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Simplify each expression, assuming it is defined. State the implicit domain of the related function(s) before and after the simplification. Determine if the related functions are identical everywhere -- or if that's not the case, where the behavior of the related functions differ.
$\displaystyle{\frac{3x^2 - 12}{x^2 + 4x + 4} \cdot \frac{x+2}{x-2}}$
$\displaystyle{\frac{x^2 - 4x - 21}{x^2 -9x + 18} \div \frac{x^2 - 49}{x-3}}$
$\displaystyle{\frac{4b}{b^2+6b+5} + \frac{2b}{b^2-1}}$
$\displaystyle{\frac{2}{2x^2 -5x -3} \, - \, \frac{1}{3x^2 -10x + 3}}$
$\displaystyle{\frac{x}{x^2+9x+20} - \frac{4}{x^2+7x+12}}$
$\displaystyle{\frac{x}{x^2-x-6} - \frac{1}{x+2} + \frac{2}{3-x}}$
$\displaystyle{\frac{1 - \displaystyle{\frac{6}{x} + \frac{5}{x^2}}}{\displaystyle{\frac{1}{x} - \frac{5}{x^2}}}}$
$\displaystyle{\frac{\displaystyle{\frac{x}{1+x}} + \displaystyle{\frac{2}{x}}}{\displaystyle{\frac{1}{x}} - \displaystyle{\frac{1}{x+1}}}}$
$\displaystyle{\frac{\displaystyle{\frac{5}{m}} - \displaystyle{\frac{2}{m+1}}}{\displaystyle{\frac{3}{m+1}} + \displaystyle{\frac{1}{m}}}}$
$\displaystyle{\frac{4x^3(x^2+4)^2 - x^4 \cdot 2 (x^2 + 4) \cdot 2x}{[(x^2+4)^2]^2}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{3(x^2-4)}{(x+2)^2} \cdot \frac{x+2}{x-2}}\\\\ &=& \displaystyle{\frac{3(x-2)(x+2)}{(x+2)^2} \cdot \frac{x+2}{x-2}}\\\\ &=& \displaystyle{\frac{(x-2)(x+2)^2}{(x-2)(x+2)^2} \cdot 3}\\\\ &=& \fbox{$3$} \end{array}}}$
Implicit domain before: $\{x \in \mathbb{R} \ | \ x \neq \pm 2\}$
Implicit domain after: $\mathbb{R}$
The behavior of the related functions differ at $x = \pm 2$.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{x^2 - 4x - 21}{x^2 -9x + 18} \cdot \frac{x-3}{x^2 - 49}}\\\\ &=& \displaystyle{\frac{(x-7)(x+3)}{(x-6)(x-3)} \cdot \frac{x-3}{(x-7)(x+7)}}\\\\ &=& \displaystyle{\frac{(x-7)(x-3)}{(x-7)(x-3)} \cdot \frac{x+3}{(x-6)(x+7)}}\\\\ &=& \fbox{$\displaystyle{\frac{x+3}{x^2+x-42}}$} \end{array}}}$
Implicit domain before: $\{x \in \mathbb{R} \ | \ x \neq -7, 3, 6, \textrm{ or } 7\}$
Implicit domain after: $\{x \in \mathbb{R} \ | \ x \neq -7,6\}$
The behavior of the related functions differ at $x = 3$ and $x = 7$.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{ \frac{4b}{(b+1)(b+5)} + \frac{2b}{(b+1)(b-1)}}\\\\ &=& \displaystyle{ \frac{4b(b-1)}{(b+1)(b+5)(b-1)} + \frac{2b(b+5)}{(b+1)(b-1)(b+5)}}\\\\ &=& \displaystyle{ \frac{(4b^2 - 4b) + (2b^2+10b)}{(b+1)(b+5)(b-1)}}\\\\ &=& \displaystyle{ \frac{6b^2 + 6b}{(b+1)(b+5)(b-1)}}\\\\ &=& \displaystyle{ \frac{6b(b+1)}{(b+1)(b+5)(b-1)}}\\\\ &=& \displaystyle{ \frac{b+1}{b+1} \cdot \frac{6b}{(b+5)(b-1)}}\\\\ &=& \fbox{$\displaystyle{\frac{6b}{b^2+4b-5}}$} \end{array}}}$
Implicit domain before: $\{b \in \mathbb{R} \ | \ b \neq -5, -1, \textrm{ or } 1\}$
Implicit domain after: $\{b \in \mathbb{R} \ | \ b \neq -5, 1\}$
