Home Page Textbooks Nelson Functions 11 Page 134: Chapter Self-Test

Nelson Functions 11

1st Edition

Chris Kirkpatrick, Marian Small

ISBN: 9780176332037

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Page 134: Chapter Self-Test

Exercise 1

Step 1

1 of 2

Combine similar terms. When subtracting expressions, change the sign of the subtrahend then add.

$a-(b-c+d)=a+(-b+c-d)$

$$
begin{align*}
bold{a);;;} &(-x^2+2x+7)+(2x^2-7x-7)\ &= (-x^2+2x^2)+(2x-7x)+(7-7)\
&=x^2-5x\\
bold{b);;;} &(2m^2-mn+4n^2)-(5m^2-n^2)+(7m^2-2mn)\
&= (2m^2-mn+4n^2)+(-5m^2+n^2)+(7m^2-2mn)\
&=(2m^2-5m^2)+(-mn-2mn)+(4n^2+n^2)\
&=-3m^2-3mn+5n^2end{align*}
$$

Result

2 of 2

a) $x^2-5x$

b) $-3m^2-3mn+5n^2$

Exercise 2

Step 1

1 of 5

a) Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $2(12a-5)(3a-7)
,$ is equivalent to

$$
begin{align*}
&
2[12a(3a)+12a(-7)-5(3a)-5(-7)]
\&=
2[36a^2-84a-15a+35]
\&=
2[36a^2-99a+35]
.end{align*}
$$

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

$$
begin{align*}
&
2(36a^2)+2(-99a)+2(35)
\&=
72a^2-198a+70
.end{align*}
$$

Step 2

2 of 5

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

$$
begin{align*}
&
2x^2y(4xy^2)+2x^2y(5x^2y)-3xy^2(4xy^2)-3xy^2(5x^2y)
\&=
10x^4y^2+(8x^3y^3-15x^3y^3)-12x^2y^4
\&=
10x^4y^2-7x^3y^3-12x^2y^4
.end{align*}
$$

Step 3

3 of 5

c) Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $(4x-1)(5x+2)(x-3)
,$ is equivalent to

$$
begin{align*}
&
[4x(5x)+4x(2)-1(5x)-1(2)](x-3)
\&=
[20x^2+8x-5x-2](x-3)
\&=
[20x^2+3x-2](x-3)
\&=
(20x^2+3x-2)(x-3)
.end{align*}
$$

Using the Distributive Property, the expression above is equivalent to

$$
begin{align*}
&
20x^2(x)+20x^2(-3)+3x(x)+3x(-3)-2(x)-2(-3)
\&=
20x^3-60x^2+3x^2-9x-2x+6
\&=
20x^3+(-60x^2+3x^2)+(-9x-2x)+6
\&=
20x^3-57x^2-11x+6
.end{align*}
$$

Step 4

4 of 5

d) 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, $(3p^2+p-2)^2
,$ is equivalent to

$$
begin{align*}
&
(3p^2)^2+(p)^2+(-2)^2+2(3p^2)(p)+2(3p^2)(-2)+2(p)(-2)
\&=
9p^4+p^2+4+6p^3-12p^2-4p
\&=
9p^4+6p^3+(p^2-12p^2)-4p+4
\&=
9p^4+6p^3-11p^2-4p+4
.end{align*}
$$

Result

5 of 5

a) $72a^2-198a+70$

b) $10x^4y^2-7x^3y^3-12x^2y^4$

c) $20x^3-57x^2-11x+6$

d) $9p^4+6p^3-11p^2-4p+4$

Exercise 3

Step 1

1 of 2

For any value of $a,$ the function, $g(x)=(3x-a)^2$ will never be equal to $f(x)=9x^2+4$ since

$$
begin{align*}
g(x)&=(3x-a)^2
\&=
(3x)^2-2(3x)(a)+(a)^2
\&=
9x^2-(6x)a+a^2
\&ne
9x^2+4
\&=
f(x)
.end{align*}
$$

Result

2 of 2

None; see explanation

Exercise 4

Step 1

1 of 3

$bold{a);}$ We are given the expression for $h(n)$ and we need to evaluate
$h(n+1)-h(n)$

Use the special product $(x+y)^3=x^3+3x^2y+3xy^2+y^3$

$$
begin{align*} h(n)&=(2n+1)^3\&=(2n)^3+3(2n)^2(1)+3(2n)(1^2)+1^3\&=8n^3+12n^2+6n+1\\
h(n+1)&=[2(n+1)+1]^3\&=(2n+2+1)^3\&=(2n+3)^3\&=(2n)^3+3(2n)^2(3)+3(2n)(3^2)+3^3\&=8n^3+36n^2+54n+27end{align*}
$$

