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Q1. If the two zeroes of the quadratic polynomial 7x² - 15x - k are reciprocals of each other, the value of k is:

• 1) 5
• 2)
• 3) -7
• 4) 7

Solution

We know that if the roots are reciprocal of each other leads to product of the roots being equal to 1. The product of roots =

Q2. Find the zeros of the polynomial f(x) = 6x² - 3 

Solution

 f(x) = 6x² - 3        = 3(2x² - 1)        =  The zeros of f(x) are given by f(x) = 0 That is  = 0 Hence the zeros of f(x) = 6x² - 3 are: and .

Q3. If and are the zeroes of the polynomial 4x² + 3x + 7, then is equal to:

• 1)
• 2)
• 3)
• 4)

Solution

Q4. If α, ß are zeroes of polynomial f(x) = x² + px + q then polynomial having and as its zeroes is :

• 1) x² - px + q
• 2) x² + qx + p
• 3) px² + qx + 1
• 4) qx² + px + 1

Solution

Q5. Find the zeroes of the quadratic polynomial 4x² - 7.

Solution

4x² - 7  = (2x)² -  =  are the zeroes of the given polynomial.

Q6. Form a quadratic polynomial whose one of the zeroes is -15 and sum of the zeroes is 42.

Solution

One of the zero = -15 Sum of the zeroes = 42 Other zero = 42 + 15 = 57 Product of the zeroes = 57 (-15) = -855 The quadratic polynomial is x² - (sum of zeroes)x + product of zeroes i.e., x² - 42x - 855

Q7. If (x + 1) is a factor of x² - 3ax + 3a - 7, then the value of a is:

• 1) -2
• 2) 1
• 3) -1
• 4) 0

Solution

Given, (x + 1) is a factor of x² - 3ax + 3a - 7.

Q8. What must be added to f(x) = 6x⁵ + 5x⁴ + 11x³ - 3x² + x + 5 so that it may be exactly divisible by g(x) = 3x² - 2x + 4?

Solution

Q9. If the polynomial p(x) is divisible by x - 4, and 2 is a zero of p(x), then find p(x).

• 1) x² – 6x+ 8
• 2) x² + 6x – 8
• 3) x² + 6x + 8
• 4) x² – 6x – 8

Solution

Since, 2 is a zero of p(x), (x - 2) is a factor of p(x). Also, given that (x - 4) is a factor of p(x). Using Division Algorithm, we get p(x) = (x - 2) (x - 4) = x² - 6x+ 8

Q10. From a quadratic polynomial whose one of the zeroes is -15 and sum of the zeroes is 42.

Solution

One of the zero = -15 Sum of the zeroes = 42 Let the other zero be x. -15 + x = 42  x = 42 + 15 = 57 Product of the zeroes = 57 (-15) = -855 The required quadratic polynomial is x² - 42x - 855

Q11. The remainder on dividing x³ + 2x² + kx + 3 by x - 3 is 21. Sanju was asked to find the quotient. He was a little puzzled and was thinking how to proceed. His classmate gunjan helped him by suggesting that he should first find k and then proceed further. Explain how the question was solved. 

Solution

Let p(x) = x³ + 2x² + kx + 3 Using Remainder theorem, we have: p(3) = 3³ + 2 3² + 3k + 3 = 21 k = -9 Thus, p(x) = x³ + 2x² - 9x + 3 When p(x) is divided by (x - 3), the quotient is x² + 5x + 6.

Q12. If (x + 1) is a factor of 2x³ + ax² + 2bx + 1, then find the values of a and b given that 2a - 3b = 4.

• 1) a = 2, b = 5
• 2) a = 2, b = 2
• 3) a = 5, b = 5
• 4) a = 5, b = 2

Solution

Since, (x + 1) is a factor of f(x) = 2x³ + ax² + 2bx + 1, f(-1) = 0 -2 + a - 2b + 1 =0 a - 2b = 1...(1) Given, 2a - 3b = 4...(2) Solving (1) and (2), we get, a = 5 b = 2

Q13. If the polynomial f(x) = 3x⁴ + 3x³ - 11x² - 5x + 10 is completely divisible by 3x² - 5, find all its zeroes.

