Quartic polynomial is a type of polynomial equation. Polynomial equations are algebraic expressions. Each polynomial expression may consist of variables, exponents and constants. These variables, exponents and constants are related by addition, subtraction, multiplication and division. There are certain restrictions in a polynomial. The restrictions include if there is a fraction, and then no variable can be present as the denominator. There cannot be any negative exponent as exponent may incluse $0,\ 1,\ 2,\ 3$ ...etc. Number of terms in a polynomial could be infinite. When the highest power of the variable in a polynomial equation is $4$, it is known as the quartic polynomial.

Quartic polynomial is known as the fourth order polynomial. The standard form to represent a quartic polynomial is $y$ = $ax^4\ +\ bx^3\ +\ cx^2\ +\ dx\ +\ e$. The number $o$ roots that a quartic polynomial can have are $0,\ 1,\ 2,\ 3$ or $4$ roots. Quartic polynomial is also known as biquadratic polynomial. There are a number of methods to solve a quartic polynomial equation such as Ferrari’s solution or by radical. In Ferrari’s solution the equation is first changed into depressed form and then it is solved. Certain basic characteristics of a quartic polynomial are s follows:

Quartic polynomial can have zero to four number of roots.

Quartic polynomial can have one, two or three extrema.

Quartic polynomial can have zero, one or two inflection points.

There is no general symmetry.

Quartic polynomial function requires minimum five number of information to be solved.

Roots of a quartic polynomial are found out using radical.
The roots of a quartic equation $ax^4\ +\ bx^3\ +\ cx^2\ +\ dx\ +\ e$ can be found out implementing the below mentioned formulas. The roots are $x1,\ x2,\ x3$ and $x4$ where

$X1,\ x2$ = $\frac{-b}{4a}$ $–\ S\ \pm$ $\frac{1}{2}$ $\sqrt{(-4S^2\ -\ 2p\ +\ \frac{q}{S})}$

$X3,\ x4$ = $\frac{-b}{4a}$ $+\ S\ \pm$ $\frac{1}{2}$ $\sqrt{(-4S^2\ -\ 2p\ -\ \frac{q}{S})}$

The four unknown variables $p,\ q$ and $S$ stand for

$‘p$ = $\frac{(8ac\ –\ 3b^2)}{8a^2}$

$‘q$ = $\frac{(b^3\ –\ 4abc\ +\ 8a^2d)}{8a^3}$

$S$ = $\frac{1}{2}$ $\sqrt{\{ \frac{-2}{3p} + \frac{1}{3a} (\frac{Q + \Delta_0}{Q})} \}$

Where, $Q$ = $\sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}$

And $\Delta_0,\ \Delta_1$ are the discriminant

$\Delta_0$ = $c^2\ –\ 3bd\ +\ 12ae$

$\Delta_1$ = $2c^3\ –\ 9bcd\ +\ 27b^2e\ +\ 27ad^2\ –\ 72ace$

$\Delta_1^2\ -\ 4 \Delta_0^3$ = $-27 \Delta$
Quartic polynomials can be factorized by a number of methods. It can either be expressed as the product of two quadratic equations, where the quadratic functions are further factorized by applying middle term factorization or by quadratic formula. There are times when a quartic polynomial might not ne expressed as the perfect product of two quadratic equation, In that case the best way to factorize a quartic polynomial shall be using factor method. Let us illustrate the method taking up an example:

$P(x)$ = $x^4\ -\ 3x^3\ -\ 2x^2\ +\ 12x\ -\ 8$

Using remainder theorem, let us try $x$ = $1$

$P(1)$ = $1^4\ -\ 3 \times\ 1^3\ -\ 2\ \times\ 1^2\ +\ 12\ \times\ 1\ -\ 8$

$P(1)$ = $1\ -\ 3\ -\ 2\ +\ 12\ -\ 8$

$P(1)$ = $0$

As the remainder is zero, so $x$ = $1$ or $x - 1$ is a factor of this quartic equation.

Next step we shall use the synthetic division method to find out the cubic polynomial.

Factoring Quartic Polynomial

The cubic polynomial is $x^3\ -\ 2x^2\ -\ 4x\ +\ 8$

We shall again use the remainder theorem to find out the next factor. Let us try $x$ = $2$

$P(2)$ = $2^3\ -\ 2\ \times\ 2^2\ -\ 4\ \times\ 2\ +\ 8$

$P(2)$ = $8\ -\ 8\ -\ 8\ +\ 8$

$P(2)$ = $0$

As the remainder is zero, so $x$ = $2$ or $x - 2$ is also a factor of this quartic equation.

Next step we shall use the synthetic division method to find out the quadratic equation

Factor Quartic Polynomial

The quadratic equation we get is $x^2 - 4$ which being a perfect square can be factorized to $(x + 2)\ \times\ (x - 2)$

So, the complete factorization of the quartic polynomial $x^4\ -\ 3x^3\ -\ 2x^2\ +\ 12x\ -\ 8$ is $(x - 1)\ \times\ (x - 2)^2\ \times\ (x + 2)$
Quartic Function Graph

Since the quartic polynomial has positive leading coefficient and even degree polynomial so according to the rule of polynomial graph and looking at the end behaviour the graph rises to both the left and the right side with a number of dips in between.
Example 1: 

Factorize the quartic polynomial: $x^4\ -\ 81$

Solution: 

The given quartic polynomial satisfies the factorization rule of difference of squares $a^2\ -\ b^2$ = $(a + b)\ \times\ (a - b)$.
So, $x^4\ -\ 81$ = $(x^2)^2\ -\ 9^2$

Here, $a$ = $x^2$ and $b$ = $9$

$x^4\ -\ 81$ = $(x^2 - 9)\ \times\ (x^2 + 9)$

We can further factorize it using the same factorization rule of difference of squares.

$(x^2 - 9)$ = $x^2\ -\ 3^2$

Here, $a$ = $x$ and $b$ = $3$

Thus the complete factorization would be $x^4\ -\ 81$ = $(x + 3)\ \times\ (x - 3)\ \times\ (x^2 + 9)$
Example 2:

Factorize the quartic polynomial $x^4\ -\ 10x^3\ +\ 21x^2\ +\ 40x\ -\ 100$ using rational root test

Solution: 

Using rational root test we find out the possible roots of the quartic polynomial. The value of $p$ is $100$ and that of $q$ is $1$. 

So, $\frac{p}{q}$ = $\pm$ $\frac{100}{1}$ = $\pm 100$. The factors are $\pm\ (1, 2, 5, 10, 20, 25, 50, 100)$. Let us choose $5$ and test whether its a root or not. 

$P(5)$ = $5^4\ -\ 10\ \times\ 5^3\ +\ 21\ \times\ 5^2\ +\ 40\ \times\ 5\ -\ 100$ = $0$

Therefore, $x - 5$ is a factor of the quartic polynomial. Using synthetic division method we try to find out the further roots.

Quartic Polynomial Example

The polynomial equation we get is $x^3\ -\ 5x^2\ -\ 4x\ +\ 20$. Factorizing it further we get:

$x^3\ -\ 5x^2\ -\ 4x\ +\ 20$

$x^2\ (x - 5)\ -\ 4\ (x - 5)$

$(x^2 - 4)\ (x - 5)$

$(x + 2)\ (x - 2)\ (x - 5)$

Thus, the roots of the quartic polynomial $x^4\ -\ 10x^3\ +\ 21x^2\ +\ 40x\ -\ 100$ are $-5,\ -2,\ 2$ and $5$.