SEBA Class 10 Mathematics Chapter 2 Polynomial

SEBA Class 10 Mathematics Chapter 2 Polynomial Solutions, SEBA Class 10 Maths Textbook Notes in English Medium, SEBA Class 10 Mathematics Chapter 2 Polynomial Notes in English to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 10 Mathematics Chapter 2 Polynomial Notes and select needs one.

SEBA Class 10 Mathematics Chapter 2 Polynomial

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 10 Mathematics Chapter 2 Polynomial Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 10 Mathematics Chapter 2 Polynomial Solutions for All Subject, You can practice these here.

Polynomial

Chapter – 2

1. The graphs of y= p(x) are given in fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Ans: (i) No zeroes.

(ii) 1

(iii) 3

(iv) 2 

(v)  4

(vi) 3

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients

(i) X² – 2x – 8

Ans: x² – 2x – 8 = x² – 4x + 2x – 8 = x(x – 4) + 2(x – 4) = (x + 2)(x – 4)

Zeroes are -2 and 4.

Sum of the zeroes = (-2) + (4) = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x²)

Product of the zeroes = (-2)(4) = -8 =(-8)/1 = (Constant term)/Coefficient of x²).

(ii) 4s² – 4s +1

Ans: 4s² – 4s +1 = (2s – 1)²

= 4s² – 2s – 2s + 1

= 2s(2s – 1) – 1(2s – 1)

= (2s – 1) (2s – 1)

∴ The two zeroes are ½, ½ 

Sum of the zeroes = ½ + ½ = 1 = -(-4)/4 = (Coefficient of x)/(Coefficient of x²)

Product of two zeroes = (½) (½) = ¼ (Constant term)/(Coefficient of x²) .

(iii) 6ײ – 3 – 7x 

Ans: 6x² – 7x – 3

= 6x² – 9x + 2x – 3

= 3x(2x-3) + 1(2x-3)

= (2x – 3) (3x + 1) putting 2x – 3 = 0 and 3x + 1 = 0

We get x = 3/2 and x= -⅓  Zeroes of the quadratic polynomial

p(x) = 6x² – 7x – 3

Sum of the two zeroes = (3/2) + (-⅓) = 3/2 = -⅓ = 7/6

= -(-7)/6 =  -(Coefficient of x)/(coefficient of x²) 

Product of the two zeroes = (3/2) × -(⅓) = -½ = (-3/6)

= (Constant term)/(Coefficient of x²) 

(iv) 4u² + 8u

Ans: 4u²  + 8u = 4u (u + 2)

It gives the two zeroes, 0 and -2 of the polynomial p(u) = 4u² + 8u + u 

Sum of the two zeroes 0 + (-2) = -2 = -(8)/4 = -(Constant term u)/(Coefficient of u²) 

Product of the two zeroes = (0) (-2) = 0 = (Constant term)/(Coefficient of u²) 

(v) t² – 15 

Ans: T² – 15  = t² (√15) = ( t + √15) (t – √15)

If gives the two zeroes of the polynomial p(t) = t² + 0t -15 are -√15 and √15. 

Sum of the two zeros = (-√15) + √15 = 0  = -(Coefficient of t)/(Coefficient of t²)                                                                                

Product of the two zeroes

= (-√15) × (√15) = -15 = (-15)/1 = (Constant term)/(Coefficient of t²)

(vi) 3x² – x – 4

Ans: 3x² – x – 4 = 3x² – 4x + 3x- 4

= 3x(x + 1) – 4(x + 1) = (x +1) (3x – 4)

Putting x + 1 = 0 and 3x – 4 = 0, we get x = -1 and x = 4/3 

i.e ., the two zeroes of the quadratic polynomial p(x) = 3x² – x – 4 are – 1 and 4/3 

Sum of the two zeroes = (-1) + (4/3) = ⅓ = – (-1)/3 = -(Coefficient of x)/(Coefficient of x²) 

Product of the two zeroes = (-1) × 4/3 = -4/3 = (Constant term)/(Coefficient of x²).

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Ans:

Ans: 

(iii) 0, √5

Ans: 

(iv) 1, 1

Ans:

(v) ¼, ¼ 

Ans: 

(vi) 4.1 

Ans:

3. Find the quadratic polynomials whose zero are: 

(i) (-4 and 3/2) 

(ii) 5 and 2

Ans: Let ɑ = 5 and β = 2 

Sum of zeroes = ( α + β ) = (5 + 2) = 7

Product of zeroes = (α β) = (5 × 2) = 10

∴ required of polynomial x² – (α + β)x + α β

= x² – 7x + 10 

(iii) ⅓ and -1

(iv) 3/2 and -2 

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x³ – 3x²  + 5x – 3,  g(x) = x² – 2

Ans: 

(ii) p(x) = x⁴ – 3x² + 4x + 5,  g(x) = x² + 1 – x 

Ans:  

(iii) p(x) = x⁴ – 5x + 6,  g(x) = 2 – x²

Ans: 

(iv) p(x) = 2x⁴ + 2x³ – 2x² – 9x – 12,  g(x) = x² – 3

Ans:

(v) p(x) = x⁶ + 3x² + 10,  g(x) = x³ + 1

Ans:

∴ Quotient q(x) = x³ – 1

Remainder r(x) = 3x² + 11

(vi) p(x) = 2x⁵ – 5x⁴ + 7x³ + 4x² – 10x² – 11,  g(x) = x³ + 2

2. Check whether the first polynomial is a factor of the second polynomial by second polynomial by:

(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12

Ans: 

(ii) x² + 3x + 1, 3x⁴ + 5x³– 7x² + 2x + 2

Ans:

(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Ans: 

Here, remainder = 2 ≠ 0.

Hence, x³ – 3x +1 does not divide the polynomial x⁵ – 4x³ + x² + 3x + 1

3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes

Ans:

4. On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2  and -2x + 4, respectively. Find g(x).

Ans: 

5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

Ans:

(ii) deg q(x) = deg r(x)

Ans: 

(iii) deg r(x) = 0 

Ans:

6. (i) If one zero of the polynomial 3x² – x² – 3x + 1 is 1, then find all other zeros.

Ans: 

(ii) If two zeros of polynomial x⁴ + x³ – 9x² – 3x + 18 are √3 and -√3, then find all the other zeros.

Ans:  

(iii) If two zeros of polynomial x⁴ + 2x³ – 26x² – 54x – 27 are  3√3 and -3√3,  then find all the others.

Ans: 

7. (i) On dividing the polynomial 6x⁴ – 11x³ – 7x² – 15x – 50 by another polynomial 

3x + 7 the remainder is found as -15. Find the quotient. 

(ii) On dividing a polynomial by x² – 2, the quotient is found as 2x² + 5x – 2 and the remainder as -x + 14. Find the polynomial.

Ans:

1. Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also verify the relationship between the zeros and the coefficients in each case:

(i)  2x³ + x² – 5x + 2; ½, 1, -2 

(ii) x³ – 4x² + 5x – 2; 2, 1, 1

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

3. If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.

Ans:

Ans:

5. If the polynomial x² – 6x + 16x² – 25x + 10 is divided by another polynomial

x² – 2x + k, the remainder comes out to be x + a, find k and a. 

Ans: By division of algorithm, we have

 Dividend = Divisor × Quotient + Remainder

⇒ Dividend – Remainder = Divisor x Quotient.

⇒ Dividend – Remainder is always divisible by the divisor.

It is given that f(x) = x⁴ – 6x³ + 16x² – 25x + 10 when divided by x² – 2x + k leaves, x + a as remainder.

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