Solution

Since, 3x² - 5 divides f(x) completely (3x² - 5) is a factor of f(x) 3 (x² - ) is a factor of f(x) is a factor of f(x) are zeroes of f(x)     (x² + x - 2) is a factor of f(x) (x² + 2x - x - 2) is a factor of f(x) (x + 2) (x - 1) is a factor of f(x) -2 and 1 are zeroes of f(x) Thus, all the zeroes of f(x) are, -2 and 1.

Q14. The sum and the product of the zeroes of a quadratic polynomial are  respectively, then the polynomial is:

• 1) 2x² - x - 1
• 2) 2x²- x + 1
• 3) 2x² + x + 1
• 4) 2x² + x - 1

Solution

Q15. The number of zeroes for the polynomial y = p (x) from the given graph is:

• 1) 3
• 2) 2
• 3) 0
• 4) 1

Solution

The number of zeroes of the polynomial y=p(x) from the given graph=number of points at which the graph intersects x axis. Since the graph intersects the x-axis at only one point, the number of zeroes of the polynomial is one.

Q16. Check whether x² + 3x + 1 is a factor of 3x⁴ + 5x³ - 7x² + 2x + 2.

Solution

Remainder = 0 Thus, x² + 3x + 1 is a factor of 3x⁴ + 5x³ - 7x² + 2x + 2.

Q17. If the squared difference of the zeros of the quadratic polynomial f (x) = x² + px + 45 is equal to 144, find the value of p.

Solution

Let the zeros of quadratic polynomial f (x) = x² + px + 45 be α and β α + β = -p αβ = 45 (α - β)² = 144  (α - β)² - 4αβ = 144  p² - 445 = 144  p² = 144 + 180 = 324  p² = 324   or p =  Hence p = 18

Q18. Sum and the product of zeroes of the polynomial x² +7x +10 is

• 1) -7 and 10
• 2) 7 and -10
• 3)
• 4)  

Solution

Q19. The number of polynomials having zeroes -2 and 5 is:

• 1) 2
• 2) 3
• 3) 1
• 4) more than 3

Solution

The polynomials having -2 and 5 as the zeroes can be written in the form k(x + 2) (x - 5), where k is a constant. Thus, number of polynomials with roots -2 and 5 are infinitely many, since k can take infinitely many values.

Q20. What must be added to f(x) = 2x⁴ + 6x³ - 4x + 8, so that the resulting polynomial is divisible by g(x) = x² - x + 1.

• 1) 6x + 2
• 2) −6x + 2
• 3) −6x + 2
• 4) 6x - 2

Solution

By the division algorithm, f(x) = g(x) q(x) + r(x)f(x) + [−r(x)] = g(x) q(x) + r(x) Remainder r(x) = −6x + 2Adding -r(x) = 6x - 2 to f(x) gives the polynomial which is divisible by g(x).

Q21. If α and β are the roots of the polynomial f (x) = x² - 2x + 3, find a polynomial whose roots are α + 2, β + 2.

Solution

Since α and β are the roots of the polynomial f (x) = x² - 2x + 3 α + β = 2 αβ = 3 Let S and P be the sum and product of the zeros of the required polynomial. Then, S = α + 2 + β + 2 = α + β + 4 = 2 + 4 = 6 P = (α + 2)(β + 2)+ 4 = αβ + 2(α + β) = 3 + 22 + 4 = 11 Hence the required polynomial g(x) is given by g (x) = k(x² - Sx + P) g (x) = k(x² - 6x + 11), where k is any nonzero real number.

Q22. Obtain all the zeros of the polynomial f(x) = 2x⁴ + x³ - 14x² - 19x - 6, if two of its zeros are -2 and -1.

