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

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

$p(x) = x^2$

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 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:

$p(x)=ax^3 + bx^2 + cx + d$

Figure:

$p(x)=x^3-20x^2-30x-4$

The Quartic

The general form of the quartic polynomial is

$p(x) = ax^4 + bx^3 + cx^2 + dx + e$

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: $y = a + bx$

2. Subtract $a$ from both sides.

3. $y-a = bx$

4. $bx=y-a$

5. Divide by $b$ on both sides if $b$ is not equal to 0.

6. $x=\displaystyle\frac{1}{b}(y-a)$

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

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:

$$\begin{aligned} x &= \displaystyle\frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \\ x &= \displaystyle\frac{3 \pm \sqrt{9 - 8}}{2} \\ x &= \displaystyle\frac{3 \pm \sqrt{1}}{2} \\ x &= \displaystyle\frac{3 + 1}{2} \text{ or } \displaystyle\frac{3 - 1}{2} \\ x &= \displaystyle\frac{4}{2} \text{ or } \displaystyle\frac{2}{2} \\ x &= 2 \text{ or } 1 \end{aligned}$$

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

$x_{3,4} = \pm \sqrt{3}$