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SEBA Class 10 Mathematics 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) No zeroes.
(ii) 1
(iii) 3
(iv) 2
(v) 4
(vi) 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 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
| Exercise 2.3 |
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:
| Exercise 2.4 |
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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