Solution

Q23. Check whether x² - x + 1 is a factor of x³ - 3x² + 3x - 2

Solution

Quotient = x - 2, Remainder = 0 Thus, x² - x + 1 is a factor of x³ - 3x² + 3x - 2.

Q24. If the sum of the zeros of quadratic polynomial f (x) = kt² + 2t + 3k is equal to their product, find the value of k.

Solution

Let the zeros of quadratic polynomial f (x) = kt² + 2t + 3k be α and β

Q25. On dividing the polynomial p(x) = 9x⁴ - 4x² + 4 by the polynomial g(x) = 3x² + x - 1, the remainder is ax - b. Find a and b.

Solution

Remainder = -x + 4 Comparing this with ax - b, we get, a = -1 b = -4

Q26. If α, β, γ are zeroes of polynomial 6x³ + 3x² - 5x + 1, then find the value of α^-1 + β^-1 + γ^-1

Solution

Given p(x) = 6x³ + 3x² - 5x + 1 a = 6, b = 3, c = -5, d = 1 α, β and γ are zeroes of p(x).

Q27. Verify that -2 and -5 are the zeros of the polynomial x² + 7x + 10.  

Solution

A real number k is said to be a zero of polynomial p(x), if p (k) = 0 p (-2) = (-2)² + 7  (-2) + 10          = 4 – 14 + 10 p (-2) = 0   p (-5) = (-5)² + 7  (-5) + 10          = 25 – 35 + 10 p (-5) = 0 Hence -2 and -5 are the zeros of the given polynomial.

Q28. Find the quotient and remainder when x⁵ +3x⁴ - 5x³ + 14x² + 39x - 11 is divided by 4x + x² - 2.

Solution

Rearranging the terms of divisor in descending order of degree, we get   Clearly, the degree of remainder 9x + 5 is less than the degree of divisor. x² + 4x - 2. ∴ Quotient, q(x) = x³ - x² + x + 8 and remainder, r(x) = 9x + 5.

Q29. Using the division algorithm, find the divisor if the quotient is p - 3, the remainder is p - 4 and the dividend is p² + 3p - 7.

• 1) p + 3
• 2) p - 1
• 3) p - 3
• 4) p - 4

Solution

f(x) = p² + 3p - 7, g(x) = ?, r(x) = p - 4 and q(x) = p - 3 f(x) = q(x) g(x) + r(x) p² + 3p - 7 = (p - 3)g(x) + p - 4 ∴ (p - 3)g(x) = p² + 3p - 7 - p + 4 ∴ (p - 3)g(x) = p² + 2p - 3 ∴ g(x) = ∴ g(x) = ∴ g(x) = ∴ g(x) = ∴ g(x) = p - 1

Q30. On dividing 3x³ - 2x² + 5x - 5 by a polynomial p(x), the quotient and the remainder are x² - x + 2 and -7 respectively. Find p(x).

Solution

Q31. If two of the zeroes of the polynomial p(x) = 5x⁴ - 5x³ - 33x² + 3x + 18 are, find the other two zeroes.

Solution

Since are zeroes of p(x) are factors of p(x) is a factor of p(x) Or, (5x² - 3) is a factor of p(x) p(x) = (5x² - 3) (x² - x - 6) = (x - 3) (x + 2) Other zeroes of p(x) are 3 and -2.

Q32. If the remainder on division of x³ + 2x² + kx + 3 by x - 3 is 21, then find the quotient and value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx - 18.

Solution

We have the following terms: Dividend: f(x) = x³ + 2x² + kx + 3, Divisor: g(x) = x - 3 and remainder, r (x) = 21 Using the remainder thermo, we have the following expression: f(3) = 21   The polynomial p(x) is x³ + 2x² - 9x + 3. Now, on long division, we get   Thus, x³ + 2x² - 9x + 3 = (x - 3 ) (x² + 5x + 6) + 21 ∴ The quotient = x² + 5x + 6 Clearly, x³ + 2x² - 9x - 18 = (x - 3 ) (x² + 5x + 6)  = (x - 3 ) (x + 2)(x + 3) Therefore, the zeroes of x³ + 2x² - 9x - 18 are 3, -2 and -3.

