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Home Page Textbooks Nelson Functions 11 Page 82: Getting Started
Nelson Functions 11
Nelson Functions 11
1st Edition
Chris Kirkpatrick, Marian Small
ISBN: 9780176332037
Textbook solutions

All Solutions

Page 82: Getting Started

Exercise 1
Step 1
1 of 3
The degree of a polynomial is the highest degree of its monomials with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.
Step 2
2 of 3
begin{tabular}{|c| c| c|}
hline
Polynomial & Type & Degree\
hlinehline
$3x^{2}+2x-1$ & trinomial &2 \
$4x-7$ & binomial & 1\
$3$ & monomial & 0\
$7+2x^{2}$ & binomial & 2\
$2xy$ & monomial & 2\
$x^{2}-3xy+y^{2}$ & trinomial & 2\
hline
end{tabular}
Result
3 of 3
begin{table}[]
begin{tabular}{l|l|l|}
cline{2-3}
& Type & Degree \ cline{2-3}
a) & binomial & 1 \ cline{2-3}
b) & monomial & 0 \ cline{2-3}
c) & binomial & 2 \ cline{2-3}
d) & monomial & 2 \ cline{2-3}
e) & trinomial & 2 \ cline{2-3}
end{tabular}
end{table}
Exercise 2
Step 1
1 of 5
a)

$$
begin{align*}
(2x+3)+(7x-5)&=2x+3+7x-5 tag{text{clear the parentheses }}\
&=2x+7x+3-5 tag{text{group the like terms}}\
&=9x-2\
end{align*}
$$

Step 2
2 of 5
b)

$$
begin{align*}
(4x^{2}-7x+1)-(2x^{2}-3x+10)&=4x^{2}-7x+1-2x^{2}+3x-10 tag{text{clear the parentheses and shange sign into second parentheses}}\
&=4x^{2}-2x^{2}-7x+3x+1-10 tag{text{group like terms}}\
&=2x^{2}-4x-9\
end{align*}
$$

Step 3
3 of 5
c)

$$
begin{align*}
(2x-3)(4x+5)&=\
&=8x^{2}+10x-12x-15 tag{text{distribute each term of the first polynomial to every term of the second polynomial. }}\
&=8x^{2}-2x-15
end{align*}
$$

Step 4
4 of 5
d)

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

Result
5 of 5
a) $9x-2$

b) $2x^2-4x-9$

c) $8x^2-2x-15$

d) $4x^2-4x+1$

Exercise 3
Solution 1
Solution 2
Step 1
1 of 3
Factoring is the opposite of expanding.

$textbf{Factoring}$ a polynomial means to write it as a product of two or more polynomials.

$textbf{Expanding}$ a means multiply using the distributive property.

Step 2
2 of 3
For example,

$(3x+1)(2x-3)overset{text{expanding$longrightarrow$ }}= 6x^{2}-7x-3$

and:

$$
(3x+1)(2x-3)underset{text{$longleftarrow$ factoring}}= 6x^{2}-7x-3
$$

Result
3 of 3
check the answer
Step 1
1 of 2
Factor 20. To factor that we get $4*5$ which is a product of two numbers
To factor means to express as a product. An example is shown on the left
Step 2
2 of 2
Expand 3(2x-4). We want to multiply the three through the parenthesis and remove them. This will yield $6x-12$
Expanding means to multiply something through. An example is shown on the left
Exercise 4
Step 1
1 of 7
a) Factoring the $GCF=
2xy^3
,$ the given expression, $6xy^3-8x^2y^3
,$ is equivalent to

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

Step 2
2 of 7
b) Using the factoring of trinomials in the form $x^2+bx+c,$ the
expression

$$
begin{align*}
a^2-7a+10
end{align*}
$$
has $c=
10$ and $b=
-7
.$

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

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

Step 3
3 of 7
c) Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression

$$
begin{align*}
12n^2+7n-10
end{align*}
$$
has $ac=
12(-10)=-120$ and $b=
7
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-8,15
right}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to

$$
begin{align*}
12n^2-8n+15n-10
.end{align*}
$$

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

$$
begin{align*}
(12n^2-8n)+(15n-10)
.end{align*}
$$

Factoring the $GCF$ in each group results to

$$
begin{align*}
4n(3n-2)+5(3n-2)
.end{align*}
$$

Factoring the $GCF=
(3n-2)$ of the entire expression above results to

$$
begin{align*}
(3n-2)(4n+5)
.end{align*}
$$

Step 4
4 of 7
d) Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the given
expression,
$9-25x^2
,$ is equivalent to

