| --- |
| source: ncert |
| topic: algebra |
| class_level: class_9 |
| chapter: ch2 |
| difficulty: intermediate |
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| |
| Class 9 | Ch2: Polynomials |
| POLYNOMIAL: p(x) = a_n*x^n + a_(n-1)*x^(n-1) + ... + a_1*x + 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). |
