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Solving Polynomial Equations

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  a1, a2, ......, an 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) = axn + an-1 xn - 1 + an-2 xn - 2 + .................+ a1 x + a0 
and where n is a natural number and the polynomial of degree n.

Polynomial equations are classified as:

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


Solving Linear and Quadratic Equations

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

 

Question 2: Solve quadratic equation 2x2 + 3x - 2 = 0
Solution:
Given quadratic equation 2x2 + 3x - 2 = 0

=> 2x2 + 3x - 2 = 0

Factors the left hand side of the equation,

=> 2x2 + 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

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Formulas for solving polynomials does not exist beyond for degree 3. The cubic formula itself is rather cumbersome which is not easy to use and generally not used.
The general methods used in solving the polynomials are:
  • Factoring
Some Higher degree polynomials can be factored using simple factoring techniques or can be reduced to quadratic forms using substitutions.
  • Using Synthetic division
The possible rational roots or solutions of the polynomials are listed using the rational root theorem. The possible roots are then checked using  factor theorem or factor theorem and the actual solutions are determined.
  • Estimating the solutions
The non rational roots can be estimated using Des Carte's rule of signs and lower and upper bound theorems. The estimates of the solutions can also be made with the help of graphing calculator using the zero and intersect features.

Solving a Polynomial Using Synthetic Division

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Synthetic division is a long division process generally used, however, not for dividing out factors but for finding zeroes of polynomials.
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. 

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:


After dividing given expressions by x + 1 and x - 2, quotient is 2x2 + 9x + 9 

Now factors of 2x2 + 9x + 9, 

=> 2x2 + 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 = $\frac{-3}{2}$.