A polynomial equation is of the form $a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .......... + a_{n}x^{n}=0$ where not all a

_{1}, a_{2}, ......, a_{n}are equal to zero and n is a positive integer. Any number that makes P(x) = 0 when substitute for x is called a zero of the polynomial function. The highest degree present in the polynomial determines the degree of the polynomial.A polynomial equation is an equation that can be written in the form P(x) = 0.

Where **P(x) = a _{n }x^{n} + a_{n-1} x^{n - 1} + a_{n-2} x^{n - 2} + .................+ a_{1} x + a**

_{0 }

and where n is a natural number and the polynomial of degree n.

Polynomial equations are classified as:

- Linear equation containing a polynomial of degree 1
- Quadratic equation with a polynomial of degree 2
- Higher degree polynomial equation whose degree is 3 or more.

## Solving Linear and Quadratic Equations

**Linear equations in one variable**are solved using properties of equalities to undo the operations done and to isolate the variable. Systems of linear equations are solved using graphical or algebraical methods. The commonly used algebraic methods are substitution and elimination methods.

**Quadratic equations,**an equation contains a variable term with the exponent of 2,

**are solved using various methods like factoring, taking square root over the equation, completing the square, quadratic formula, and graphical method.**

## Solved Examples

**Question 1:**Solve for x, 4x + 2 = 1

**Solution:**

Given linear equation, 4x + 2 = 1

=> 4x + 2 = 1

Subtract 2 from both sides

=> 4x + 2 - 2 = 1 - 2

=> 4x = -1

Divide each side by 4

=> $\frac{4x}{4} = {-1}{4}$

=> x = $\frac{-1}{4}$.

=> 4x + 2 = 1

Subtract 2 from both sides

=> 4x + 2 - 2 = 1 - 2

=> 4x = -1

Divide each side by 4

=> $\frac{4x}{4} = {-1}{4}$

=> x = $\frac{-1}{4}$.

**Question 2:**Solve quadratic equation 2x

^{2}+ 3x - 2 = 0

**Solution:**

Given quadratic equation 2x

=> 2x

Factors the left hand side of the equation,

=> 2x

=> 2x(x + 2) - (x + 2) = 0

=> (2x - 1)(x +2) = 0

=> either 2x - 1 = 0 or x +2 = 0

=> x = $\frac{1}{2}$ or x = -2

=> Solutions of equation are x = 0.5, -2.

^{2}+ 3x - 2 = 0=> 2x

^{2}+ 3x - 2 = 0Factors the left hand side of the equation,

=> 2x

^{2}+ 4x - x - 2 = 0=> 2x(x + 2) - (x + 2) = 0

=> (2x - 1)(x +2) = 0

=> either 2x - 1 = 0 or x +2 = 0

=> x = $\frac{1}{2}$ or x = -2

=> Solutions of equation are x = 0.5, -2.

## Method For Solving Higher Degree Polynomial Equations

**The general methods used in solving the polynomials are:**

**Factoring**

**Using Synthetic division**

**Estimating the solutions**

## Solving a Polynomial Using Synthetic Division

It is the shortcut method, that we can use to divide a polynomial by a binomial of the form x - k. Synthetic division method is used to find the solution of polynomial equations with higher degrees.

Solved Example

**Question:**Find the solution of the polynomial equation $2x^{4} + 7x^{3} - 4x^{2} - 27x -18 = 0$

**Solution:**

Let P(x) = $2x^{4} + 7x^{3} - 4x^{2} - 27x - 18 = 0$

Using rational root theorem, the list of rational zeros are long.

=> P(-1) = $2(-1)^{4} + 7(-1)^{3} - 4(-1)^{2} - 27(-1) - 18$

= 2 - 7 - 4 + 27 - 18

= 0

=> x = -1 is zero of the polynomial

again

=> P(2) = $2(2)^{4} + 7(2)^{3} - 4(2)^{2} - 27(2) - 18$

= 32 + 56 - 16 - 54 - 18

= 0

=> x = 2 is zero of the polynomial

After dividing given expressions by x + 1 and x - 2, quotient is 2x

Now factors of 2x

=> 2x

which gives the other two solutions as x = -3 and x = $\frac{-3}{2}$.

Hence the solutions for the polynomial are x = -1, x = 2, x = -3 and x = $\frac{-3}{2}$.

Using rational root theorem, the list of rational zeros are long.

**Step 1:**Let us try**x = -1**and**x = 2**to check whether they are solutions of the equations.=> P(-1) = $2(-1)^{4} + 7(-1)^{3} - 4(-1)^{2} - 27(-1) - 18$

= 2 - 7 - 4 + 27 - 18

= 0

=> x = -1 is zero of the polynomial

again

=> P(2) = $2(2)^{4} + 7(2)^{3} - 4(2)^{2} - 27(2) - 18$

= 32 + 56 - 16 - 54 - 18

= 0

=> x = 2 is zero of the polynomial

Step 2:Step 2:

After dividing given expressions by x + 1 and x - 2, quotient is 2x

^{2}+ 9x + 9Now factors of 2x

^{2}+ 9x + 9,=> 2x

^{2}+ 9x + 9 = (x + 3)(2x + 3)which gives the other two solutions as x = -3 and x = $\frac{-3}{2}$.

Hence the solutions for the polynomial are x = -1, x = 2, x = -3 and x = $\frac{-3}{2}$.