source: ncert
topic: algebra
class_level: class_9
chapter: ch2
difficulty: intermediate
Class 9 | Ch2: Polynomials POLYNOMIAL: p(x) = a_n*x^n + a_(n-1)x^(n-1) + ... + a_1x + a_0 where coefficients are real numbers. Degree: highest power of variable. Linear(degree 1), Quadratic(degree 2), Cubic(degree 3). Zero polynomial: p(x)=0. Degree not defined. Constant polynomial: p(x)=c. Degree=0. ZEROES OF POLYNOMIAL: value of x where p(x)=0. Linear polynomial ax+b has zero at x=-b/a. Quadratic can have 0, 1 or 2 zeroes. Cubic can have 1, 2 or 3 zeroes. REMAINDER THEOREM: When p(x) is divided by (x-a), remainder = p(a). No need to do long division — just substitute x=a. p(x)=x^3-6x^2+11x-6 divided by (x-2): remainder=p(2)=8-24+22-6=0. So (x-2) is a factor. FACTOR THEOREM: (x-a) is a factor of p(x) if and only if p(a)=0. To factorise: find a value 'a' where p(a)=0, then divide by (x-a). FACTORISING CUBIC POLYNOMIALS: Try factors of constant term. If p(a)=0, divide p(x) by (x-a) to get quadratic. Factorise quadratic. ALGEBRAIC IDENTITIES: (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx. (x+y)^3 = x^3+3x^2y+3xy^2+y^3 = x^3+y^3+3xy(x+y). (x-y)^3 = x^3-3x^2y+3xy^2-y^3 = x^3-y^3-3xy(x-y). x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx). Special case: if x+y+z=0 then x^3+y^3+z^3=3xyz. SOLVED EXAMPLES: Example 1: Find remainder when x^3-6x^2+11x-6 divided by x-2. p(2)=8-24+22-6=0. Remainder=0. So (x-2) is a factor. Example 2: Factorise x^3-23x^2+142x-120. Try x=1: 1-23+142-120=0. So (x-1) is factor. Divide: x^3-23x^2+142x-120 = (x-1)(x^2-22x+120) = (x-1)(x-10)(x-12). COMMON MISTAKES: Degree of zero polynomial is undefined, not zero or -1. Remainder theorem uses SUBSTITUTION, not division, for finding remainder quickly. Factor theorem: (x-a) is factor iff p(a)=0 (note: x MINUS a, not x PLUS a).