Therefore,
$$
begin{align*} h(n+1)-h(n)&=(8n^3+36n^2+54n+27)-(8n^3+12n^2+6n+1\
&=(8n^3+36n^2+54n+27)+(-8n^3-12n^2-6n-1)\
&=(8n^3-8n^3)+(36n^2-12n^2)+(54n-6n)+(27-1)\
&=24n^2+48n+26end{align*}
$$

Step 2

2 of 3

$bold{b);;}$ This is $h(6)-h(5)$ or $h(n+1)-h(n)$ when $n=5$

$$
begin{align*} h(n+1)-h(n)&=24n^2+48n+26\
h(5+1)-h(5)&=24(5^2)+48(5)+26\
&=866end{align*}
$$

Therefore, Clyde feels 866 more pain on day 6 than on day 5.

Result

3 of 3

a) $h(n+1)-h(n)=24n^2+48n+26$

b) $h(6)-h(5)=866$

Exercise 5

Step 1

1 of 7

a) The given expression, $3m(m-1)+2m(1-m)
,$ is equivalent to

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

Factoring the $GCF=
(m-1)
,$ the expression above is equivalent to

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

Step 2

2 of 7

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

$$
begin{align*}
x^2-27x+72
end{align*}
$$
has $c=
72$ and $b=
-27
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-3,-24
right}.$ Using these two numbers, the
expression
above is equivalent to

$$
begin{align*}
(x-3)(x-24)
.end{align*}
$$

Step 3

3 of 7

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

$$
begin{align*}
15x^2-7xy-2y^2
end{align*}
$$
has $ac=
15(-2)=-30$ and $b=
-7
.$

The two numbers with a product of $c$ and a sum of $b$ are $left{
-10,3
right}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to

$$
begin{align*}
15x^2-10xy+3xy-2y^2
.end{align*}
$$

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

$$
begin{align*}
(15x^2-10xy)+(3xy-2y^2)
.end{align*}
$$

Factoring the $GCF$ in each group results to

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

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

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

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,
$(2x-y+1)^2-(x-y-2)^2
,$ is equivalent to

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

Step 5

5 of 7

e) Grouping the first and second terms and the third and fourth terms, the given expression, $5xy-10x-3y+6
,$ is equivalent to

$$
begin{align*}
(5xy-10x)-(3y-6)
.end{align*}
$$

Factoring the $GCF$ in each group results to

$$
begin{align*}
5x(y-2)-3(y-2)
.end{align*}
$$

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

$$
begin{align*}
(y-2)(5x-3)
.end{align*}
$$

Step 6

6 of 7

f) Grouping the last three terms, the given expression, $p^2-m^2+6m-9
,$ is equivalent to

$$
begin{align*}
p^2-(m^2-6m+9)
.end{align*}
$$

Since the last three terms form a perfect square trinomial, the expression above is equivalent to

$$
begin{align*}
p^2-(m-3)^2
.end{align*}
$$

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

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

Result

7 of 7

a) $m(m-1)$

b) $(x-3)(x-24)$

c) $(3x-2y)(5x+y)$

d) $(3x-2y-1)(x+3)$

e) $(y-2)(5x-3)$

f) $(p+m-3)(p-m+3)$

Exercise 6

Step 1

1 of 2

Grouping the first and second terms and the third and fourth terms, the given equation, $y=x^3-4x^2-x+4
,$ is equivalent to

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

Factoring the $GCF,$ the equation above is equivalent to

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

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

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

Since the $x$-intercept is the value of $x$ when $y$ is zero, then

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

Equating each factor to zero (Zero Product Property) and then solving for $x,$ then the $x$-intercepts are ${ -1,1,4 }
.$

Result

2 of 2

$x$-intercepts: ${ -1,1,4 }$

Exercise 7

Step 1

1 of 5

a) Multiplying by the reciprocal of the divisor, the given expression, $dfrac{4a^2b}{5ab^3}divdfrac{6a^2b}{35ab}
,$ is equivalent to

$$
begin{align*}
&
dfrac{4a^2b}
{5ab^3}cdotdfrac{35ab}{6a^2b}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{cancel{2}cdot2cancel{a^2b}}
{cancel{5}cancel{a}cancel{b}(b^2)}cdotdfrac{cancel{5}(7)cancel{a}cancel{b}}{cancel{2}cdot3cancel{a^2b}}
\&=
dfrac{14}{3b^2}
.end{align*}
$$

