Polynomials do cover a lot of portion in maths. The highest power of the variables in a polynomial is termed as its degree. In mathematical language, by quintic function, one means to refer a polynomial of degree 5.

The general form of a quintic function is given below:**$f (x)$ = $a x^{5}$ + $b x^{4}$ + $c x^{3}$ + $d x^{2}$ + $e x$ + $f$ ; $a \neq$ 0**

Where; $a, b, c, d, e$ and $f$ are constant terms, and may belong to the field of real numbers, rational number or complex numbers. The constant "a" must **not** be equal to zero, otherwise the polynomial will be of degree 4 or quartic polynomial.

When $f (x)$ is set to zero; provided that $a \neq$ 0, one gets **quintic equation **as illustrated below:**$a x^{5}$ + $b x^{4}$ + $c x^{3}$ + $d x^{2}$ + $e x$ + $f$ = 0**

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The graph of a quintic function looks quite similar to a cubic function or any odd-degee polynomial. Quintic functions have four local maxima and minima. Since, the quintic equation has positive leading coefficient and odd highest power ; therefore, according to the rule of polynomial graphing, its graph rises to right side and falls to left side.

**The graph of quintic function is shown below:**

**Following are the properties that are possessed by quintic function:**

**1)** It can have **five roots**, since it is a polynomial of degree five. Though, one or more roots may be zero; but a quintic function has at most five roots.

**2)** There are at most **four local extrema **in the graph of quintic function. In other words, this function has zero to four local maxima or minima, when graphically represented.

**3) **Quintic function does not characterize any general symmetry. Its graph is **asymmetric**.

**4) **The highest number of **inflection points **can be from **one to three **in quintic function graph.

**5)** The roots of quintics are quite difficult to solve by radicals. But, there are few quintic equations that are solvable.

**There are few methods by which quintic functions can be solved. These are listed below:**

**1)** Few quintic equations may be solved with the help of **simple factorization techniques**. Though, all cannot be solved by this method.

**2)** Quintics may be solved by using **Jacobi theta function **which is an elliptic analog of exponential function.

**3)** Quinitics, sometimes, may be solved using **differential equations**.

**4)** **Tschirnhaus transformation** was introduced by mathematician **Ehrenfried Tschirnhausen** in late 16$^{th}$ century. It is actually a mapping on polynomials and which field theory. This transformation may be used to solve an irreducible quintic with roots to rational function.

**5)** Even **graphical method **may also be used to find the roots of a quintic function or equation.

All the quintic equations are not easy to deal with. In fact, many of them are not solvable. Here, let us have a look at some examples of simple solvable quintics.

**Examples 1 :** Find the roots of $m^{5}$ - 3 = 0.

**Solution :** $m^{5}$ - 3 = 0

$m^{5}$ = 3

$m$ = $\sqrt[5]{3}$

$m$ = 1.2457**(approx)**

**Example 2 :** Solve $x^{5}+2x^{4}+x^{3} = 0$

**Solution :** $x^{5}+2x^{4}+x^{3} = 0$

$x^{3}(x^{2}+2x+1) = 0$

$x^{3}(x+1)^{2} = 0$

$x^{3} = 0$ and $(x+1)^{2}=0$

$x$ = 0, 0, 0, -1, -1

**Example 3 :** Calculate the roots of following quintic: $x^{5}-x^{4}-x+1=0$

**Solution : **$x^{5}-x^{4}-x+1=0$

$x^{4}(x-1)-1(x-1)=0$

$(x^{4}-1)(x-1)=0$

$(x^{2}+1)(x^{2}-1)(x-1)=0$**......... using the identity $a^{2}-b^{2}=(a+b)(a-b)$**

$(x^{2}+1)(x+1)(x-1)(x-1)=0$

which gives

$(x^{2}+1)=0 \Rightarrow x = \pm 1$

$x$ + 1 = 0 $\Rightarrow$ $x$ = - 1

$x$ - 1 = 0 $\Rightarrow$ $x$ = 1

$x$ - 1 = 0 $\Rightarrow$ $x$ = 1

So, $x$ = 1, 1, 1, -1, -1

All the quintic equations are not easy to deal with. In fact, many of them are not solvable. Here, let us have a look at some examples of simple solvable quintics.

$m^{5}$ = 3

$m$ = $\sqrt[5]{3}$

$m$ = 1.2457

$x^{3}(x^{2}+2x+1) = 0$

$x^{3}(x+1)^{2} = 0$

$x^{3} = 0$ and $(x+1)^{2}=0$

$x$ = 0, 0, 0, -1, -1

$x^{4}(x-1)-1(x-1)=0$

$(x^{4}-1)(x-1)=0$

$(x^{2}+1)(x^{2}-1)(x-1)=0$

$(x^{2}+1)(x+1)(x-1)(x-1)=0$

which gives

$(x^{2}+1)=0 \Rightarrow x = \pm 1$

$x$ + 1 = 0 $\Rightarrow$ $x$ = - 1

$x$ - 1 = 0 $\Rightarrow$ $x$ = 1

$x$ - 1 = 0 $\Rightarrow$ $x$ = 1

So, $x$ = 1, 1, 1, -1, -1

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