Add content for numbers-to-indices/polynomials

Add actual content to the skeleton of the
`numbers-to-indices/polynomials` section.

Signed-off-by: Eggert Karl Hafsteinsson <[email protected]>
Signed-off-by: Teodor Dutu <[email protected]>
Signed-off-by: Razvan Deaconescu <[email protected]>

chapters/numbers-to-indices/polynomials/reading/read.md

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+# Polynomials
+
+## The General Polynomial
+
+The general polynomial:
+
+$$p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\dots +a_{n}x^{n}$$
+
+The simplest:
+
+$$p(x)=a$$
+
+### Details
+
+:::note Definition
+
+A **polynomial** describes a specific function consisting of linear combinations of positive integer powers of the explanatory variable.
+
+:::
+
+The general form of a polynomial is:
+
+$$p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\dots +a_{n}x^{n}$$
+
+The simplest of these is the constant polynomial
+
+$$p(x)=a$$
+
+## The Quadratic
+
+The general form of the quadratic (parabola) is:
+
+$$p(x) = ax^2 + bx + c$$
+
+The simplest quadratic is
+
+$$p(x) = x^2$$
+
+![Fig. 6](../media/6_2_The_quadratic.png)
+
+Figure: Parabolas: Quadratic functions
+
+### Details
+
+The quadratic polynomial of the form $p(x) = ax^2 + bx + c$ describes a parabola when points $(x,y)$ with $y = p(x)$ are plotted.
+The simplest parabola is $p(x) = x^2$ (Fig. a) which is always non-negative $p(x)\geq 0$ and $p(x)=0$ only when $x=0$.
+
+:::note Note
+
+Note that $p(-x) = p(x)$ since $(-x)^2= x^2$.
+
+:::
+
+If the leading coefficient is negative, then the parabola is concave (fig. b) but if it's positive the parabola is convex (fig. a).
+This is sometimes used to describe a response function.
+
+## The Cubic
+
+The general form of a cubic polynomial is:
+
+$$p(x)=ax^3 + bx^2 + cx + d$$
+
+![Fig. 7](../media/6_3_The_Cubic.png)
+
+Figure:
+
+$$p(x)=x^3-20x^2-30x-4$$
+
+## The Quartic
+
+The general form of the quartic polynomial is
+
+$$p(x) = ax^4 + bx^3 + cx^2 + dx + e$$
+
+![Fig. 8](../media/6_4_The_Quartic.png)
+
+Figure: The general shape.
+
+Here we used the following equation
+
+$$y=x^4-x^3-7x^2+x+6$$
+
+## Solving the Linear Equation
+
+If the value of $y$ is given and we know that $x$ and $y$ are on a specific line so that $y = a + bx$, then we can find the value of $x$.
+
+### Details
+
+If a value of $y$ is given, and we know that $x$ and $y$ lie on a specific straight line, so that $y = a + bx$, then we can find the value of $x$ by considering $y = a+bx$ as an equation to be solved for $x$, since $y$, $a$ and $b$ are all known.
+
+The general solution is found through the following steps:
+
+1. Equation: $y = a + bx$
+
+1. Subtract $a$ from both sides.
+
+1. $y-a = bx$
+
+1. $bx=y-a$
+
+1. Divide by $b$ on both sides if $b$ is not equal to 0.
+
+1. $x=\displaystyle\frac{1}{b}(y-a)$
+
+## Roots of the Quadratic Equation
+
+The general solution of $ax^2 + bx + c = 0$ is given by $x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
+
+### Details
+
+Suppose we want to solve $ax^2 + bx + c = 0$, where $a \neq 0$.
+
+The general solution is given by the formula
+
+$$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$
+
+if $b^2 - 4ac \geq 0$.
+
+On the other hand, if $b^2-4ac<0$, the quadratic equation has no real solution.
+
+### Examples
+
+:::info Example
+
+Solve $x^2 - 3x + 2 = 0$
+
+Putting this into the context of the formulation $ax^2+bx+c=0$, the constants are: $a = 1, b = -3, c = 2$.
+
+Inserting this into the formula for the roots gives:
+
+$$
+\begin{aligned}
+  x &= \displaystyle\frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \\
+  x &= \displaystyle\frac{3 \pm \sqrt{9 - 8}}{2} \\
+  x &= \displaystyle\frac{3 \pm \sqrt{1}}{2} \\
+  x &= \displaystyle\frac{3 + 1}{2} \text{ or } \displaystyle\frac{3 - 1}{2} \\
+  x &= \displaystyle\frac{4}{2} \text{ or } \displaystyle\frac{2}{2} \\
+  x &= 2 \text{ or } 1
+\end{aligned}
+$$
+
+:::
+
+:::info Example
+
+Find the roots of the following polynomial
+
+$$3x^{4} + 14x^{2} + 15$$
+
+We can use the quadratic equation to solve for the roots of this polynomial if we substitute a variable for $x^{2}$.
+Let's use the letter $a$:
+
+$$3a^{2} + 14a + 15$$
+
+We then plug the constants in to the quadratic equation.
+
+$$x = \displaystyle\frac{-(14) \pm \sqrt{14^{2} - 4 \cdot 3 \cdot 15}}{2 \cdot 3}$$
+
+which simplifies to:
+
+$$\displaystyle\frac{-(14) \pm \sqrt{196 - 180}}{6}$$
+
+which equals $-\displaystyle\frac{5}{3}$ (using the $+$ sign) and $-3$ (using the $-$ sign).
+
+Then, since we substituted a for $x^2$ we need to take the square root of these values to get the roots of the polynomial.
+
+So,
+
+$$x_{1,2} = \pm \sqrt{-\displaystyle\frac{5}{3}}$$
+
+and
+
+$$x_{3,4} = \pm \sqrt{3}$$
+
+:::