Since the denominator cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{14}{3b^2}
text{, }ane0text{, }bne0
.$

Step 2

2 of 5

b) The factored form of the given expression, $dfrac{x-2}{x^2-x-12}timesdfrac{2x-8}{x^2-4x+4}
,$ is

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

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

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

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

Step 3

3 of 5

c) The factored form of the given expression, $dfrac{5}{t^2-7t-18}+dfrac{6}{t+2}
,$ is

$$
begin{align*}
&
dfrac{5}{(t-9)(t+2)}+dfrac{6}{t+2}
.end{align*}
$$

Multiplying the fractions by an expression equal to $1$ which will make the denominators equal to the $LCD=
(t-9)(t+2)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{5}{(t-9)(t+2)}+dfrac{6}{t+2}cdotdfrac{t-9}{t-9}
\&=
dfrac{5}{(t-9)(t+2)}+dfrac{6t-54}{(t-9)(t+2)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{5+6t-54}{(t-9)(t+2)}
\\&=
dfrac{6t-49}{(t-9)(t+2)}
.end{align*}
$$

Since the denominator cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{6t-49}{(t-9)(t+2)}
text{, }tne{ -2,9 }
.$

Step 4

4 of 5

d) The factored form of the given expression, $dfrac{4x}{6x^2+13x+6}-dfrac{3x}{4x^2-9}
,$ is

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

Multiplying the fractions by an expression equal to $1$ which will make the denominators equal to the $LCD=
(3x+2)(2x+3)(2x-3)
,$ the expression above is equivalent to

$$
begin{align*}
&
dfrac{4x}{(3x+2)(2x+3)}cdotdfrac{2x-3}{2x-3}-dfrac{3x}{(2x+3)(2x-3)}cdotdfrac{3x+2}{3x+2}
\&=
dfrac{8x^2-12x}{(3x+2)(2x+3)(2x-3)}-dfrac{9x^2+6x}{(3x+2)(2x+3)(2x-3)}
.end{align*}
$$

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

$$
begin{align*}
&
dfrac{8x^2-12x-9x^2-6x}{(3x+2)(2x+3)(2x-3)}
\\&=
dfrac{-x^2-18x}{(3x+2)(2x+3)(2x-3)}
\\&=
dfrac{-x(x+18)}{(3x+2)(2x+3)(2x-3)}
.end{align*}
$$

Since the denominator cannot be equal to zero, then the simplified form and the restrictions of the given expression are $dfrac{-x(x+18)}{(3x+2)(2x+3)(2x-3)}
text{, }xneleft{ -dfrac{3}{2},-dfrac{2}{3},dfrac{3}{2} right}
.$

Result

5 of 5

a) $dfrac{14}{3b^2}
text{, }ane0text{, }bne0$

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

c) $dfrac{6t-49}{(t-9)(t+2)}
text{, }tne{ -2,9 }$

d) $dfrac{-x(x+18)}{(3x+2)(2x+3)(2x-3)}
text{, }xneleft{ -dfrac{3}{2},-dfrac{2}{3},dfrac{3}{2} right}$

Exercise 8

Step 1

1 of 2

The functions are equivalent because both functions simply to the same expression. You could further proof this statement is correct by graphing the two functions. Since they are the same they would lie on top of one another and you would only see one curve.

Result

2 of 2

Yes

Exercise 9

Step 1

1 of 2

Let $a,$ $a+1,$ and $a+2$ be the consecutive numbers. Then,

$$
begin{align*}
&
dfrac{1}{a}+dfrac{1}{a+1}+dfrac{1}{a+2}
\\&=
dfrac{(a+1)(a+2)+a(a+2)+a(a+1)}{a(a+1)(a+2)}
\\&=
dfrac{(a^2+3a+2)+(a^2+2a)+(a^2+a)}{a(a+1)(a+2)}
\\&=
dfrac{3a^2+6a+2}{a(a+1)(a+2)}
text{ (Expression $1$)}
.end{align*}
$$

Three times the product of the first and third denominators, plus $2$ results to the expression

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

Since the numerator of Expression $1$ is the same as Expression $2,$ and the denominator of the sum is the product of the three denominators, then Roman’s conjecture is true.

Result

2 of 2

TRUE

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All Solutions

Page 134: Chapter Self-Test

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