Q33. If one solution of the equation 3x² = 8x + 2k + 1 is seven times the other. Find the solutions and the value of k.

Solution

3x² = 8x + 2k + 1 3x² - 8x - 2k - 1 = 0 Let α is one zero. Then, β = 7α is the other zero α + 7α = 8/3 8α = 8/3 α = 1/3 β = 7/3

Q34. Find all other zeroes of the polynomial p (x) = 2x³ + 3x² - 11x - 6, if one of its zero is -3.

Solution

p (x) = 2x³ + 3x² - 11x - 6 One zero is -3. So, (x + 3) is a factor of p (x). Quotient = 2x² - 3x - 2 = (2x + 1) (x - 2) Thus, the other zeroes of p(x) are 2 and. Hence, all zeroes of p (x) is -3, 2 and.

Q35. Divide (6 + 23x + 25x² + 6x³) by (3 + 7x + 2x²) and verify the division algorithm.

Solution

Quotient = 2 + 3x Remainder = 0 Verification: It can be seen that: (3 + 7x + 2x²) (2 + 3x) + 0 = 6 + 23x + 25x² + 6x³ Dividend= Divisor x Quotient + Remainder Hence, the division algorithm is verified.

Q36. On dividing 3x³ - 2x² + 5x - 5 by a polynomial p(x), the quotient and remainder are x² - x + 2 and -7 respectively. Find p(x).

Solution

_{By division algorithm,}

Q37. Find all zeroes of polynomial 4x⁴ - 20x³ + 23x² + 5x - 6 if two of its zeroes are 2 and 3.

Solution

Given 2 and 3 are the zeroes of the polynomial. Thus(x - 2) (x - 3) are factors of this polynomial 4x⁴ - 20x³ + 23x² + 5x - 6 = (x² - 5x + 6) (4x² - 1) Thus, 4x⁴ - 20x³ + 23x² +5x-6=(x - 2) (x - 3) (2x - 1) (2x + 1) Hence, 2, 3, are the zeroes of the given polynomial.

Q38. Divide x⁴ - 3x² + 4x + 5 by x² - x + 1 and verify that Dividend = Divisor  Quotient + Reminder.

Solution

Arranging the terms in descending order, we get   Clearly, the degree of remainder 8 is zero, which is less than the degree of x² - x + 1. Quotient = x² + x - 3 and remainder = 8. Verification: Divisor  Quotient + Reminder = (x² - x + 1) (x² + x - 3) + 8 = x⁴ + x³ - 3x² - x³ - x² + 3x + x² + x - 3 + 8 = x⁴ + x³- x³ - 3x² - x² + x² + 3x + x -3 +8 = x⁴ - 3x² + 4x + 5 = dividend

Q39. Find the zeroes of x² + 5x -  and verify the relation between the zeroes and coefficients of the polynomial.

Solution

Hence, the relation between the zeroes and coefficients of the polynomial is verified.

Q40. Divide x⁴ - 3x² + 4x + 5 by x² - x + 1, find quotient and remainder.