The behavior of the related functions differ at $x = -1$.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{2}{(2x+1)(x-3)} \, - \, \frac{1}{(3x-1)(x-3)}}\\\\ &=& \displaystyle{\frac{2(3x-1)}{(2x+1)(x-3)(3x-1)} \, - \, \frac{2x+1}{(3x-1)(x-3)(2x+1)}}\\\\ &=& \displaystyle{\frac{(6x-2) - (2x+1)}{(x-3)(2x+1)(3x-1)}}\\\\ &=& \fbox{$\displaystyle{\frac{4x-3}{(x-3)(2x+1)(3x-1)}}$} \end{array}}}$
Implicit domain before: $\{x \in \mathbb{R} \ | \ x \neq \frac{1}{3}, -\frac{1}{2}, 3\}$
Implicit domain after: $\{x \in \mathbb{R} \ | \ x \neq \frac{1}{3}, -\frac{1}{2}, 3\}$
The related functions share the same behavior everywhere. They are identical.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{x}{(x+4)(x+5)} - \frac{4}{(x+3)(x+4)}}\\\\ &=& \displaystyle{\frac{x(x+3)}{(x+4)(x+5)(x+3)} - \frac{4(x+5)}{(x+3)(x+4)(x+5)}}\\\\ &=& \displaystyle{\frac{(x^2+3x) - (4x+20)}{(x+3)(x+4)(x+5)}}\\\\ &=& \displaystyle{\frac{x^2-x-20}{(x+3)(x+4)(x+5)}}\\\\ &=& \displaystyle{\frac{(x-5)(x+4)}{(x+3)(x+4)(x+5)}}\\\\ &=& \displaystyle{\frac{x+4}{x+4} \cdot \frac{x-5}{(x+3)(x+5)}}\\\\ &=& \fbox{$\displaystyle{\frac{x-5}{(x+3)(x+5)}}$} \end{array}}}$
Implicit domain before: $\{x \in \mathbb{R} \ | \ x \neq -5,-4,-3\}$
Implicit domain after: $\{x \in \mathbb{R} \ | \ x \neq -5,-3\}$
The behavior of the related functions differ at $x=-4$.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{x}{(x-3)(x+2)} - \frac{1}{x+2} - \frac{2}{x-3}}\\\\ &=& \displaystyle{\frac{x}{(x-3)(x+2)} - \frac{x-3}{(x+2)(x-3)} - \frac{2(x+2)}{(x-3)(x+2)}}\\\\ &=& \displaystyle{\frac{x - (x-3) - 2(x+2)}{(x-3)(x+2)}}\\\\ &=& \displaystyle{\frac{x - x + 3 - 2x - 4}{(x-3)(x+2)}}\\\\ &=& \displaystyle{\frac{-2x-1}{(x-3)(x+2)}}\\\\ &=& \fbox{$\displaystyle{\frac{-2x-1}{x^2-x-6}}$} \end{array}}}$
Implicit domain before: $\{x \in \mathbb{R} \ | \ x \neq -2, 3\}$
Implicit domain after: $\{x \in \mathbb{R} \ | \ x \neq -2, 3\}$
The related functions share the same behavior everywhere. They are identical.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{\displaystyle{ \left( 1 - \frac{6}{x} + \frac{5}{x^2} \right)}}{\displaystyle{\left( \frac{1}{x} - \frac{5}{x^2} \right)}} \cdot \frac{x^2}{x^2}}\\\\ &=& \displaystyle{\frac{x^2 - 6x + 5}{x-5}}\\\\ &=& \displaystyle{\frac{(x-5)(x-1)}{x-5}}\\\\ &=& \displaystyle{\frac{x-5}{x-5} \cdot (x-1)}\\\\ &=& \fbox{$\displaystyle{x-1}$} \end{array}}}$
Implicit domain before: $\{x \in \mathbb{R} \ | \ x \neq 0,5\}$; Implicit domain after: $\mathbb{R}$
Note that in determining the implicit domain before simplification, we need to check where the overall denominator might be zero. Combining that overall denominator into a single fraction by getting common denominators first will help towards that end: $$\frac{1}{x} - \frac{5}{x^2} = \frac{1}{x} \cdot \frac{x}{x} - \frac{5}{x^2} = \frac{x-5}{x}$$ When a fraction is zero, its numerator must be zero. This happens when $x=5$ above. This is why $x=5$ is excluded from the implicit domain of the related function before simplification.
