Polynomial for Class 10(TEN) by GOLAM ROBBANI AHMED TGT MATHEMATIC AT K V HPCL JAGIROAD

POLYNOMIALS

PREPARED BY GOLAM ROBBANI AHMED

FOLLOWINGS ARE NOT POLYNOMIAL

3X4 + 5X2 – 7X + 1

THE POLYNOMIAL ABOVE IS IN STANDARD FORM. STANDARD FORM OF A POLYNOMIAL - MEANS THAT THE DEGREES OF ITS MONOMIAL TERMS DECREASE FROM LEFT TO RIGHT.

Polynomial

Degree

Name using

Degree

Number of

Terms

Name using number of

terms 7x + 4 1 Linear 2 Binomial

3x2 + 2x + 1 2 Quadratic 3 Trinomial 4x3 3 Cubic 1 Monomial

9x4 + 11x 4 Fourth degree 2 Binomial 5 0 Constant 1 monomial

STATE WHETHER EACH EXPRESSION IS A POLYNOMIAL. IF IT IS, IDENTIFY IT.1) 7y - 3x + 4 trinomial2) 10x3yz2 monomial3)

not a polynomial

2

5 72

yy

FIND THE DEGREE OF X5 – X3Y2 + 41. 02. 23. 34. 55. 10

3) PUT IN ASCENDING ORDER IN TERMS OF Y: 12X2Y3 - 6X3Y2 + 3Y - 2X-2x + 3y - 6x3y2 + 12x2y3

4) Put in ascending order:5a3 - 3 + 2a - a2

-3 + 2a - a2 + 5a3

WRITE IN ASCENDING ORDER IN TERMS OF Y:X4 – X3Y2 + 4XY – 2X2Y3

1. x4 + 4xy – x3y2– 2x2y3

2. – 2x2y3 – x3y2 + 4xy + x4 3. x4 – x3y2– 2x2y3 + 4xy

4. 4xy – 2x2y3 – x3y2 + x4

Remainder TheoramIf p(x) a polynomial divided by a divisor g(x) of order 1(ex. x-a)And r(x) be the remainder then we can writeP(x)=g(x).q(x) +r(x)P(x)= (x-a).q(x) + r(x)P(a)=r(a)Since g(x) is order of 1 hence r(x) will be order of zero(N.B: the order of remainder is always less than divisor.)Since the order of r(x) is zero hence it is constant.Suppose r(a)= rTherefore p(a)= r

Q. On dividing the polynomial 4x² - 5x² - 39x² - 46x – 2 by the polynomial g(x) the quotient is x² - 3x – 5 and the remainder is -5x + 8.Find the polynomial g(x).

Ans: p(x) = g(x).q(x)+r(x) let p(x) = 4x4 – 5x³ – 39x²– 46x – 2 q(x) = x²– 3x – 5 and r (x) = -5x + 8

now p(x) – r(x) = g(x).q(x) 4x² - 5x² - 39x² - 46x – 2 –(-5x+8)= (x² - 3x – 5 ).g(x) ∴ g(x) = 4x2 + 7x + 2

Q. If the squared difference of the zeros of the quadratic polynomial x² + px + 45 is equal to 144 , find the value of p.

Ans: Let two zeros are α and β where α > β According given condition (α - β)2 = 144 Let p(x) = x² + px + 45 α + β = − p

αβ = 45 now (α + β)² = (α - β)² + 4 αβ (-P)² = 144 +180=324 Solving this we get p = ± 18

Q. If α & ß are the zeroes of the polynomial 2x² ─ 4x + 5, then find the value of a. α² + ß² b. 1/ α + 1/ ß c. (α ─ ß)² d. 1/α² + 1/ß² e. α² + ß² Q. Obtain all the zeros of the polynomial p(x) = 3x4 ─ 15x³ + 17x² +5x ─6 if two zeroes are ─1/√3 and 1/√3Q. If one zero of the polynomial 3x² - 8x +2k+1 is seven times the other, find the zeros and the value of kQ.If two zeros of the polynomial f(x) = x4 - 6x3 - 26x2 + 138x – 35 are 2±√3.Find the other zeros.

Q. Find all zeroes of f(x) = 3x4+6x³-2x²-10x-5, if two of its zeroes are √ 5/3 and -√5/3.

Since √ 5/3 and -√5/3.are zeroes of f(x). Therefore (x- √ 5/3 )(x+-√5/3) = (x²- 5/3) is a factor of f(x).After dividing f(x) by (x²- 5/3), we get

As (x²+2x+1) = (x+1)²

Therefore for obtaining other zeroes of f(x) ,we put (x+1)² = 0 ,

Therefore x= -1 and -1

Hence zeroes of f(x) is are √ 5/3 and -√5/3.1 and -1.

Alternative Method:Say the other roots are α and βTherefore √ 5/3 -√5/3+α +β=-6/3=-2α +β=-2……………….(i)Again√ 5/3 (-√5/3).α.β=-5/3

α.β=5/3Using the algebraic identity α -β=………….… (ii)Solving (i) and (ii) we can find the other two roots

Find zeros third degree polynomial equation, x³ - 2x² + 2x - 1 = 0Let P(x) = , x³ - 2x² + 2x – 1P(1) = (1)3 - 2(1)2 + 2(1) - 1 = 0Since P(1) = 0, x - 1 is the factor of P(x).Long Division Method:

Þ x3 - 2x2 + 2x - 1 = (x - 1)(x2 + 6x + 5)

Þ x2 + 6x + 5 = x2 + 5x + x + 5

= x(x + 5) + (x + 5)

= (x + 1)(x + 5)Þ x3 - 2x2 + 2x - 1 = (x - 1)(x + 1)(x

+ 5) = 0

Hence Zeros of the cubic equation are 1, -1, -5.

HOTs

Question :If two zeroes of the polynomial are find other zeros

= x² + 4 − 4x − 3= x² − 4x + 1 is a factor of the given polynomial For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing

= =

Or x = 7 or −5 Hence, 7 and −5 are also zeroes of this polynomial.