Solution

Quotient = x² + x - 3 Remainder = 8

Q41. If are the zeroes of f(x) = px² - 2x + 3p and = then the value of p is:

• 1)
• 2)
• 3)
• 4) -

Solution

Q42. If α and β are the roots of the polynomial f (x) = 6x² + x - 2, find the value of

Solution

Since α and β are the roots of the polynomial f (x) = 6x² + x - 2

Q43. Find the quotient and reminder using division algorithm: f(x) = 15x³ - 20x² + 13x - 12, g(x) = 2 - 2x + x²  

Solution

We have the following equations: f(x) = 15x³ - 20x² + 13x - 12 and g(x) = 2 - 2x + x² is x² - 2x + 2 Clearly, degree of f(x) = 3 and degree of g(x) = 2. Therefore, the degree of quotient q(x) = 3 - 2 = 1 and degree of remainder r(x) is less than the degree of g(x), i.e. 2. Let quotient, q(x) = ax + b and remainder, r(x) = cx + d. Using division algorithm, we have the following equation: p(x) = g(x) × q(x) + r(x) ∴15x³ - 20x² + 13x - 12 = (x² - 2x + 2) (ax + b) + (cx + d) = ax³ + bx² - 2ax² - 2bx + 2ax + 2b + cx + d = ax³ + (b - 2a)x² + (2a + c - 2b) x + (2b + d) Comparing the coefficient of same powers of x on both sides, we get a = 15  [Comparing the coefficient of x³] b - 2a = -20  [Comparing the coefficient of x²] 2a + c - 2b = 13  [Comparing the coefficient of x] 2b + d = -12  [Comparing the constant terms ] Solving the above equation, we get the following values: a = 15, b = 10, c = 3 and d = -32 Hence, quotient, q(x) = 15x + 10 and remainder, r(x) = 3x - 32.        

Q44. It is being given that 1 is one of the zeros of the polynomial 7x - x³ - 6. Find its other zeros.

Solution

It is given that 1 is a zero of the polynomial -x³ + 7x - 6. So, (x - 1) is a factor of -x³ + 7x - 6. Now, we have: (-x³ + 7x - 6) (x - 1) = (-x² - x + 6) - (x² + x - 6) = -(x + 3) (x - 2) Other zeros are -3 and 2.

Q45. Find the zeroes of the polynomial 100x² - 81.

Solution

Since a² - b² = (a + b)(a - b) 100x² - 81 = (10x + 9)(10x - 9) Thus, the required zeroes are  and 

Q46. Which of the following functions are not polynomials. Give reason for your answer. i)  (ii) x² + x + 3 (iii) (iv)

Solution

In (i) the powers of x are 3, -2, -1, and 0. Since -2, -1 are not non-negative integers, therefore it is not a polynomial. In (ii) the powers of x are 2, 1, 0. Since 2, 1 and 0 are positive integers, therefore it is a polynomial. In (iii) the powers of the variable u are , 1, and 0. Since  is not non-negative integer, therefore it is not a polynomial. In (iv) the powers of x are 3, 0, 1 and 0. As 3, 0, 1, 0 are positive integers, therefore it is a polynomial.

Q47. What must be subtracted from the polynomial 8x⁴ + 14x³ + x² + 7x + 8 so that the resulting polynomial is exactly divisible by 4x² - 3x + 2?

Solution

Thus, when 6x + 2 is subtracted from the given polynomial 8x⁴ + 14x³ + x² + 7x + 8, then it will be divisible by 4x² - 3x + 2.

Q48. Find all the zeroes of p(x) = x³ - 9x² - 12x + 20, if (x + 2) is a factor of p(x).

Solution

Given, x + 2 is a factor of p(x). p(x) = (x + 2) (x² - 11x + 10) = (x + 2) (x² - 10x - x + 10) = (x + 2) (x (x - 10) - (x - 10)) = (x + 2) (x - 1) (x - 10) Zeroes of p(x) are -2, 1, 10.

Q49. If (x + a) is a factor of two polynomials x² + px + q and x² + mx + n, then prove that: 

Solution

(x + a) is a factor of x² + px + q (-a)² + p(-a) + q = 0 a² - ap + q = 0. a² = ap - q ..(i) (x + a) is a factor of x² + mx + n (-a)² + m (-a) + n = 0 a² - am + n = 0 a²= am - n ...(ii) From (i) and (ii), ap - q = am - n ap - am = q - n a(p - m) = q - n a =

Q50. What must be added to x³ - 4x² + x - 6 so that x² + 2x - 3 becomes its factor?

