kendriya vidyalaya

MATHS PROJECT WORK

1.INTRODUCTION2.POLYNOMIALS IN ONE VARIABLE3.ZERO POLYNOMIAL4.REMAINDER THEOREM 5. FACTOR THEOREM 6. FACTORISATION OF POLYNOMIALS7.ALGEBRIC IDENTITIES

INTRODUCTIONWe know about algebricexpressions, their additions, subtraction, multiplication and division in earlier classes. Wehave studied how to factorise some algebraic expressions.

Some identities are-:(a+b)²=a²+b²+2ab(a-b)²=a²+b²-2aba²-b²=(a+b)(a-b)

2.POLYNOMIALS IN ONE VARIABLE

VARIABLE:- A variable is denoted by a symbol that can take any realvalue.We use letters like x, y, z to

denote variables.

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.

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 x0.

DEGREE:-The highest power of variable in the polynomial is known as degree of the polynomial. For ex:-5x2 +3 ,here the degree is 2.

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.

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.

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)+RemainderNow, 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.

5. FACTOR THEOREM:-Let p(x) be a polynomial of degree n≥1 and a

be any real constant thenIf p(a) =0,then (x-a) is a factor of p(x).

P(x)=(x-a) X q(x) +p(a)

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) = 2x3 + x2 − 2x − 1, g(x) = x + 1(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2(iii) p(x) = x3 − 4 x2 + x + 6, g(x) = x − 3

Answer :

(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p(−1) must be zero.

p(x) = 2x3 + x2 − 2x − 1p(−1) = 2(−1)3 + (−1)2 − 2(−1) − 1

= 2(−1) + 1 + 2 − 1 = 0Hence, 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) = x3 +3x2 + 3x + 1

p(−2) = (−2)3 + 3(−2)2 + 3(−2) + 1= − 8 + 12 − 6 + 1

= −1

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) mustbe 0.

p(x) = x3 − 4 x2 + x + 6p(3) = (3)3 − 4(3)2 + 3 + 6

= 27 − 36 + 9 = 0Hence, g(x) = x − 3 is a factor of the given

polynomial.

7.ALGEBRIC IDENTITIES 1.(x + a)(x - b) = x2 + (a - b) x - ab2. (x - a)(x + b) = x2 + (b - a) x - ab 3.(x - a)(x - b) = x2 - (a + b)x + ab 4.(a + b)3 = a3 + b3 +3ab (a + b) 5.(a - b)3 = a3 - b3 - 3ab (a - b)

z)2 + (z -x)2]

6.(x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2xz

7.(x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz – 2xz

8. (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz

9. (x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz

10.x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz -xz)

11. x2 + y2 = 12 [(x+y)2 + (x-y)2]12. (x+a) (x+b) (x+c) = x3+ (a + b + c )

x2 + (ab + bc +ca) x + abc13. x3 + y3 = (x + y) ( x2 - xy + y2)14. x3 - y3 = (x - y) ( x2 + xy + y2)15. x2 + y2 + z2 -xy -yz -zx = 12

[(x-y)2 + (y -

SUBMITTED TO P.K. SRIVASTAVA SIR

{THANX SIR FOR LIGHTING US WITH UR KNOWLEDGE}

PREPARED BY:- ANUSHKA DUBEY IX B 10

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