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

Step 5
5 of 7
e) Using the factoring of trinomials in the form $x^2+bx+c,$ the
expression

$$
begin{align*}
x^2+5x+8
end{align*}
$$
has $c=
8$ and $b=
5
.$

There are no two integers whose product is $8$ and whose sum is $5.$ Hence, the given expression is not factorable using integers.

Step 6
6 of 7
f) Using the factoring of trinomials in the form $x^2+bx+c,$ the
expression

$$
begin{align*}
y^2-5y-36
end{align*}
$$
has $c=
-36$ and $b=
-5
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-9,4
right}.$ Using these two numbers, the
expression
above is equivalent to

$$
begin{align*}
(y-9)(y+4)
.end{align*}
$$

Result
7 of 7
a) $2xy^3(3-4x)$

b) $(x-2)(x-5)$

c) $(3n-2)(4n+5)$

d) $(3+5x)(3-5x)$

e) not factorable

f) $(y-9)(y+4)$

Exercise 5
Step 1
1 of 5
a) Converting the terms to similar fractions using the $LCD=
12
,$ the given expression, $dfrac{3}{4}+dfrac{1}{6}
,$ simplifies to

$$
begin{align*}
&
dfrac{3}{4}cdotdfrac{3}{3}+dfrac{1}{6}cdotdfrac{2}{2}
\\&=
dfrac{9}{12}+dfrac{2}{12}
\\&=
dfrac{9+2}{12}
\\&=
dfrac{11}{12}
.end{align*}
$$

Step 2
2 of 5
b) Converting the terms to similar fractions using the $LCD=
10
,$ the given expression, $dfrac{-2}{5}-dfrac{1}{10}
,$ simplifies to

$$
begin{align*}
&
dfrac{-2}{5}cdotdfrac{2}{2}-dfrac{1}{10}
\\&=
dfrac{-4}{10}-dfrac{1}{10}
\\&=
dfrac{-4-1}{10}
\\&=
dfrac{-5}{10}
\\&=
-dfrac{1}{2}
.end{align*}
$$

Step 3
3 of 5
c) Cancelling the common factors between the numerator and the denominator, the given expression, $left( dfrac{-12}{25} right)left( dfrac{-10}{9} right)
,$ is equivalent to

$$
begin{align*}
&
left( dfrac{cancel3(-4)}{cancel5(5)} right)left( dfrac{cancel5(-2)}{cancel3(3)} right)
\\&=
left( dfrac{-4}{5} right)left( dfrac{-2}{3} right)
.end{align*}
$$

Multiplying the numerators and the denominators, the expression above is equivalent to

$$
begin{align*}
dfrac{8}{15}
.end{align*}
$$

Step 4
4 of 5
d) Multiplying by the reciprocal of the divisor, the given expression, $left( dfrac{4}{3} right)divleft( dfrac{-2}{15} right)
,$ is equivalent to

$$
begin{align*}
&
left( dfrac{4}{3} right)cdotleft( dfrac{15}{-2} right)
.end{align*}
$$

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

$$
begin{align*}
&
left( dfrac{cancel2(2)}{cancel3} right)cdotleft( dfrac{cancel3(5)}{cancel2(-1)} right)
\\&=
left( dfrac{2}{1} right)left( dfrac{5}{-1} right)
.end{align*}
$$

Multiplying the numerators and the denominators, the expression above is equivalent to

$$
begin{align*}
&
dfrac{10}{-1}
\\&=
-10
.end{align*}
$$

Result
5 of 5
a) $dfrac{11}{12}$

b) $-dfrac{1}{2}$

c) $dfrac{8}{15}$

d) $-10$

Exercise 6
Step 1
1 of 5
a) Cancelling the common factors between the numerator and the denominator, the given expression, $left( dfrac{2x^2}{3} right)left( dfrac{5x^3}{4} right)
,$ is equivalent to

$$
begin{align*}
&
left( dfrac{cancel2x^2}{3} right)left( dfrac{5x^3}{cancel2(2)} right)
\\&=
left( dfrac{x^2}{3} right)left( dfrac{5x^3}{2} right)
.end{align*}
$$