The behavior of the related functions differ at $x=0$ and $x=5$.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{ \displaystyle{\left( \frac{x}{1+x} + \frac{2}{x} \right) }}{\displaystyle{ \left( \frac{1}{x} - \frac{1}{x+1} \right)}} \cdot \frac{x(x+1)}{x(x+1)}}\\\\ &=& \displaystyle{\frac{x^2 + 2(x+1)}{(x+1) - x}}\\\\ &=& \displaystyle{\frac{x^2 + 2x + 2}{1}}\\\\ &=& \fbox{$\displaystyle{x^2+2x+2}$} \end{array}}}$
Implicit domain before: $\{x \in \mathbb{R} \ | \ x \neq -1, 0\}$. Implicit domain after: $\mathbb{R}$
Even though in this particular problem it won't end up restricting the domain -- when determining the implicit domain before simplification, we must check to see where the overall denominator might be zero. We can do this more easily if we express that overall denominator as a single fraction, which requires getting common denominators: $$\frac{1}{x} - \frac{1}{x+1} = \frac{1}{x} \cdot \frac{x+1}{x+1} - \frac{1}{x+1} \cdot \frac{x}{x} = \frac{(x+1) - x}{x(x+1)} = \frac{1}{x(x+1)}$$ As a fraction equaling zero must have zero in its numerator, we see here the overall denominator can never be zero.
The behavior of the related functions differ at $x=-1$ and $x=0$.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{\displaystyle{\left( \frac{5}{m} - \frac{2}{m+1} \right)}}{\displaystyle{\left( \frac{3}{m+1} + \frac{1}{m} \right)}} \cdot \frac{m(m+1)}{m(m+1)}}\\\\ &=& \displaystyle{\frac{5(m+1) - 2m}{3m + (m+1)}}\\\\ &=& \displaystyle{\frac{5m + 5 - 2m}{3m + m + 1}}\\\\ &=& \fbox{$\displaystyle{\frac{3m+5}{4m+1}}$} \end{array}}}$
Implicit domain before: $\{m \in \mathbb{R} \ | \ m \neq -1, -\frac{1}{4}, \textrm{ or } 0\}$
Implicit domain after: $\{m \in \mathbb{R} \ | \ m \neq -\frac{1}{4}\}$
Note that in determining the implicit domain before simplification, we must check to see where the overall denominator might be zero. We can do this more easily if we express that denominator as a single fraction, which requires getting common denominators:
$$\frac{3}{m+1} + \frac{1}{m} = \frac{3}{m+1} \cdot \frac{m}{m} + \frac{1}{m} \cdot \frac{m+1}{m+1} = \frac{3m + (m+1)}{m(m+1)} = \frac{4m+1}{m(m+1)}$$ As a fraction equal to zero must have zero in its numerator, and noting this happens above when $4m+1=0$ and consequently when $m=-\frac{1}{4}$, we exclude this value of $m$ from the implicit domain of the related function before simplification.The behavior of the related functions differ at $m=0$ and $m=-1$.
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{4x^3(x^2+4)[(x^2+4)-x^2]}{(x^2+4)^4}}\\\\ &=& \displaystyle{\frac{x^2+4}{x^2+4} \cdot \frac{4x^3(x^2 + 4 - x^2)}{(x^2+4)^3}}\\\\ &=& \displaystyle{\frac{4x^3(4)}{(x^2+4)^3}}\\\\ &=& \fbox{$\displaystyle{\frac{16x^3}{(x^2+4)^3}}$} \end{array}}}$
Implicit domain before: $\mathbb{R}$
Implicit domain after: $\mathbb{R}$
The related functions share the same behavior everywhere. They are identical.
Simplify each expression, assuming it is defined.
