Section 6.1 A first look at polynomial equations
Nyx is making a blanket out of 3’’ \(\times\) 3’’ squares of fabric in the middle and long strips of fabric for the border. They have decided to make the blanket 7 squares longer than it is wide. So, for example, if they use 10 squares across, they will use \(10+7=17\) squares down. They will do a 4’’ border on the top and bottom of the blanket but only a 2.5’’ border on the left and right.
\begin{tikzpicture}[scale=0.1, line join=miter] \colorlet{babyblue}{cyan!35!white} \colorlet{quiltpink}{pink!80!red} \colorlet{sagegreen}{green!50!gray!60!white} \foreach \x in {0,3,...,27} { \foreach \y in {0,3,...,48} { \pgfmathparse{int(mod(\x/3 + \y/3, 4))} \ifcase\pgfmathresult \fill[babyblue] (\x,\y) rectangle (\x+3,\y+3); \or \fill[white] (\x,\y) rectangle (\x+3,\y+3); \or \fill[quiltpink] (\x,\y) rectangle (\x+3,\y+3); \or \fill[sagegreen] (\x,\y) rectangle (\x+3,\y+3); } } \fill[sagegreen] (-2.5,0) rectangle (0,51); \fill[sagegreen] (30,0) rectangle (32.5,51); \fill[sagegreen] (-2.5,-4) rectangle (32.5,0); \fill[sagegreen] (-2.5,51) rectangle (32.5,55); \draw[thick, step=3cm] (0,0) grid (30,51); \draw[thick] (-2.5,0) rectangle (0,51); \draw[thick] (30,0) rectangle (32.5,51); \draw[thick] (-2.5,-4) rectangle (32.5,0); \draw[thick] (-2.5,51) rectangle (32.5,55); \node[below, font=\bfseries] at (15, -4) {Width}; \node[left, font=\bfseries, rotate=90, anchor=south] at (-2.5, 25.5) {Length}; \end{tikzpicture}
Subsection Different sizes of blanket
What size blanket will Nyx make if they use 10 squares across, as shown in the picture? The width of the blanket will include the left border (2.5’’), the width of the 10 squares ( \(10 \times 3"\) ), and the right border (2.5’’), so the blanket will be
\begin{equation*}
2.5+10\times3+2.5 = 35"
\end{equation*}
wide. Notice that we add the width of the border twice so instead of adding 2.5’’ twice, we can add \(2 \times 2.5" = 5"\) once. We can then simplify our calculation to
\begin{equation*}
10\times3+5 = 35"
\end{equation*}
wide.
The length of the blanket will include the top border (4’’), the length of the \(10+7=17\) squares ( \((10+7) \times 3"\) ), and the right border (4’’), so the blanket will be
\begin{equation*}
4+(10+7)\times 3+4 = 59"
\end{equation*}
long. As before, instead of adding 4’’ twice, we can add \(2 \times 4"=8"\) once. We can then simplify our calculation to
\begin{equation*}
(10+7)\times 3+8 = 59"
\end{equation*}
long. The blanket will be \(35" \times 59"\) which is a good size for a baby’s crib.
Let’s do a few more examples. Let’s underline the number of squares to help us see a pattern. If Nyx uses only 2 squares, then
\begin{equation*}
\text{width: }\underline{2}\times3+5 = 11"
\end{equation*}
and
\begin{equation*}
\text{length: }(\underline{2}+7)\times3+8 = 35"
\end{equation*}
The blanket will be \(11" \times 35"\) which might be a good size for a doll.
If Nyx uses 20 squares, then
\begin{equation*}
\text{width: }\underline{20}\times3+5 = 65"
\end{equation*}
and
\begin{equation*}
\text{length: }(\underline{20}+7)\times3+8 = 86"
\end{equation*}
The blanket will be \(65" \times 86"\) which would fit a twin bed.
Subsection Linear models for blanket dimensions
First, the variables are
\begin{equation*}
S = \text{ number of squares (squares)}\sim \text{indep}
\end{equation*}
\begin{equation*}
W = \text{ width of blanket (inches)}\sim \text{dep}
\end{equation*}
\begin{equation*}
L = \text{ length of blanket (inches)}\sim \text{dep}
\end{equation*}
We can summarize our examples in a table.
| \(S\) | 2 | 10 | 20 |
| \(W\) | 11 | 35 | 65 |
| \(L\) | 35 | 59 | 86 |
When we graph both dependences, we see that the points fall on a line, so we have linear functions.
\begin{tikzpicture}[xscale=0.5, yscale=0.05] \foreach \y in {10,20,...,90} { \draw[gray!30, very thin] (0,\y) -- (20,\y); } \foreach \x in {1,2,...,20} { \draw[gray!30, very thin] (\x,0) -- (\x,90); } \draw[thick, -\gt ] (0,0) -- (21,0) node[right] {$S$}; \draw[thick, -\gt ] (0,0) -- (0,93) node[above] {{$W$}, {$L$}}; \foreach \x in {1,2,...,20} { \draw (\x, 1) -- (\x, -1) node[below] { \x}; } \foreach \y in {10,20,...,90} { \draw (0.2, \y) -- (-0.2, \y) node[left] { \y}; } \draw[very thick, blue] plot[domain=0:20] (\x, {3*\x + 5}); \draw[very thick, red] plot[domain=0:19.66] (\x, {3*\x + 31}); \end{tikzpicture}
Subsection Equations for the width and length of blanket
Let’s write equations for the width and for the length of Nyx’s blanket, each a function of the number of squares they use.