Multiplying the numerators and the denominators, the expression above is equivalent to

$$
begin{align*}
&
dfrac{5x^{2+3}}{6}
\\&=
dfrac{5x^{5}}{6}
.end{align*}
$$

Step 2
2 of 5
b) Multiplying by the reciprocal of the divisor, the given expression, $left( dfrac{3x}{2} right)divleft( dfrac{x^3}{5} right)
,$ is equivalent to

$$
begin{align*}
&
left( dfrac{3x}{2} right)cdotleft( dfrac{5}{x^3} right)
.end{align*}
$$

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

$$
begin{align*}
&
left( dfrac{3cancel{x}}{2} right)cdotleft( dfrac{5}{cancel{x(x^2)}} right)
\\&=
left( dfrac{3}{2} right)left( dfrac{5}{x^2} right)
.end{align*}
$$

Multiplying the numerators and the denominators, the expression above is equivalent to

$$
begin{align*}
&
dfrac{15}{2x^2}
.end{align*}
$$

Step 3
3 of 5
c) Using the Product Rule of the laws of exponents which is given by $x^mcdot x^n=x^{m+n},$ the given expression, $(2x^2y^3)(4xy^2)
,$ is equivalent to

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

Step 4
4 of 5
d) Using the Quotient Rule of the laws of exponents which states that $dfrac{x^m}{x^n}=x^{m-n},$ the given expression, $(25x^5y^3)div(5x^2y)
,$ is equivalent to

$$
begin{align*}
&
dfrac{25}{5}x^{5-2}y^{3-1}
\\&=
5x^{3}y^{2}
.end{align*}
$$

Result
5 of 5
a) $dfrac{5x^{5}}{6}$

b) $dfrac{15}{2x^2}$

c) $8x^3y^5$

d) $5x^{3}y^{2}$

Exercise 7
Step 1
1 of 7
a) For any value of $x$, the given function, $f(x)=x
,$ is defined. Hence, the domain is the set of all real numbers.
Step 2
2 of 7
b) For any value of $x$, the given function, $g(x)=2x^2
,$ is defined. Hence, the domain is the set of all real numbers.
Step 3
3 of 7
c) The radicand of the given function, $m(x)=sqrt{x}
,$ should be nonnegaative. That is

$$
begin{align*}
xge0
.end{align*}
$$

Hence, the domain is ${x|xge0}
.$

Step 4
4 of 7
d) The denominator of the given function, $h(x)=dfrac{1}{x}
,$ should not be zero. That is

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

Hence, the domain is ${x|xne0}
.$

Step 5
5 of 7
e) The denominator of the given function, $f(x)=dfrac{3}{x-4}
,$ should not be zero. That is

$$
begin{align*}
x-4&ne0
\
x&ne4
.end{align*}
$$

Hence, the domain is ${x|xne4}
.$

Step 6
6 of 7
f) The radicand of the given function, $n(x)=sqrt{x+10}
,$ should be nonnegative. That is,

$$
begin{align*}
x+10ge0
\
xge-10
.end{align*}
$$

Hence, the domain is ${x|xge-10}
.$

Result
7 of 7
a) domain: all real numbers

b) domain: all readomain: l numbers

c) domain: ${x|xge0}$

d) domain: ${x|xne0}$

e) domain: ${x|xne4}$

f) domain: ${x|xge-10}$

Exercise 8
Step 1
1 of 1
Definition of Polynomials:

An expression that takes the form $a_0x^n+a_1x^{n-1}+a_2x^{n-2}+…+a_n
,$ where all the exponents are nonnegative integers and all the coefficients are real numbers.

Characteristics:

The exponents are nonnegative integers.

Examples:

$5$,

$x^2+1$,

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

Non-examples:

$dfrac{5}{x}$,

$x^{-5}$,

$sqrt{x}$.

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