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 v = 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)/1 = (Coefficient term)/Coefficient of x².

(ii) 4s² – 4s +1

Ans: 4s² – 4s +1 = ( 2s – 1)²

∴ The two zeroes are ½, ½ 

Sum of the zeroes = ½ + ½ = -(-4)/4 = (Coefficient of ×) /(Coefficient of ײ)

Product of two zeroes = (½) (⅓) =¼  (Coefficient)/(coefficient of ײ) .l

(iii) 6ײ-3-7x = 6x² – 7x – 3 = 6x – 9x +2x-3 = (2x-3) (3x+1) Putting 2x-3 = 0 and 3x +1 = 0

Ans: We get x = 3/2 and x= -1)2 Zeroes of the quadratic polynomial 2 p(x)=6x²-7x-3

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

= -(-7)/6 = (Coefficient of ×)/ (coefficient of ײ) 

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

=(Constant term) / (coefficient of ײ) 

(iv) 4u² + 8u = 4u (u + 2)

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

Sum of the two zeroes  0 + (-) = -2 =(8)/4 = =(Constant term) / (coefficient of ײ) 

Product of the two zeroes = (0) (-2) = 0 = (Constant term) / (coefficient of ײ) 

(v) t²-15 = t² (√15) =( t²√15) (t-√15)

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

-(Coefficient of t)/Coefficient of t²)

 Product of the two zeroes

= (-√15) × (√15) = -15 = (-15) = constant 

Sum of the two zeroes = Product of the two zeroes (-√15) + √15=0= Coefficient of term) /(Coefficient of t²)

(vi) 3x² – × – 4  3x² = – 4× + 3×- 4

Ans: 3x² – × – 4  3x² = – 4× + 3×- 4

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

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

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

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

Product of the two zeroes = (-) × (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: 

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 a = 5 and β = 2 

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

Product of zeroes   =  α + 𝛃  = (5×2) x + a

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

= x² – 7 x 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 

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

Ans:  

(iii) p ( x ) = x⁴  = 5x + 4x + 5g (x) =  g (x) = 2- x²

Ans: 

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

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

(vi) p(x) =  2 x 5 – 5 x 4 + 7 x 3 + 4 x 2 – 10 x 2 – 11 g (x) = g(x) = 2 x 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 + 3 x + 1, 3x⁴ + 5 x³-7x² + 2x +2

Ans:

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

Ans: 

3. Obtain all other zeroes of 3×4 + 6x³- 2x² – 10 – 5, if two of its zeroes

Ans: 

4. On dividing x3x2 = x+ by a polynomial g (x) the quotient and reminded were x – and – 2x – 2x + 4 respectively . find (x) g

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 3x ² – x² – 3x + 1 is 1, then find all other zeros.

Ans: 

(ii) If two zeros of polynomial x ⁴ + x 3 – 9x ² – 3x + 18 are sqrt(3) and – sqrt(3) then find all the other zeros.

Ans:  

(iii) If two zeros of polynomial ⁴ + 2x³ – 26x² – 54x – 27 are root 3,√ 3and – root 3,√ 3 then find all the others.

Ans: 

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

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

Ans: 

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.

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 = Division x Quotient.

Dividend – Remainder is always divisible by the division.

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

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