Polynomial and polynomial equation
A polynomial is an algebraic expression of the form
A polynomial is an algebraic expression of the form
 $ax^n + bx^{n1} + \cdots + rx + s$
Examples
$4x^3x^25$ \gap[10] has degree 3. Degree 3 polynomials are called cubics.
\\
\\ $x^41$ \gap[10] has degree 4. Degree 4 polynomials are called cuartics. \\
\\ $4x^5x$ \gap[10] has degree 5. Degree 5 polynomials are called quintics.
Some for you to do
A polynomial equation of degree $n$ is an equation that can be written in the form
 $ax^n + bx^{n1} + \cdots + rx + s = 0. \quad (a \neq 0) \quad \qquad$ Degree n polymonial = 0
Examples
$4x^3x^25 = 0$ \gap[10] is a degree 3 polymonial equation. Degree 3 polynomial equations are called cubic equations.
\\
\\ $x^41 = 0$ \gap[10] is a degree 4 polymonial equation. Degree 4 polynomial equations are called quartic equations.
\\
\\ $4x^5x = 0$ \gap[10] is a degree 5 polymonial equation. Degree 5 polynomial equations are called quintic equations.
Solving cubic equations
By definition, a cubic equation can be written in the form 
$ax^3 + bx^2 + cx + d = 0 \qquad$ $a, b, c,$ %and $d$ are fixed numbers and $a \neq 0.$
Solution of ax^{3} + bx^{2} + cx + d = 0 when one solution is rational
Easy case: d = 0: Eg. x^{3} − 2x^{2} + x = 0

In this case, $x$ is a common factor:
$ax^3+bx^2+cx = x(ax^2+bx+c)$ \t \gap[40] %Eg. $x^3+2x^2+x = x(x^2+2x+1)$
The equation is therefore $x(ax^2+bx+c)= 0$ and its solutions are $x = 0$ and the solutions of the quadratic $ax^2+bx+c = 0,$ if any.
Examples
The cubic equation $3x^35x^2+2x = 0$ has $a = 3, b = 5, c = 2, d = 0,$

$3x^35x^2+2x = 0 \quad \Rightarrow \quad x(2x^25x+2) = 0 \quad \Rightarrow \quad x(2x1)(x2) = 0$
$\quad \Rightarrow \quad x = 0,\ \ x = \frac{1}{2}, \ \ x = 2$
Harder case: d ≠ 0: Eg. 6x^{3} + 7x^{2} − x − 2 = 0
\t 1. Try putting $x=\pm\frac{\text{Factor %of }d}{\text{Factor %of }a}$.
\\ \t \gap[40] %Eg. $x=\pm\frac{\text{Factor %of }2}{\text{Factor %of }6} = \pm\frac{1}{1}$ %or $\pm\frac{1}{6}$ %or $\pm\frac{1}{3}$ %or $\pm\frac{2}{3}$ %or $\pm\frac{2}{1}$
\\ \t !10! If there is a rational solution, it will be one of these values; call it $r.$ \\ \t
\\ \t !10! 2. $(x  r)$ is then a factor of the lefthand side. To get the other factor, you can either divide the given cubic by $(xr)$ (the other factor will then be the quotient), or use the following formula: \\
\\ \t \gap[40] $ax^3+bx^2+cx+d = (xr)(ax^2 \ + \ [ar+b]x \ + \ [ar^2+br+c])$
\\ \t
\\ \t !10! 3. Now use factorization or the quadratic formula on the second factor to obtain the remaining solutions (if any).
Examples
The cubic equation $3x^3+5x^25x+1 = 0$ has $a = 3, b = 5, c = 5, d = 1,$

1. The only factor of $d = 1$ is $1$ and the factors of $a = 3$ are $1$ and $3$.
Taking all possible ratios $\pm\frac{\text{Factor %of }d}{\text{Factor %of }a}$ gives $x=\pm\frac{\text{Factor %of }1}{\text{Factor %of }3} = \pm\frac{1}{1}$ %or $\pm\frac{1}{3}$
$x = 1:$ \t \gap[10] $3(1)^3+5(1)^25(1)+1 = 4 \neq 0 $ ✘
\\ $x = 1:$ \t \gap[10] $3(1)^3+5(1)^25(1)+1 = 8 \neq 0 $ ✘
\\ $x = \frac{1}{3}:$ \t \gap[10] $3\left(\frac{1}{3}\right)^3+5\left(\frac{1}{3}\right)^25\left(\frac{1}{3}\right)+1 = 0 $ ✔
Thus, $x = \sfrac{1}{3}$ is one solution. 2. $\ \left(x\sfrac{1}{3}\right)$ is therefore a factor of the lefthand side $3x^3+5x^25x+1.$ To get other solutions (if any), we first factor the lefthand side using the formula above:
\\ $ax^3+bx^2+cx+d$ \t $= (xr)(ax^2 \ + \ [ar+b]x \ + \ [ar^2+br+c])$
\\ $3x^3+5x^25x+1$ \t $= \left(x\sfrac{1}{3}\right)\left(3x^2 \ + \ \left\[3\left(\sfrac{1}{3}\right)+5\right\]x \ + \ \left\[3\left(\sfrac{1}{3}\right)^2+5\left(\sfrac{1}{3}\right)5\right\]\right)$
\\ \t $= \left(x\sfrac{1}{3}\right)\left(3x^2 \ + \ [1+5]x \ + \ \left\[\sfrac{1}{3}+\sfrac{5}{3}5\right\]\right)$
\\ \t $= \left(x\sfrac{1}{3}\right)(3x^2 \ + \ 6x \  \ 3)$
3. The second factor is  $3x^2 + 6x  3 = 3(x^2+2x1).$
 $x = \frac{b\pm\sqrt{b^24ac}}{2a} = \frac{2\pm\sqrt{8}}{2} = 1 \pm \sqrt{2}.$
 $x = \frac{1}{3},\quad x = 1+\sqrt{2},$ and $x = 1\sqrt{2}$
Solution of higherorder polynomial equations
Logically speaking, our next step should be a discussion of quartics, then quintics, and so on forever. Well, we've got to stop somewhere, and cubics may be as good a place as any. On the other hand, since we've gotten so far, we ought to at least tell you what is known about higher order polynomials.
Quartics
Just as in the case of cubics, there is a formula to find the solutions of quartics. See, for instance, this article.
Quintics and beyond:
All good things must come to an end, we're afraid. It turns out that there is no "quintic formula." In other words, there is no single algebraic formula or collection of algebraic formulas that will give the solutions to all quintics. This question was settled by the Norwegian mathematician Niels Henrik Abel in 1824 after almost 300 years of controversy about this question. (In fact, several notable mathematicians had previously claimed to have devised formulas for solving the quintic, but these were all shot down by other mathematicians—this being one of the favorite pastimes of practitioners of our art.) The same negative answer applies to polynomial equations of degree 6 and higher. It's not that these equations don't have solutions, just that they can't be found using algebraic formulas. However, there are certain special classes of polynomial equations that can be solved with algebraic methods. The way of identifying such equations was discovered around 1829 by the French mathematician Évariste Galois.
Now try the rest of the exercises in Section 0.4 of or , or move ahead to the next tutorial by pressing on the sidebar.
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