polynomials class 9th maths presentation

polynomials class 9th ppt maths

1. kendriya vidyalaya

2. MATHS PROJECT WORK

3. 1.INTRODUCTION 2.POLYNOMIALS IN ONE VARIABLE 3.ZERO POLYNOMIAL 4.REMAINDER THEOREM 5. FACTOR THEOREM 6. FACTORISATION OF POLYNOMIALS 7.ALGEBRIC IDENTITIES

4. INTRODUCTION We know about algebric expressions, their additions, subtraction, multiplication and division in earlier classes. We have studied how to factorise some algebraic expressions .

5. Some identities are-: ( a+b ) ²=a ²+b²+2ab (a-b)²=a²+b²-2ab a²-b²=( a+b )(a-b)

6. 2.POLYNOMIALS IN ONE VARIABLE VARIABLE:- A variable is denoted by a symbol that can take any real value.We use letters like x, y, z to denote variables.

7. 2x,3y,-x,-1/2x are all algebraic expressions. All these expressions are of the form (a constant).x. So we can say it as (a constant)X(a variable) and we and we don’t know what the constant is. In such cases, we write the constant as a, b, c etc. So the expression will be ax.

8. TERM :- In the polynomial x² +2x , x² & 2x are called the terms. COEFFICIENT:-Each term of a polynomial has a coefficient. So, in -x³+4x²+7x-2,the coefficient of x³ is -1, the coefficient of x² is 4, the coefficient of x is 7 and -2 is the coefficient of x .

9. DEGREE:-The highest power of variable in the polynomial is known as degree of the polynomial. For ex:-5x 2 +3 ,here the degree is 2.

10. CONSTANT POLYNOMIAL:-A polynomial containing one term only, consisting of a constant is called a constant polynomial. The degree of a non zero constant polynomial is zero. Eg :-3, -5, 7/8,etc., are all constant polynomials.

11. 3.ZERO POLYNOMIAL :-A polynomial consisting one term only, namelyzero only, is called a zero polynomial. The degree of a zero polynomial is not defined.

12. 4.REMAINDER THEOREM :- We know that, when a natural number n is divided by a natural number m less than or equal to n, the remainder is either 0 or a natural number r<m. Example:23 when divided by 5 gives the quotient 4 and the remainder 3. Here,we can express 23 as 23=(5x4)+3 i.e.,Dividend =(Divisor X Quotient)+Remainder Now, we extend the above phenomenon for the division of a polynomial p(x). Then,we can find polynomial q(x) and r(x) such that:- p(x)=g(x) X q(x)+ r(x),where r(x)=0 or degree or r(x),degree of g(x).Division of a polynomial by a linear polynomial.

13. 5. FACTOR THEOREM:- Let p(x) be a polynomial of degree n≥1 and a be any real constant then If p(a) =0,then (x-a) is a factor of p(x). P(x)=(x-a) X q(x) +p(a)

14. 6. FACTORISATION OF POLYNOMIALS EXAMPLE Question 2: Use the Factor Theorem to determine whether  g ( x ) is a factor of  p ( x ) in each of the following cases: ( i ) p ( x ) = 2 x 3  +  x 2  − 2 x  − 1,  g ( x ) =  x  + 1 (ii) p ( x ) =  x 3  + 3 x 2  + 3 x  + 1,  g ( x ) =  x  + 2 (iii) p ( x ) =  x 3  − 4  x 2  +  x  + 6,  g ( x ) =  x  − 3 Answer :

15. ( i ) If  g ( x ) =  x  + 1 is a factor of the given polynomial  p ( x ), then  p (−1) must be zero. p ( x ) = 2 x 3  +  x 2  − 2 x  − 1 p (−1) = 2(−1) 3  + (−1) 2  − 2(−1) − 1 = 2(−1) + 1 + 2 − 1 = 0 Hence,  g ( x ) =  x  + 1 is a factor of the given polynomial. (ii) If  g ( x ) =  x  + 2 is a factor of the given polynomial  p ( x ), then  p (−2) must be 0. p ( x ) =  x 3  +3 x 2  + 3 x  + 1 p (−2) = (−2) 3  + 3(−2) 2  + 3(−2) + 1 = − 8 + 12 − 6 + 1 = −1

16. As  p (−2) ≠ 0, Hence,  g ( x ) =  x  + 2 is not a factor of the given polynomial. (iii) If  g ( x ) =  x  − 3 is a factor of the given polynomial  p ( x ), then  p (3) must be 0. p ( x ) =  x 3  − 4  x 2  +  x  + 6 p (3) = (3) 3  − 4(3) 2  + 3 + 6 = 27 − 36 + 9 = 0 Hence,  g ( x ) =  x  − 3 is a factor of the given polynomial.

17. 7.ALGEBRIC IDENTITIES 1. (x + a)(x - b) = x 2  + (a - b) x - ab 2. (x - a)(x + b) = x 2  + (b - a) x - ab 3.(x - a)(x - b) = x 2  - (a + b)x + ab 4.(a + b) 3  = a 3  + b 3  +3ab (a + b) 5.(a - b) 3  = a 3  - b 3  - 3ab (a - b) z) 2  + (z -x) 2 ]

18. 6.(x + y + z) 2  = x 2  + y 2  + z 2  + 2xy +2yz + 2xz 7.(x + y - z) 2  = x 2  + y 2  + z 2  + 2xy - 2yz – 2xz 8. (x - y + z) 2  = x 2  + y 2  + z 2  - 2xy - 2yz + 2xz 9. (x - y - z) 2  = x 2  + y 2  + z 2  - 2xy + 2yz - 2xz

19. 10.x 3  + y 3  + z 3  - 3xyz = (x + y + z) (x 2  + y 2   + z 2  - xy - yz - xz ) 11. x 2  + y 2  = 12 [( x+y ) 2  + (x-y) 2 ] 12. ( x+a ) ( x+b ) ( x+c ) = x 3 + (a + b + c ) x 2  + ( ab + bc +ca) x + abc 13. x 3  + y 3  = (x + y) ( x 2  - xy + y 2 ) 14. x 3  - y 3  = (x - y) ( x 2  + xy + y 2 ) 15. x 2  + y 2  + z 2  - xy - yz - zx = 12  [(x-y) 2  + (y -

20. SUBMITTED TO P.K. SRIVASTAVA SIR {THANX SIR FOR LIGHTING US WITH UR KNOWLEDGE}

21. PREPARED BY:- ANUSHKA DUBEY IX B 10

22. THE END THANK YOU