

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

a) $6x^2-38x+40$

b) $27x^3-27x^2+9x-1$

c) $-2x^4+12x^3-34x^2+48x-32$

d) $11x^2-25x+11$

e) $-30x+30$

f) $-x^3+3x^2y-3xy^2+y^3$

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

$$
begin{equation*} A=lw end{equation*}
$$

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

If the length is increased by 2 $(implies l+2)$ and width is decreased by 1 $(implies w-1)$, the new area $A_2$ and $P_2$ would be

$$
begin{align*} A_2 &=(l+2)(w-1)\&=l(w-1)+2(w-1)\&=lw-l+2w-2\\
P_2 &=2[(l+2)+(w-1)]\&=2(l+w+1)end{align*}
$$

The change in perimeter is $P_2-P$

$$
begin{align*}
P_2-P&=2(l+w+1)-2(l+w)\&=2[l+w+1-(l+w)]\&=boxed{2}end{align*}
$$

The change in area is $A_2-A$

$$
begin{align*} A_2-A&=(lw-l+2w-2)-(lw)\ &=boxed{-l+2w-2}
end{align*}
$$

Therefore, perimeter is increased by 2 and the area is changed by $(-l+2w-2)$

a) 2

b) $2w-l-2$

Factoring the $GCF=
(x-2)
,$ the given expression, $x(x-2)-3(x-2)
,$ is equivalent to

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

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

$$
begin{align*}
x^2-11x+28
end{align*}
$$
has $c=
28$ and $b=
-11
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-4,-7
right}.$ Using these two numbers, the
expression
above is equivalent to

$$
begin{align*}
(x-4)(x-7)
.end{align*}
$$

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

$$
begin{align*}
3a^2-10a-8
end{align*}
$$
has $ac=
3(-8)=-24$ and $b=
-10
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-12,2
right}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to

$$
begin{align*}
3a^2-12a+2a-8
.end{align*}
$$

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

$$
begin{align*}
(3a^2-12a)+(2a-8)
.end{align*}
$$

Factoring the $GCF$ in each group results to

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

Factoring the $GCF=
(a-4)$ of the entire expression above results to

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

d) Factoring the $GCF=
3
,$ the given $text{
expression,
}
30x^2-9x-3
,$ is equivalent to

$$
begin{align*}
3(10x^2-3x-1)
.end{align*}
$$

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

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

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

$$
begin{align*}
3(10x^2-5x+2x-1)
.end{align*}
$$

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

$$
begin{align*}
3[(10x^2-5x)+(2x-1)]
.end{align*}
$$

Factoring the $GCF$ in each group results to

$$
begin{align*}
3[5x(2x-1)+(2x-1)]
.end{align*}
$$

Factoring the $GCF=
(2x-1)$ of the entire expression above results to

$$
begin{align*}
&
3[(2x-1)(5x+1)]
\&=
3(2x-1)(5x+1)
.end{align*}
$$
d) Factoring the $GCF=
2
,$ the given expression, $2x^2-8y^2+8x+8
,$ is equivalent to

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

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,
$16-25x^2
,$ is equivalent to

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

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,
$4(2-a)^2-81
,$ is equivalent to

$$
begin{align*}
&
[2(2-a)]^2-[9]^2
\&=
[2(2-a)+9][2(2-a)-9]
\&=
[4-2a+9][4-2a-9]
\&=
(-2a+13)(-2a-5)
.end{align*}
$$

a) $(x-2)(x-3)$

b) $(x-4)(x-7)$

c) $(a-4)(3a+2)$

d) $2(x^2-4y^2+4x+4)$

e) $(4+5x)(4-5x)$

f) $(-2a+13)(-2a-5)$

a) Grouping the first and second terms and the third and fourth terms, the given expression, $2n-6m+5n^2-15mn
,$ is equivalent to

$$
begin{align*}
(2n-6m)+(5n^2-15mn)
.end{align*}
$$

Factoring the $GCF$ in each group results to

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

Factoring the $GCF=
(n-3m)$ of the entire expression above results to

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

b) The given expression, $y^2+9-6y-x^2
,$ is equivalent to

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

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

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

c) The given expression, $y-b-(y-b)^2
,$ is equivalent to

$$
begin{align*}
(y-b)-(y-b)^2
.end{align*}
$$

Factoring the $GCF=
(y-b)
,$ the expression above is equivalent to

$$
begin{align*}
&
(y-b)[1-(y-b)]
\&=
(y-b)[1-y+b]
\&=
(y-b)(1-y+b)
.end{align*}
$$

d) Factoring the $GCF=
2
,$ the given expression, $2x^2-8y^2+8x+8
,$ is equivalent to

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

e) Grouping the first and second terms and the third and fourth terms, the given expression, $w^2+wb-aw-ab
,$ is equivalent to

$$
begin{align*}
(w^2+wb)-(aw+ab)
.end{align*}
$$

Factoring the $GCF$ in each group results to

$$
begin{align*}
w(w+b)-a(w+b)
.end{align*}
$$

Factoring the $GCF=
(w+b)$ of the entire expression above results to

$$
begin{align*}
(w+b)(w-a)
.end{align*}
$$

f) Grouping the first and second terms and the third and fourth terms, the given expression, $ab+b^2+6a+6b
,$ is equivalent to

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

Factoring the $GCF$ in each group results to

$$
begin{align*}
b(a+b)+6(a+b)
.end{align*}
$$

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

$$
begin{align*}
(a+b)(b+6)
.end{align*}
$$

a) $(n-3m)(2+5n)$

b) $(y-3+x)(y-3-x)$

c) $(y-b)(1-y+b)$

d) $2(x^2-4y^2+4x+4)$

e) $(w+b)(w-a)$

f) $(a+b)(b+6)$

We know that $5(5)=25$ and that $5(4)=20$ Therefore: $(5x+2)^2$

$b=sqrt{(5x+2)^2}$

$$
b=5x+2
$$

The area of a square is:

$A=b^2$

$b=sqrt{A}$

Therefore to find the lengths of the bases take the square root of the area

$P=4(5x+2)$

$$
P=20x+8
$$

The equation for the perimeter of a square is:

$$
P=4b
$$

$$
20x+8
$$

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

$$
begin{align*}
3n^2-11n+k
end{align*}
$$
has $ac=
3(k)=3k$ and $b=
-11
.$

Hence, $k$ is an integer such that $3k$ has integer factors whose sum is $-11$.

$k$ is an integer such that $3k$ has integer factors whose sum is $-11$