First, the width. In our example, Nyx used \(S=10\) squares across and the width of the blanket was
\begin{equation*}
\underline{10}\times 3 + 5 = 35"
\end{equation*}
Our equation is
\begin{equation*}
\text{width: }S \times 3 + 5 = W.
\end{equation*}
Let’s write the dependent variable on the left-hand side of the equation, and use algebraic notation to write \(3S\) instead of \(S \times 3\text{.}\) Our equation is then
\begin{equation*}
W = 3S + 5
\end{equation*}
Next, the length. In our example, Nyx used \(S=10\) squares across and the length of the blanket was
\begin{equation*}
(\underline{10}+7) \times 3+8 = 59".
\end{equation*}
Our equation is
\begin{equation*}
(S+7) \times 3+8 = L,
\end{equation*}
which we can rewrite as
\begin{equation*}
L = 3(S+7)+8.
\end{equation*}
Subsection The distributive property
But, wait! We could have calculated the length differently. We can think of the length as having three parts: \(3S\) which is the length of the \(S\) squares, \(3 \times 7 = 21\) which is the length of the additional 7 squares, and \(2 \times 4 = 8\) which is the length of the top and bottom borders. Breaking into these pieces we get a new equation for the length:
\begin{equation*}
L = 3S + 21 + 8
\end{equation*}
which can be simplified further to
\begin{equation*}
L = 3S + 29
\end{equation*}
Does this new equation make sense? The 3’’/square represents the length of each square Nyx uses. The 29’’ is the part of the length of the 7 squares and top/bottom borders combined. Okay.
Uh oh! That means we have two different looking equations for the length
\begin{equation*}
L = 3(S+7)+8 \text{ and }L=3S + 29.
\end{equation*}
There must be some reason why these equations are actually the same.
The distributive property is the secret. It says that
\begin{equation*}
3(S+7)=3 \times (S+7) = (3 \times S) + (3 \times 7) = 3S + 21.
\end{equation*}
Here’s a picture illustrating the distributive property.
Going back we can use the distributive property to write
\begin{equation*}
L = 3(S+7) + 10 = 3S +21 + 8 = 3S + 29
\end{equation*}
which is exactly the other version of the equation. Whew.
Subsection Area of blanket
Nyx has decided after they finish the blanket, they will make it the top layer of a light comforter. For that, Nyx will use plain fabric for the bottom layer and use some fiberfill to get a thickness of 2 inches in between. How much fiberfill do they need?
To start, Nyx needs to know the area of the blanket. The area is the length times the width. For the baby blanket that was 10 squares across, the width was 35’’ and the length was 59’’ so the area was \(35 \times 59 = 2065\) square inches.
We can write an equation for the area of the blanket using the variables
\begin{equation*}
S = \text{ number of squares (squares) }\sim \text{indep}
\end{equation*}
\begin{equation*}
A = \text{ area of blanket (square inches)}\sim \text{ dep}
\end{equation*}
We know that the area is the length times the width, so putting in our formulas we get
\begin{equation*}
A = (3S+29)(3S+5)
\end{equation*}
Subsection The FOIL property
We can simplify the equation for area using the distributive property.
\begin{align*}
A \amp= (3S+29)(3S+5)\\
\amp= (3S)(3S+5) + (29)(3S+5) \\
\amp= \underbrace{(3S)(3S)}_{\text{First}} + \underbrace{(3S)(5)}_{\text{Outside}} + \underbrace{(29)(3S)}_{\text{Inside}} + \underbrace{(29)(5)}_{\text{Last}}
\end{align*}
Let’s simplify each term separately. The first product is
\begin{equation*}
(3S)(3S) = (3 \times 3)(S \times S) = 9S^{2}
\end{equation*}
The outside product is
\begin{equation*}
(3S)(5) = (3 \times 5)S = 15S
\end{equation*}
The inside product is
\begin{equation*}
(29)(3S) = (29 \times 3)S = 87S
\end{equation*}
And the last product is
\begin{equation*}
(29)(5) = 29 \times 5 = 145
\end{equation*}
Maybe you’ve heard of FOIL? That’s short for First-Outside-Inside-Last which can help you remember how this multiplying out works.