Solution

(x³ - 4x² + x - 6)(x² + 2x - 3) Remainder = 16x - 24 Thus, the required expression that must be added is 16x - 24

Q51. If α, β are the zeros of the polynomial 25p² - 15p + 2, find a quadratic polynomial whose zeros are  and 

Solution

α and β are two roots of 25p² - 15p + 2  _{αβ =}  For the required polynomial, Sum of the zeros = Product of zeroes = a = 8, b = -30, c = 25 Therefore, required polynomial is 8x² - 30x + 25.

Q52. The graph of the polynomial p(x) intersects the x-axis three times in distinct points, then which of the following could be an expression for p(x):

• 1) 3x + 3
• 2) 4 - 4x - x² + x³
• 3) 3x² + 3x - 3
• 4) x² - 9

Solution

Since, the graph of the polynomial p(x) intersects the x-axis at three distinct points, it will have 3 distinct roots. Hence, it will be a cubic polynomial. Thus, amongst the given alternatives the polynomial p(x) can be 4 - 4x - x² + x³.

Q53. If a and b are the zeroes of the polynomial x² - 5x + k such that a - b = 1, find the value of k.

Solution

_{x² - 5x + k}  .....(i) α - β = 1 ...(ii) Solving (i) and (ii) we get α = 3 and β = 2 Hence, k = αβ = 3 × 2 = 6  

Q54. On dividing the polynomial 4x⁴ - 5x³ - 39x² - 46x - 2 by the polynomial g(x) the quotient and remainder are x² - 3x - 5 and -5x + 8 respectively. Find g(x).

Solution

p(x) = 4x⁴ - 5x³ - 39x² - 46x - 2 q(x) = x² - 3x - 5 r(x) = -5x + 8 g(x) = = g(x) = 4x² + 7x + 2

Q55. On dividing the polynomial p(x) = 5x⁴ - 4x³ + 3x² - 2x + 1 another polynomial g(x) = x² + 2, if the quotient is ax² + bx + c, find a, b and c.

Solution

Comparing the quotient obtained with ax² + bx + c, we get, a = 5 b = -4 c = -7

Q56. The graph of y = p(x) is given below. The number of zeroes of p(x) are :

• 1) 0
• 2) 3
• 3) 2
• 4) 4

Solution

The number of zeroes of a polynomial p(x) is the number of times its graph intersects or touches the x-axis. The given graph touches the x-axis twice, so the number of zeroes of the polynomial y = p(x) is 2.

Q57. Obtain all the zeroes of the polynomial f(x) = x⁴ - 7x³ + 10x² + 14x - 24, if two of its zeroes are + and

Solution

Since, and are zeroes of f(x). (x - ) and (x + ) are factors of f(x) (x - ) (x + ) is a factor is f(x) (x² - 2) is factor of f(x) We have,   (x² - 7x + 12) is a factor of f(x) (x - 3) (x - 4) is a factor of f(x) (x - 3) and (x - 4) are factors of f(x) x = 3 and x = 4 are other zeroes of f(x) Hence, all the zeroes of f(x) are, - 3 and 4.

Q58. If the polynomial x⁴ - 6x³ + 16x² - 25x + 10 is divided by another polynomial x² - 2x + k, the remainder comes out to be x + a, find the values of k and a.

Solution

If the polynomial x⁴ - 6x³ + 16x² - 25x + 10 on division by x² - 2x + k leaves remainder (x + a). Then, polynomial (x⁴ - 6x³ + 16x² - 25x + 10) - (x + a) on division by x² - 2x + k leaves remainder zero. On long division of polynomial (x⁴ - 6x³ + 16x² - 25x + 10) - (x + a) = x⁴ - 6x³ + 16x² - 26x + 10 -a by x² - 2x + k, the remainder obtained is (-10 + 2k ) x + ( 10 - a - 8k + k²). (-10 + 2k )x + ( 10 - a - 8k + k²) = 0 -10 + 2k = 0 and 10 - a - 8k + k² = 0 k = 5 and a = -5

Q59. Find the zeroes of the polynomial

Solution

  The zeros are 

Q60. Find the zeroes of the polynomial x² + 3x - 10 and verify the relation between its zeroes and coefficient.  

