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\begin{center}\huge MATH 1011 - EXAM 3 - FALL 2016\\
\huge SOLUTION \\
\end{center}
\begin{flushleft}\vspace{0.5in}
Thursday 13 October 2016\\
Instructor: Tom Cuchta
\end{flushleft}
\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}
\end{coverpages}

\begin{questions}
\question[15] Factor completely.
\begin{parts}
\part[5] $24xy+48x$ \\
\textit{Solution:} Since $24$ and $x$ are common factors and $48=2 \cdot 24$, we factor them out and {\color{blue} \href{http://www.wolframalpha.com/input/?i=factor+24xy%2B48x}{see}}
$$24xy+48x=24x(y+2).$$
\part[5] $6a^3b+12a^2b^2$ \\
\textit{Solution:} Since $6$, $a^2$, and $b$ are common factors and $12=6 \cdot 2$, we factor them out and {\color{blue} \href{http://www.wolframalpha.com/input/?i=factor+6a%5E3b%2B12a%5E2b%5E2}{see}}
$$6a^3b+12a^2b^2=6a^2b(a+2b).$$ 
\part[5] $12x^3y + 2xy^3 + 6xy$ \\
\textit{Solution:} Since $2$, $x$, and $y$ are common factors, factor them out and {\color{blue} \href{http://www.wolframalpha.com/input/?i=factor+12x%5E3y%2B2xy%5E3%2B6xy}{see}} \\
$$12x^3y + 2xy^3 + 6xy=2xy(6x^2+y^2+3).$$
\end{parts}

\question[15] Simplify and write with no negative exponent.
\begin{parts}
\part[5] $15^{-1}$ \\
\textit{Solution:} $\dfrac{1}{15}$ 
\part[5] $(2a^{-4}b^3)^2$ \\
\textit{Solution:} 
$$(2a^{-4}b^3)^2=2^2 \left( a^{-4} \right)^2 \left( b^3 \right)^2 = 4a^{-8}b^6 = \dfrac{4b^6}{a^8}.$$
\part[5] $\left( \dfrac{2x^2y^{-2}}{z} \right)^2  \left( \dfrac{z^2}{x^2y^2} \right)^3$  \\
\textit{Solution:} 
$$\begin{array}{ll}
\left( \dfrac{2x^2y^{-2}}{z} \right)^2 \left( \dfrac{z^2}{x^2y^2} \right)^3 &= \left( \dfrac{2^2 \left( x^2 \right)^2 \left(y^{-2} \right)^2}{z^2} \right) \left( \dfrac{\left(z^2\right)^3}{\left( x^2 \right)^3 \left( y^2 \right)^3} \right) \\
&= \left( \dfrac{4x^4 y^{-4}}{z^2} \right) \left( \dfrac{z^6}{x^6 y^6}  \right) \\
&=  4 x^{4-6} y^{-4-6} z^{6-2} \\
&= 4x^{-2} y^{-10} z^4 \\
&= \dfrac{4z^4}{x^2 y^{10}}.
\end{array}$$
\end{parts}
\question[5] 
\begin{parts}
\part[2] Expand the scientific notation into a normal decimal: $2.63 \times 10^{-6}$ \\
\textit{Solution:} {\color{blue} \href{http://www.wolframalpha.com/input/?i=2.63*10%5E(-6)}{$0.00000263$}}
\part[3] Write using scientific notation: $0.000000000782$ \\
\textit{Solution:} {\color{blue} \href{http://www.wolframalpha.com/input/?i=0.000000000782}{$7.82 \times 10^{-10}$}}
\end{parts}

\question[10] Factor by grouping.
\begin{parts}
\part[5] $x^2+7x+2x+14$ \\
\textit{Solution:} $$\begin{array}{ll}
(x^2+7x)+(2x+14) &= x(x+7)+2(x+7) \\
&= (x+2)(x+7)
\end{array}$$
\part[5] $8x^2+6x+4x+3$ \\
\textit{Solution:} $$\begin{array}{ll}
(8x^2+6x) + (4x+3) &= 2x(4x+3) + (4x+3) \\ 
&= (2x+1)(4x+3)
\end{array}$$
\vfill
\end{parts}