Adding everything up and combining the like terms we get
\begin{equation*}
A = 9S^{2} + 15S + 87S + 145 = 9S^{2} + (15+87)S + 145 = 9S^{2}+102S + 145
\end{equation*}
Here’s a picture illustrating the FOIL property.
\begin{tikzpicture}[line width=1.5pt, scale =.5] \draw (0,0) rectangle (18,7); \draw (3,0) -- (3,7); \draw (0,4) -- (18,4); \draw[\lt -, \gt =stealth, line width=0.8pt] (0, 8.25) -- (5.5, 8.25); \node at (7, 8.25) {$3S+29$}; \draw[-\gt , \gt =stealth, line width=0.8pt] (8.5, 8.25) -- (18, 8.25); \node at (1.5, 7.5) {$3S$}; \node at (10.5, 7.5) {$29$}; \draw[\lt -, \gt =stealth, line width=0.8pt] (-2.25, 0) -- (-2.25, 3.5); \node at (-2.25, 4) {$3S+5$}; \draw[-\gt , \gt =stealth, line width=0.8pt] (-2.25, 4.5) -- (-2.25, 7); \node[left] at (-0.25, 5.5) {$3S$}; \node[left] at (-0.25, 2) {$5$}; \node at (1.5, 5.5) {$9S^2$}; \node at (1.5, 2) {$15S$}; \node at (10.5, 5.5) {$87S$}; \node at (10.5, 2) {$145$}; \end{tikzpicture}
That was a lot of algebra. Here’s a quicker version of what we just did so you can see it all in one place.
\begin{align*}
\amp \\
A \amp= (3S+29)(3S+5) \\
\amp= \underbrace{(3S)(3S)}_{\text{First}} + \underbrace{(3S)(5)}_{\text{Outside}} + \underbrace{(29)(3S)}_{\text{Inside}} + \underbrace{(29)(5)}_{\text{Last}} \\
\amp= 9S^2+15S + 87S + 145 \\
\amp= 9S^2+102S + 145
\end{align*}
Let’s double check, when \(S=10\) we get
\begin{equation*}
A = 9\times 10^{2} + 102 \times 10 + 145 = 2065
\end{equation*}
as before.
Subsection A quadratic model
As before, we make make a table and graph. Using the same values for \(S\) as before we get
| \(S\) | 2 | 10 | 20 |
| \(A\) | 385 | 2065 | 5785 |
We can see that the function is not linear because the graph is not a line.
Subsection Fiberfill
To get a fluffy feel without being too heavy, Nyx has decided to use a premium cluster fiberfill to get 2″ thick. (Nyx will stitch the fill along every fabric join and at intervals along the border, in case you were wondering, because nobody likes a lumpy blanket.) The clerk at the fabric store said that Nyx will need 14 ounces of fiberfill for every 1000 square inches of area. Notice that
\begin{equation*}
\frac{14 \text{ ounces}}{1000 \text{ square inches}}= 0.14 \text{ ounces per square inch}
\end{equation*}
For example, for a baby quilt that’s \(35" \times 59"\text{,}\) Nyx will need \(0.14\) times the area. That is, they will need
\begin{equation*}
0.14 \times2065 = 289.1
\end{equation*}
ounces of fiberfill.
We can write an equation for the amount of fiberfill as a function of the number of squares using
\begin{equation*}
S = \text{ number of squares (squares)}\sim \text{indep}
\end{equation*}
\begin{equation*}
F = \text{ weight of fiberfill (ounces)}\sim \text{dep}
\end{equation*}
The equation is
\begin{equation*}
F = 0.14 (3S + 29)(3S + 5) = 0.14(9S2 + 102S + 145) = 1.26S^{2} + 14.28S + 20.3
\end{equation*}
in case you were curious.
Subsection Another quadratic model
Let’s make a table and a graph for the fiberfill. First, we have
| \(S\) | 2 | 10 | 20 |
| \(F\) | 53.9 | 289.1 | 822.5 |
Subsection Polynomial Equations
Our equations for length and width were linear equations
\begin{equation*}
W=3S+5
\end{equation*}
\begin{equation*}
L=3(S+7)+8 \text{ or }L = 3S + 29
\end{equation*}
because they fit the standard linear equation template:
\begin{equation*}
\text{dep }= \text{ start }+ \text{ slope }* \text{ indep}
\end{equation*}
Our equations for the area and fiberfill were quadratic equations
\begin{equation*}
A=(3S+31)(3S+5)\text{\quad or \quad}A = 9S^{2}+108S + 155
\end{equation*}
\begin{equation*}
F = 0.14 (3S + 29)(3S + 5) \text{\quad or \quad}F= 1.26S^{2} + 14.28S + 20.3
\end{equation*}
because they fit the standard quadratic equation template:
\begin{equation*}
\text{dep }= a*\text{indep}^{2}+b*\text{indep}+c
\end{equation*}
All of these equations are examples of polynomial equation, meaning they can be written as the sum of multiples of powers of the independent variable.
-
How to use geometry to find an area?
-
When to use the distributive property?
-
What FOIL stands for and how to multiply out expressions in parentheses?
If you’re not sure, work the rest of exercises and then return to these questions. Or, ask your instructor or a classmate for help.
