SEBA Class 10 Maths Chapter 2 Polynomial

SEBA Class 10 Maths Chapter 2 Polynomial Solutions English Medium As Per New Syllabus. to each chapter is provided in the list so that you can easily browse through different chapters SEBA Class 10 Maths Chapter 2 Polynomial Notes and select need one. SEBA Class 10 Maths Chapter 2 Polynomial Question Answers Download PDF. SEBA Class 10 General Mathematics Solutions.

SEBA Class 10 Maths Chapter 2 Polynomial

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Polynomial

Chapter: 2

Exercise 2.1

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) In the graph of y = p(x) since no point intersects at x-axis, therefore the number of zeroes of p(x) is zero (0).

(ii) In the graph of y = p(x) since one point intersects at x-axis, therefore the number of zeroes of p(x) is one (1).

(iii) In the graph of y = p(x) since three points intersect at x-axis, therefore the number of zeroes of p(x) is three (3).

(iv) In the graph of y = p(x) since two points intersect at x-axis, therefore the number of zeroes of p(x) is two (2).

(v) In the graph of y = p(x) since four points intersect at x-axis, therefore the number of zeroes of p(x) is four (4).

(vi) In the graph of y = p(x) , since three points intersect at x-axis, therefore the number of zeroes of p(x) is three (3).

Exercise 2.2

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 1/2, 1/2 

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

Product of two zeroes = (1/2) (1/2) = 1/4 (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) = 1/3 = – (-1)/3 = -(Coefficient of x)/(Coefficient of x²) 

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

(vii) x² + 7x + 12

Ans: Let  p(x) = x² + 7x + 12

∴  p(x) = x² + (3 + 4)x + 12

= x² + 3x + 4x + 12

= x(x + 3) + 4(x + 3)

= (x + 3)(x + 4)

∴ The value of p(x) is zero, when

Either,   x + 3 = 0   or,   x + 4 = 0

⇒ x = -3      ⇒ x = -4

Therefore, the zeroes of p(x) = x² + 7x + 12 are α = -3 and β = -4

Now, Coefficient of x², a = 1

Coefficient of x, b = 7

Constant term, c = 12

(viii) x² – 4x + 3

Ans: Let p(x) = x² − 4x + 3

∴ p(x) = x² − (3+1)x + 3

= x² − 3x − x + 3

= x(x−3) − 1(x−3)

= (x−1) (x−3)

Either, x−1 = 0 or, x − 3 = 0

⇒ x = 1 ⇒ x = 3

Therefore, the zeroes of

p(x) = x² − 4x + 3 are α = 1 and β = 3

Now, Coefficient of x², a = 1

Coefficient of x,b = −4

Constant term, c = 3

(ix) x² – 6x – 7

Ans: Let p(x) = x² – 6x – 7

∴ p(x) = x² – 7x + x – 7 

= x(x – 7) + 1(x – 7)

= (x – 7) (x + 1)

∴ The value of p(x) is zero, when

Therefore, the zeroes of p(x) = x² – 6x – 7 are α = 7 and β = – 1 

Now, Coefficient of x² a = 1 

Coefficient of x, b = – 6 

Constant term, c = – 7

(x) 2x² – 5x – 7

Ans: Let p(x) = 2x² – 5x – 7

∴ p(x) = 2x² + 2x – 7x – 7

= 2x(x + 1) -7 (x + 1)

= (2x – 7) (x + 1) 

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

(i) 1/4, -1

Ans:

(ii) √2, 1/3

Ans: 

(iii) 0, √5

Ans: 

(iv) 1, 1

Ans:

(v) 1/4, 1/4 

Ans: 

(vi) 4, 1 

Ans: