MathsClass 10 MathsChapter 2: Polynomials

Polynomial FormulaPolynomial equations mean the relation between numbers and variables are explained in a pattern. Polynomial is one of the major concepts of Mathematics. In Maths, we have studied a variety of equations formed with algebraic expressions. When we talk about polynomials, it is also a form of the algebraic equation and polynomial formula. Let us get started.Polynomial FormulaWhat is a Polynomial?Polynomial is an algebraic expression, in which the variables have non-negative powers. A polynomial function is anequation, which consists of a single independent variable, where the variable can occur in the equation more than one time with different degrees of the exponent. A polynomial expression is the one, which has more than two algebraic terms. As the name suggests, Polynomial is a repetitive addition of a monomial or a binomial.Types of PolynomialPolynomial equation is of four types :Monomial: This type of polynomial contains only one term. For example, x2 , x, y, 3y, 4zBinomial: This type of polynomial contains two terms. For example, x2 – 10xTrinomial: This type of polynomial contains three terms. For example, x2 – 10x+9Quadratic Polynomial: This type of polynomial contains four terms. For example, x3+2x2 – 10x+7The general Polynomial Formula is,F(x) = anxn + bxn-1 + an-2xn-2 + …….. + rx +sIf n is a natural number: an – bn = (a – b)(an-1 + an-2b +…+ bn-2a + bn-1)If n is even (n = 2a): xn + yn = (x + y)(xn-1 – xn-2y +…+ yn-2x – yn-1)If n is odd number: xn + yn = (x + y)(xn-1 – xn-2y +…- yn-2x + yn-1)Image result for polynomial formulaPolynomial Identities(x + y)2= x2 + 2xy + y2(x – y)2= x2 – 2xy + y2x2– y2 = (x + y)(x – y)(x + a)(x + b) = x2+ (a + b)x + ab(x + y + z)2= x2 + y2 + c2 + 2xy + 2yz + 2zx(x + y)3= x3 + y3 + 3xy (x + y)(x – y)3= x3 – y3 – 3xy (x – y)x3+ y3 = (x + y)(x2 – xy + y2)x3– y3 = (x – y)(x2 + xy + y2)x3+ y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)ax2 + bx + c = 0 then x = \frac{( -b \pm \sqrt{(b2 – 4ac)} )}{2a}Solved ExamplesQ1. Solve the equation: x2 +16x + 64 = 0Solution: Factors of x2 +16x + 64We the factors of the 64 are 2, 4, 8, 16, 32, 64. The sum of middle term is 8.Now, x2 + (8 + 8) x + 8 ͯ 8=x2 +8x +8x + 8 ͯ 8= x(x + 8) + 8 (x + 8) = (x + 8) (x + 8)Q.2. Solve x3 – 7x2 + 12x = 0Solution: To factor x3−7x2+12x= x (x2−5x+6)Now, we factorise x2−7x+12We the factors of the 12 = 3, 4Now, x2 − (3 + 4) x + 3 ͯ 4=x2 -3x -4x + 3 ͯ 4= x(x – 3) – 4 (x – 3) = (x−3) (x−4)Factors of x3−7x2+12x  are  x (x−3) (x−4)