$\displaystyle{\frac{x^2+3xy}{2x^3-x^2y} \cdot \frac{4x^2-y^2}{x^2+6xy+9y^2}}$
$\displaystyle{\frac{x^2 + 3xy}{2x^3 - x^2 y} \, \div \, \frac{x^2 + 6xy + 9y^2}{4x^2 - y^2}}$
$\displaystyle{\frac{\displaystyle{\frac{x^2}{y}} + \displaystyle{\frac{y^2}{x}}}{y^2 - xy + x^2}}$
$\displaystyle{\frac{b-a^{-1}}{a - b^{-1}}}$
$\displaystyle{\frac{x^{-2} - y^{-2}}{x^{-1} + \displaystyle{\frac{1}{y}}}}$
$\displaystyle{\frac{x^3-y^3}{x^2+xy+y^2} \cdot \frac{x+y}{x^2-y^2}}$
$\displaystyle{\frac{2x}{x-y} + \frac{3y}{y-x}}$
$\displaystyle{\frac{\displaystyle{\frac{x-2}{x}+\frac{x}{x+2}}}{\displaystyle{\frac{x-2}{x+2}+\frac{1}{x}}}}$
$\displaystyle{\frac{\displaystyle{\frac{x}{y}+\frac{y}{x}}}{\displaystyle{\frac{x^2}{y^2}-\frac{y^2}{x^2}}}}$
$\displaystyle{\frac{x^2-9}{2x+3} \div \frac{x+3}{2x^2-3x-9}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{x(x+3y)}{x^2(2x-y)} \cdot \frac{(2x+y)(2x-y)}{(x+3y)^2}}\\\\ &=& \displaystyle{\frac{x(x+3y)(2x-y)}{x(x+3y)(2x-y)} \cdot \frac{2x+y}{x(x+3y)}}\\\\ &=& \fbox{$\displaystyle{\frac{2x+y}{x(x+3y)}}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{x^2 + 3xy}{2x^3 - x^2 y} \, \cdot \, \frac{4x^2 - y^2}{x^2 + 6xy + 9y^2}}\\\\ &=& \displaystyle{\frac{x(x+3y)}{x^2(2x-y)} \, \cdot \, \frac{(2x+y)(2x-y)}{(x+3y)^2}}\\\\ &=& \displaystyle{\frac{x(2x-y)(x+3y)}{x(2x-y)(x+3y)} \cdot \frac{2x+y}{x(x+3y)}}\\\\ &=& \fbox{$\displaystyle{\frac{2x+y}{x(x+3y)}}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{\displaystyle{\left( \frac{x^2}{y} + \frac{y^2}{x} \right)}}{\displaystyle{y^2 - xy + x^2}} \cdot \frac{xy}{xy}}\\\\ &=& \displaystyle{\frac{x^3 + y^3}{(y^2 - xy + x^2)xy}}\\\\ &=& \displaystyle{\frac{(x+y)(x^2-xy+y^2)}{xy(x^2-xy+y^2)}}\\\\ &=& \displaystyle{\frac{x+y}{xy} \cdot \frac{x^2-xy+y^2}{x^2-xy+y^2} }\\\\ &=& \fbox{$\displaystyle{\frac{x+y}{xy}}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{(b-a^{-1})}{(a - b^{-1})} \cdot \frac{ab}{ab}}\\\\ &=& \displaystyle{\frac{ab^2 - b}{a^2b - a}}\\\\ &=& \displaystyle{\frac{b(ab - 1)}{a(ab - 1)}}\\\\ &=& \displaystyle{\frac{b}{a} \cdot \frac{ab-1}{ab-1}}\\\\ &=& \fbox{$\displaystyle{\frac{b}{a}}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{(x^{-2} - y^{-2})}{\displaystyle{\left( x^{-1} + \frac{1}{y} \right)}} \cdot \frac{x^2 y^2}{x^2 y^2}}\\\\ &=& \displaystyle{\frac{y^2 - x^2}{xy^2 + x^2y}}\\\\ &=& \displaystyle{\frac{(y+x)(y-x)}{xy(y+x)}}\\\\ &=& \displaystyle{\frac{y+x}{y+x} \cdot \frac{y-x}{xy}}\\\\ &=& \fbox{$\displaystyle{\frac{y-x}{xy}}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{(x-y)(x^2+xy+y^2)}{x^2+xy+y^2} \cdot \frac{x+y}{(x+y)(x-y)}}\\\\ &=& \displaystyle{\frac{(x-y)(x^2+xy+y^2)(x+y)}{(x-y)(x^2+xy+y^2)(x+y)}}\\\\ &=& \fbox{$\displaystyle{1}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{2x}{x-y} - \frac{3y}{x-y}}\\\\ &=& \fbox{$\displaystyle{\frac{2x-3y}{x-y}}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{\displaystyle{\left( \frac{x-2}{x}+\frac{x}{x+2} \right)}}{\displaystyle{\left( \frac{x-2}{x+2}+\frac{1}{x} \right)}} \cdot \frac{x(x+2)}{x(x+2)}}\\\\ &=& \displaystyle{\frac{(x-2)(x+2) + x^2}{x(x-2)+(x+2)}}\\\\ &=& \displaystyle{\frac{(x^2-4) + x^2}{x^2-2x+x+2}}\\\\ &=& \displaystyle{\frac{2x^2-4}{x^2-2x+x+2}}\\\\ &=& \fbox{$\displaystyle{\frac{2x^2-4}{x^2-x+2}}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{\displaystyle{\left( \frac{x}{y}+\frac{y}{x} \right)}}{\displaystyle{\left( \frac{x^2}{y^2}-\frac{y^2}{x^2} \right)}} \cdot \frac{x^2 y^2}{x^2 y^2}}\\\\ &=& \displaystyle{\frac{x^3 y+xy^3}{x^4 - y^4}}\\\\ &=& \displaystyle{\frac{xy(x^2+y^2)}{(x^2+y^2)(x^2-y^2)}}\\\\ &=& \displaystyle{\frac{x^2+y^2}{x^2+y^2} \cdot \frac{xy}{x^2-y^2}}\\\\ &=& \fbox{$\displaystyle{\frac{xy}{x^2-y^2}}$} \end{array}}}$
$\displaystyle{{ \begin{array}[t]{rcl} &=& \displaystyle{\frac{x^2-9}{2x+3} \cdot \frac{2x^2-3x-9}{x+3}}\\\\ &=& \displaystyle{\frac{(x+3)(x-3)}{2x+3} \cdot \frac{(2x+3)(x-3)}{x+3}}\\\\ &=& \displaystyle{\frac{(x+3)(2x+3)}{(x+3)(2x+3)} \cdot (x-3)^2}\\\\ &=& \fbox{$\displaystyle{(x-3)^2}$} \end{array}}}$
The following is an important expression encountered in calculus, called the difference quotient. $$\frac{f(x+h)-f(x)}{h}$$
For each function $f$, write the expression for the corresponding difference quotient at the given value of $x$. What value of $h$ causes this difference quotient to be undefined? Under an assumption that we avoid using this value of $h$, simplify the difference quotient. What is the value of the simplified expression at that same value of $h$ that previously caused a problem?
$f(x) = x^2; \quad x = 5$
$f(x) = x^3; \quad x = 2$
$f(x) = \cfrac{1}{x}; \quad x = 2$
$f(x) = x^2 + x; \quad x = x_0$
The difference quotient $\displaystyle{\frac{(5+h)^2 - 25}{h}}$ is undefined at $h=0$.
However, if $h \neq 0$, then $\displaystyle{ \begin{array}[t]{rcl} \displaystyle{\frac{(5+h)^2 - 25}{h}} &=& \displaystyle{\frac{(25+10h+h^2) - 25}{h}}\\\\ &=& \displaystyle{\frac{10h+h^2}{h}}\\\\ &=& \displaystyle{\frac{h(10+h)}{h}}\\\\ &=& 10+h \end{array} }$
Notice $10+h = 10$ when $h=0$.
The difference quotient $\displaystyle{\frac{(2+h)^3 - 8}{h}}$ is undefined at $h=0$.
However, if $h \neq 0$, then $\displaystyle{ \begin{array}[t]{rcl} \displaystyle{\frac{(2+h)^3 - 8}{h}} &=& \displaystyle{\frac{(8 + 12h + 6h^2 + h^3) - 8}{h}}\\\\ &=& \displaystyle{\frac{12h+6h^2+h^3}{h}}\\\\ &=& \displaystyle{\frac{h(12+6h+h^2)}{h}}\\\\ &=& 12+6h+h^2 \end{array} }$
Notice $12+6h+h^2=12$ when $h=0$.
The difference quotient $\displaystyle{\frac{\cfrac{1}{2+h} - \cfrac{1}{2}}{h}}$ is undefined at $h=0$.
However, if $h \neq 0$, then $\displaystyle{ \begin{array}[t]{rcl} \displaystyle{\frac{\cfrac{1}{2+h} - \cfrac{1}{2}}{h}} &=& \displaystyle{\frac{\cfrac{1}{2+h} - \cfrac{1}{2}}{h} \cdot \frac{2(2+h)}{2(2+h)}}\\\\ &=& \displaystyle{\frac{2 - (2+h)}{2h(2+h)}}\\ &=& \cfrac{-h}{2h(2+h)}\\ &=& \cfrac{-1}{4+2h} \end{array} }$
Notice $\cfrac{1}{4+2h} = \cfrac{1}{4}$ when $h=0$.
The difference quotient $\displaystyle{\frac{[(x_0 + h)^2 + (x_0 + h)] - (x_0^2 + x_0)}{h}}$ is undefined at $h=0$.
However, if $h \neq 0$, then
$\displaystyle{ \begin{array}[t]{rcl} \displaystyle{\frac{[(x_0 + h)^2 + (x_0 + h)] - (x_0^2 + x_0)}{h}} &=& \displaystyle{\frac{[x_0^2 + 2x_0h + h^2 + x_0 + h] - (x_0^2 + x_0)}{h}}\\\\ &=& \displaystyle{\frac{x_0^2 + 2x_0h + h^2 + x_0 + h - x_0^2 - x_0}{h}}\\\\ &=& \displaystyle{\frac{2x_0h + h^2 + h}{h}}\\\\ &=& \displaystyle{\frac{h(2x_0 + h + 1)}{h}}\\\\ &=& \displaystyle{2x_0 + h + 1} \end{array}}$
Notice $2x_0 + h + 1 = 2x_0 + 1$ when $h=0$.
Use the Euclidean Algorithm to find the greatest common divisor for each pair given
$2783$ and $437$
$1143$ and $693$
$1746$ and $1261$
$610$ and $377$
$\displaystyle{ \begin{array}[t]{rcl} 2783 &=& 6 \cdot 437 + 161\\ 437 &=& 2 \cdot 161 + 115\\ 161 &=& 1 \cdot 115 + 46\\ 115 &=& 2 \cdot 46 + \fbox{23}\\ 46 &=& 2 \cdot 23 + 0 \end{array}}$
So the greatest common divisor of $2783$ and $437$ is $23$.
$\displaystyle{ \begin{array}[t]{rcl} 1143 &=& 1 \cdot 693 + 450\\ 693 &=& 1 \cdot 450 + 243\\ 450 &=& 1 \cdot 243 + 207\\ 243 &=& 1 \cdot 207 + 36\\ 207 &=& 5 \cdot 36 + 27\\ 36 &=& 1 \cdot 27 + \fbox{9}\\ 27 &=& 3 \cdot 9 + 0 \end{array}}$
So the greatest common divisor of $1143$ and $693$ is $9$.
$\displaystyle{ \begin{array}[t]{rcl} 1746 &=& 1 \cdot 1261 + 485\\ 1261 &=& 2 \cdot 485 + 291\\ 485 &=& 1 \cdot 291 + 194\\ 291 &=& 1 \cdot 194 + \fbox{97}\\ 194 &=& 2 \cdot 97 + 0 \end{array}}$
So the greatest common divisor of $1746$ and $1261$ is $97$.
$\displaystyle{ \begin{array}[t]{rcl} 610 &=& 1 \cdot 377 + 233\\ 377 &=& 1 \cdot 233 + 144\\ 233 &=& 1 \cdot 144 + 89\\ 144 &=& 1 \cdot 89 + 55\\ 89 &=& 1 \cdot 55 + 34\\ 55 &=& 1 \cdot 34 + 21\\ 34 &=& 1 \cdot 21 + 13\\ 21 &=& 1 \cdot 13 + 8\\ 13 &=& 1 \cdot 8 + 5\\ 8 &=& 1 \cdot 5 + 3\\ 5 &=& 1 \cdot 3 + 2\\ 3 &=& 1 \cdot 2 + \fbox{1}\\ 2 &=& 2 \cdot 1 + 0 \end{array}}$
So the greatest common divisor of $610$ and $377$ is $1$ (i.e., they share no common factor other than $1$).
Simplify the following, assuming it is defined: $$\frac{x^6 - 2x^5 - 24x^4 + 56x^3 - 15x^2 - 45x + 30}{x^6 + x^5 - 7x^4 + 46x^3 - 90x^2 + 66x - 12}$$ You may find the following results useful towards this end. (Of course, you are welcome to replicate these calculations on your own via long division if you wish -- but hopefully this saves you the tedium of doing so.)