Solution

We have p(x) = x² + 3x - 10 = (x + 5)(x - 2) For any zero, p(x) = 0  x² + 3x - 10 = 0  (x + 5)(x - 2) = 0  (x + 5) = 0 OR (x - 2) = 0  x = - 5 OR x = 2 The zeros of p(x) = x² + 3x - 10 are as α = -5 and β = 2 Now, sum of zeros = α + β = -5 + 2 = -3 =  Product of zeros = αβ = (-5) × 2 = - 10 =     

Q61. Find the quotient and reminder using division algorithm: f(x) = 10x⁴ + 17x³ - 62x² + 30x - 3, g(x)=2x² + 7x + 1  

Solution

We have the following equations: f(x) = 10x⁴ + 17x³ - 62x² + 30x - 3 and g(x) = 2x² + 7x + 1. Clearly, degree of f(x) = 4 and degree of g(x) = 2. Therefore, degree of quotient, q(x) = 4 - 2 = 2 and degree of remainder, r(x) is less than degree of g(x), i.e. 2. Let quotient, q(x) = ax² + bx + c and remainder, r(x) = dx + e Using division algorithm, we have the following equation: f(x) = g(x) × q(x) + r(x) ∴10x⁴ + 17x³ - 62x² + 30x - 3 = (2x² + 7x + 1) (ax² + bx + c) + (dx + e) = 2ax⁴ + 2bx³ + 2cx² + 7ax³ + 7bx² + 7cx + ax² + bx + c + dx + e = 2ax⁴ + (7a + 2b)x³ + (a + 7b + 2c)x² + (b + 7c + d)x + (c + e) Comparing the coefficient of same powers of x on both sides, we get 2a = 10  [Comparing the power of x⁴] 7a + 2b = 17 [Comparing the power of x³] a + 7b + 2c = -62 [comparing the power of x²] b + 7c + d = 30  [comparing the power of x] c + e = -3  [comparing the constant terms] Solving the above equations, we get the following values: a = 5, b = -9, c = -2, d = 53 and e = -1 Hence, quotient, q (x) = 5x² - 9x - 4 and remainder, r(x) = 53x - 1.   

Q62. Find the number of zeros in the following graphs.

Solution

In (i) the graph intersects the x-axis at two points, hence there are two zeros. In (ii) the graph intersects the x-axis at three points, hence there are three zeros.

Q63. Find the value of k for which the polynomial x⁴ + 10x³ + 25x ² + 15 x + k is exactly divisible by x + 7

Solution

Q64. Find the polynomial of least degree which should be subtracted from the polynomial so that it is exactly divisible by x² - x + 1.

Solution

Here, p(x) = x⁴ + 2x³ - 4x² + 6x - 3, g(x) = x² - x +1 On dividing p(x) by g(x) Therefore (x-1) must be subtracted from the polynomial p(x) to make it divisible by g(x).

Q65. and are zeroes of the quadratic polynomial x² - 6x + a. Find the value of 'a' if 3 + 2 = 20.

Solution

p(x) = x² - 6x + a  ...(1) Given, 3α + 2β = 20... (2) Multiplying (1) by 3, _{3α + 3β = 18} .....(3) On solving (2) and (3), we get: β = -2, α = 8 Now, αβ = a (8) (-2) = a a = -16

Q66. On dividing the polynomial p(x) by polynomial g(x) = 4x² + 3x - 2, the quotient is q(x) = 2x² + 2x - 1 and remainder is r(x) = 14x - 10. Find the polynomial p(x).

