MathAlgebraAlgebraic EquationsPolynomial Equation Top Polynomial Equations Polynomial equations are the expressions that contain monomial, binomial, trinomial and also the higher order polynomials. Polynomial equation has the lengthy term which contains the endless expression. Polynomials may have both positive and negative values. Polynomial equations have decimal values also. The process of providing the polynomial solutions is to simplify the lengthy polynomial. Polynomial equation solutions are to find the values for the given polynomial equation. A polynomial equation is an equation that can be written in the form, an xn + an - 1 xn - 1 + ......... + a1 x + a0 = 0, where a0, a1, . . . , an - 1 and an are constants. Types of Polynomial EquationsBack to Top There are three different types of polynomial equations as follows: Monomial Equations Binomial Equations Trinomial Equations Monomial Equations: The polynomial equations which has only one term is called as monomial equations. Examples for monomials are 2x = 0, -5xy = 0 Binomial Equations: The polynomial equations which has two terms is called as binomial equations. Examples for binomials are 34x2 + 17y2 = 0, 3x2 + 10 = 0. Trinomial Equations: The polynomial equations which has three terms is called as trinomial equations. Examples for binomials are 7xy - 2y + 6x = 0, 12x3 + 36 + 5x = 0. Roots of Polynomial EquationsBack to Top The polynomial expression is constructed with the help of the variables and also the constants. Polynomial is an expression which has a finite length. Polynomial equation roots are to find the values for the given polynomial equation. A polynomial of degree n will have n roots, some of which may be multiple roots. Finding Real Roots of Polynomial Equations Given below are some of the examples in finding the real roots of polynomial equations. Solved Example Question: Find the roots of the polynomial equation 5x2 - 7x + 2 = 0 Solution: Given 5x2 - 7x + 2 = 0 5x2 - 7x + 2 = 0 => 5x2 - 5x - 2x + 2 = 0 => 5x(x - 1) - 2(x - 1) = 0 => (5x - 2)(x - 1) = 0 Either 5x - 2 = 0 or x - 1 = 0 => x = 25, 1 Hence, the roots of the given polynomial equation are 1, 25. Quadratic Polynomial EquationBack to Top A quadratic equation is a polynomial equation of degree two. Quadratic equation has two solutions may be both real, or both complex. Quadratic equation is an equation of degree 2. A quadratic equation has the form ax2 + bx + c = 0, a ≠ 0 where, a, b, and c are real numbers and x is the variable. The roots of the quadratic equation can be found by factoring or by using quadratic formula. ie. x = −b±b2−4ac√2a Solved Example Question: Find the roots of the quadratic equation by using quadratic formula. x2 - 3x + 4 = 0 Solution: Given x2 - 3x + 4 = 0 Step 1: x2 - 3x + 4 = 0 ......................(1) We know that general form of quadratic equation is ax2 + bx + c = 0 By comparing equation (1), we get => a = 1, b = -3 and c = 4 Step 2: Now, x = −b±b2−4ac√2a => x = −b±b2−4ac√2a = −(−3)±(−3)2−4∗1∗4√2∗1 = 3±9−16√2 = 3±7√i2 => x = 3±7√i2. Cubic Polynomial EquationBack to Top Cubic polynomial equation is also called as third degree polynomial equation. Cubic polynomial equations are the equations having maximum degree of 3. There are three possible outcomes when solving a cubic equation. The general form of cubic equation is a0 x3 + a1x2 + a2x + a3 = 0, a ≠ 0 For example, x3 - 2x2 + 7 is a degree 3 polynomial. Solved Example Question: Find the real solution of the cubic polynomial equation, x3 - x2 + x - 1 = 0 Solution: Given x3 - x2 + x - 1 = 0 Step 1: By trial-and-error method, find one solution by substitution. Put x = -1 => (-1)3 - (-1)2 + (-1) - 1 = 0 => -1 - 1 - 1 - 1 = 0 => -4 ≠ 0 (x = - 1 is not the solution) Again, put x = 1 => 13 - 12 + 1 -1 = 0 => 1 - 1 + 1 - 1 = 0 => 0 = 0 So, x = 1 is a solution of the given equation. Thus, (x - 1) is one of the factor. Step 2: Dividing x3 - x2 + x - 1 = 0 by (x - 1) yields the quotient (x2 + 1). Thus, x3 - x2 + x - 1 = (x - 1)(x2 + 1) = 0 So that, either x - 1 = 0 or x2 + 1 = 0 Since the discriminant of the quadratic x2 + 1 is negative, we dont get any real solutions from x2 + 1 = 0. So, the only real solution is x = 1. Dividing Polynomial EquationsBack to Top Equations in the form of polynomial is called as Polynomial equation. The polynomials are the terms in the order from 0 to n. The polynomial includes monomials, binomials, trinomials and also the higher order polynomials. In the terms of the polynomials, calculate the values for the algebraic polynomial equations. Let us see with the help of the following examples how to divide two polynomials. Solved Examples Question 1: Solve x2+2x−3x2−2x−15 Solution: Given x2+2x−3x2−2x−15 x2+2x−3x2−2x−15 = x2+3x−x−3x2−5x+3x−15 = x(x+3)−(x+3)x(x−5)+3(x−5) = (x−1)(x+3)(x+3)(x−5) Cancelling common terms, we get = x−1x−5 => x2+2x−3x2−2x−15 = x−1x−5. Question 2: Check whether x - 1 is the solution for the polynomial equation x3 - 4x + 6x2 - 1 = 0. Solution: Given x3 - 4x + 6x2 - 1 = 0 Step 1: Let us assume that x - 1 is the solution of the given polynomial. Then, x - 1 = 0 => x = 1 satisfy the equation. Step 2: x3 - 4x + 6x2 - 1 = 0 => 13 - 4 * 1 + 6 * 12 - 1 = 0 => 1 - 4 + 6 - 1 = 0 => 2 ≠ 0 So, x - 1 is not the solution of the given polynomial. Solving Polynomial Equations by FactoringBack to Top Given below are some of the examples in solving polynomial equations by factoring. Solved Examples Question 1: Solve the polynomial quadratic equation x2 - 100 = 0 Solution: Given quadratic equation is x2 - 100 = 0 Adding 100 on both the sides of the equation, we get x2 = 100 Taking square root on both the sides, we get x = 10−−√ => x = ±10. Question 2: Solve the polynomial equation x2 + 40x + 300 = 0. Solution: The given polynomial equation is x2 + 40x + 300 = 0. Step 1: x2 + 40x + 300 = x2 + 10x + 30x + (10 * 30) Step 2: x2 + 40x + 300 = x(x + 10) + 30(x + 10) Step 3: x2 + 40x + 300 = (x + 10) (x + 30) Step 4: x + 10 = 0 and x + 30 = 0 Step 5: (x + 10) (x + 30) = 0 The factors for the polynomial equation x2 + 40x + 300 = 0 are (x + 10) and (x + 30). Graphing Polynomial EquationsBack to Top Graphing is an important concept to visualize the polynomial equations. There are some typical graphs of polynomial functions of odd degree that has atleast one real zero. And, the polynomial functions of even degree having zeros of the function are the x intercepts of the graph. Graphing Polynomial Equations Graphing Polynomial Equation Polynomial Equations ExamplesBack to Top Given below are some of the examples on solving polynomial equations. Solved Examples Question 1: Calculate the value of x for the polynomial equation (2x + 102) + (x + 33) + (2x + 84) = 0. Solution: The given polynomial equation is (2x + 102) + (x + 33) + (2x + 84) = 0. (2x + 102) + (x + 33) + (2x + 84) = 0 => (2x + x + 2x) + (102 + 33 + 84) = 0 => 5x + 219 = 0 => 5x + 219 - 219 = 0 - 219 => 5x = -219 => x = -43.8 The value of x for the given polynomial equation is -43.8. Question 2: Solve the polynomial equations (x2 + 22x -12) = (x2 - 10x) and find the value of x. Solution: Given polynomial equations are (x2 + 22x -12) = (x2 - 10x) Moving the right hand side values into left hand side, we get (x2 + 22x -12) - (x2 - 10x) = 0 Expanding the parenthesis values, we get x2 + 22x -12 - x2 + 10x = 0 32x - 12 = 0 Adding 12 on both the sides, we get 32x = 12 Dividing the above equation by 32, we get 32x32 = 1232 x = 38 So, the answer is 38 Polynomial Equation Practice ProblemsBack to Top Given below are some of the practice problems on polynomial equation. Practice Problems Question 1: Solve the polynomial equation x2 + 30x + 225 = 0. Question 2: Find the value for the polynomial equation (2x + 18) + (2x + 258) = 0
Posted on: Thu, 16 Oct 2014 05:24:13 +0000
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