$(x^6 - 2x^5 - 24x^4 + 56x^3 - 15x^2 - 45x + 30)\\ \hspace{0.1cm} = 1 \cdot (x^6 + x^5 - 7x^4 + 46x^3 - 90x^2 + 66x - 12) + (-3x^5 - 17x^4 + 10x^3 + 75x^2 - 111x + 42)$
$(x^6 + x^5 - 7x^4 + 46x^3 - 90x^2 + 66x - 12)\\ \hspace{0.1cm} = (-\frac{1}{3}x + \frac{14}{9})(-3x^5 - 17x^4 + 10x^3 + 75x^2 - 111x + 42) + (\frac{205}{9}x^4 + \frac{499}{9}x^3 - \frac{731}{3}x^2 + \frac{758}{3}x - \frac{232}{3})$
$(-3x^5 - 17x^4 + 10x^3 + 75x^2 - 111x + 42)\\ \hspace{0.1cm} = (-\frac{27}{205}x-\frac{17892}{42025})(\frac{205}{9}x^4 + \frac{499}{9}x^3 - \frac{731}{3}x^2 + \frac{758}{3}x - \frac{232}{3}) + (\frac{63567}{42025}x^3 + \frac{190701}{42025}x^2 - \frac{572103}{42025}x + \frac{381402}{42025})$
$(\frac{205}{9}x^4 + \frac{499}{9}x^3 - \frac{731}{3}x^2 + \frac{758}{3}x - \frac{232}{3})\\ \hspace{0.1cm} = (\frac{8615125}{572103}x - \frac{4874900}{572103})(\frac{63567}{42025}x^3 + \frac{190701}{42025}x^2 - \frac{572103}{42025}x + \frac{381402}{42025}) + 0$
We need to find (and cancel) the greatest common divisor of the polynomials in the numerator and denominator. However, with both being $6^{th}$ degree polynomials this would be hard were it not for the Euclidean Algorithm -- whose calculations are shown. These reveal that the polynomial below is (a multiple of) the greatest common divisor of both the numerator and denominator.
$(\frac{63567}{42025}x^3 + \frac{190701}{42025}x^2 - \frac{572103}{42025}x + \frac{381402}{42025})$
Factoring out the leading coefficient reveals the simpler form below that will save us some time (although this is not strictly necessary).
$\frac{63567}{42025}(x^3 + 3x^2 - 9x + 6)$
Then, we divide both numerator and denominator of the original rational expression by this common factor. To be efficient, we might notice we could divide by the $\frac{63567}{42025}$ first, and then by the $(x^3 + 3x^2 - 9x + 6)$. However, doing that first division only places a common factor of $\frac{42025}{63567}$ in both the numerator and denominator, which then promptly cancels. As such, we can ignore dividing numerator and denominator by $\frac{63567}{42025}$ altogether.
Dividing both numerator and denominator by $(x^3 + 3x^2 - 9x + 6)$ factor is more routine. These divisions produce quotients of $x^3 - 5x^2 + 5$ and $x^3 -2x^2 + 8x - 2$, respectively. As such, the simplified form we seek is $$\frac{x^3 - 5x^2 + 5}{x^3 -2x^2 + 8x - 2}$$
Argue why the set $\{A,B,C,D\}$ together with an "addition" and "subtraction" defined by the following tables forms a field. You may assume associativity holds for both of the operations shown, if it helps!
It's a strange idea -- but consider polynomials of degree $1$, but where the only possible coefficients were $0$ and $1$ and they worked like a $2$-hour clock. This gives you $4$ possible such polynomials: $0$, $1$, $x$, and $x+1$. With the $2$-hour clock arithmetic for the coefficients, we'll see some "normal" sums like $0 + (x+1) = (x+1)$ and some not-so-normal sums like $x + (x+1) = 1$ and $1 + (x+1) = x$. As for "multiplication", define it in this way: Multiply the polynomials like normal -- but if a quadratic is produces, replace $x^2$ with $x+1$ and simplify. So for example, $$x(x+1) = x^2 + x = (x+1) + x = 1$$ $$(x+1)(x+1) = x^2 + 2x + 1 = x^2 + 1 = (x+1) + 1 = x$$
Can you find a way to pair up $0$, $1$, $x$ and $x+1$ with $A$, $B$, $C$, and $D$ so that the corresponding sums and products agree?