Solution

p(x) = q(x) g(x) + r(x) (By division algorithm) = (2x² + 2x - 1) (4x² + 3x - 2) + 14x - 10 = 8x⁴ + 6x³ - 4x² + 8x³ + 6x² - 4x - 4x² - 3x + 2 + 14x - 10 = 8x⁴ + 14x³ - 2x² + 7x - 8

Q67. Sum of the two zeroes of a polynomial of degree 4 is -1 and their product is -2. If other two zeroes are and -. Find the polynomial.

Solution

Two zeroes of polynomial p(x) and and - (x - ) and (x + ) are factors of p(x) (x - ) (x + ) is a factor of p(x) x² - 3 is a factor of p(x) Since, degree of p(x) is 4, so the other factor of p(x) will also be a quadratic polynomial. Now, it is also given that the sum of the other two zeroes of p(x) is -1 and their product is -2. So, the other factor is x² - (sum of roots) x + product of roots i.e., x² + x - 2 p(x) = (x² - 3) (x² + x - 2) = x⁴ + x³ - 5x² - 3x + 6

Q68. Divide the polynomial p(x) = 3x² - x³ - 3x + 5 by g(x) = x - 1 - x² and find its quotient and remainder.

Solution

Quotient = x - 2, Reminder = 3

Q69. Divide (2x² + x - 20) by (x + 3) and verify the result by division algorithm.

Solution

Quotient = 2x - 5, remainder = -5 Verification: It can be seen that: (x + 3) (2x - 5) - 5 = 2x² + 6x - 5x - 15 - 5 = 2x² + x - 20 Dividend= Divisor x Quotient + Remainder Hence, the division algorithm is verified.

Q70. On dividing the polynomial p(x) = 5x⁴ -4x³ + 3x² - 2x + 1 by another polynomial g(x) = x² + 2, if the quotient is ax² + bx + c, find a, b and c.

Solution

Comparing with the obtained quotient with ax² + bx + c, we get, a = 5 b = -4 c = -7

Q71. By applying the division algorithm prove that the polynomial g(x) = x² + 3x  + 1 is a factor of the polynomial f(x) = 3x⁴ + 5x³  - 7x² + 2x + 2.

Solution

Q72. Find the quotient and reminder using division algorithm: f(x) = x³ - 6x² + 11x - 6, g(x) = x + 1  

Solution

We have: f(x) = x³ - 6x² + 11x - 6 and g(x) = x + 1 Clearly, degree of f(x) = 3 and degree of g(x) = 1. Therefore, the degree of quotient is q(x) = 3 - 1 = 2 and the degree of remainder is r(x) = 0 Let quotient q(x) = ax² + bx + c and remainder r(x) = k. Using division algorithm, we have f(x) = g(x) × q(x) + r(x)   Comparing the coefficient of same powers of x on both sides, we get a = 1  [Comparing the coefficient of x³] a + b = -6  [Comparing the coefficient of x²] b + c = 11  [Comparing the coefficient of x] c + k = -6  [Comparing the constant terms ] Solving the above equations, we get the following values: a = 1, b = -7, c = 18, and k = -24 ∴ Quotient is q (x) = x² - 7x + 18 and remainder is r(x) = -24.        

Q73. Find the zeroes of the quadratic polynomial 6x² - 3 - 7x, and verify the relationship between the zeroes and the coefficients.

Solution

6x² - 7x - 3 = 6x² - 9x + 2x - 3 = 3x(2x - 3) + (2x - 3) = (3x + 1)(2x - 3) Therefore, the two zeroes are and  Hence, the relationship between the zeroes and the coefficients is verified.

Q74. Find the values of a and b so that the polynomial p(x) = x⁴ + x³ + 8x² - ax + b is exactly divisible by x² - 1.

Solution

Since, remainder has to be zero, we have: 1 - a = 0 a = 1 b + 9 = 0 b = -9