\newpage
\question[20] Factor completely or state that it is prime.
\begin{parts}
\part[5] $x^2+x-6$ \\
\textit{Solution:} We seek numbers $p$ and $q$ so that 
$$\left\{ \begin{array}{ll}
p\cdot q = -6 \\
p+q=1.
\end{array} \right.$$
The numbers we need are $p=3$ and $q=-2$. Now rewrite the original polynomial and factor by grouping to {\color{blue}\href{http://www.wolframalpha.com/input/?i=factor+x%5E2%2Bx-6}{get}}
$$\left( x^2+3x \right) + \left( (-2)x+(-6) \right)=(x(x+3)+(-2)(x+3)=(x-2)(x+3).$$
\part[5] $x^2+x-31$ \\
\textit{Solution:} Prime
\part[5] $5x^2-9x-2$ \\
\textit{Solution:} We need $p$ and $q$ so that
$$\left\{ \begin{array}{ll}
p \cdot q = -10 \\
p+q = -9.
\end{array} \right.$$
The numbers we need are $p=-10$ and $q=1$. Now rewrite the original polynomial and factor by grouping to {\color{blue} \href{http://www.wolframalpha.com/input/?i=factor+factor+5x%5E2-9x-2}{get}}
$$5x^2-9x-2=(5x^2-10x)+(x-2)=5x(x-2)+(x-2)=(5x+1)(x-2).$$
\part[5] $x^4-16$ \\
\textit{(hint: this is a difference of squares)} \\
\textit{Solution:} The easiest way is to use difference of squares factoring twice to {\color{blue} \href{http://www.wolframalpha.com/input/?i=factor+x%5E4-16}{factor}}
$$x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4).$$
\end{parts}

\question[10] Perform the long division.
\begin{parts}
\part[5] $(x^2+3x+2) \div (x+2)$\\
\textit{Solution:} 
$$\polylongdiv{x^2+3x+2}{x+2}$$ \\
\part[5] $(3x^2-2x-5) \div (x+1)$ \\
\textit{Solution:}
$$\polylongdiv{3x^2-2x-5}{x+1}$$
\end{parts}

\newpage

\question[15] Solve the equation.
\begin{parts}
\part[5] $x(x+3)=0$ \\
\textit{Solution:} Recall the zero-product property which says that if $A \cdot B = 0$ the n$A=0$ or $B=0$. Since this is already factored, we see $x=0$ or $x+3=0$, i.e. $x=-3$
\part[5] $x^2-2x-3=0$ \\
\textit{Solution:} Factor the left-hand side to get
$$(x-3)(x+1)=0.$$
Using the zero-product property we get $x-3=0$ or $x+1=0$. Therefore $x=3$ or $x=-1$.
\part[5] $2x^2+x-6=0$ \\
\textit{Solution:} Factor the left-hand side to get
$$(2x-3)(x+2)=0.$$
Using the zero-product property, we get $2x-3=0$ or $x+2=0$. Therefore $x=\dfrac{3}{2}$ or $x=-2$. 
\end{parts}


\question[10] Word problems.
\begin{parts}
\part[5] The product of two consecutive numbers is $210$. Find a pair of such numbers. (note: there are two pairs that could be found) \\
\textit{Solution:} Let $x$ be the first of the two consecutive numbers. Then the product of $x$ and $x+1$ equaling $210$ means
$$x(x+1)=210,$$
or
$$x^2+x-210=0.$$
Factor the left-hand side to get
$$(x+15)(x-14)=0.$$
Solve this by using the zero-product property: $x+15=0$ or $x-14=0$, therefore $x=-15$ or $x=14$. Therefore the pair $-15, -14$ or the pair $14, 15$ obey the desired property.
\part[5] The base of a triangle is $1$ $\mathrm{cm}$ larger than its height. Its area is $6$ $\mathrm{cm}^2$. Find the length of the base and the height of the triangle. \\
\textit{Solution:} Recall that $A=\dfrac{1}{2} bh$, where $b$ represents the base of the triangle, $h$ its height, and $A$ its area. Plug in $A=6$. We are told that $b=h+1$, so the area equation becomes
$$6 = \dfrac{1}{2}(h+1)h.$$
Multiply by $2$ to remove the fraction and distribute the $h$ into the binomial to get
$$12=h^2+h.$$
Subtract by $12$ to get
$$0 = h^2+h-12.$$
Factor the right-hand side to get
$$0=(h-3)(h+4).$$
Now using the zero-product property, $h-3=0$ or $h+4=0$. Therefore $h=3$ or $h=-4$. Since $h=-4$ is not physically meaningful, we are left with the solution $h=3$. Since we saw earlier that $b=h+1$, we get $b=3+1=4$. 
\vfill
\end{parts}
\end{questions}

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