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Polynomials Test - 6

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Polynomials Test - 6
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  • Question 1
    1 / -0
    If the sum of zeros of the polynomial 12ax2 + (5 – b)x + 10 is 0 and the product of all zeros is 1, then the zeros of the polynomial are
    Solution
    12ax2 + (5 – b)x + 10
    According to the question:
    Sum of all zeros = =
    0 =
    0 = b – 5
    b = 5

    Product of all zeros = =

    12a = 10

    Now, becomes .

    To find the zeros of the polynomial, we need to make it equal to 0.
    i.e. 20x2 – 25x – 45 = 0 or 4x2 – 5x – 9 = 0
    => 4x2 – 9x + 4x – 9 = 0
    => x(4x – 9) + 1(4x – 9) = 0
    => (4x – 9)(x + 1) = 0
    => x = -1,
    Hence, the zeros of the polynomial 20x2 – 25x – 45 are -1 and
  • Question 2
    1 / -0
    If the sum of the product of the zeros taken two at a time of the polynomial f(x) = 12x3 + 2qx2 – 5px + 18 is -15 and the sum of all zeros is 10, then the value of p + q is
    Solution
    Given polynomial: 12x3 + 2qx2 – 5px + 18
    Sum of product of the zeros taken two at a time =

    p = 36
    Sum of all zeros =

    q = -60
    p + q = 36 + (-60) = -24
  • Question 3
    1 / -0
    If 8 and d are the roots of the quadratic equation 2x2 – 10rx + 18 = 0, then the value of 'r' is
    Solution
    Given equation: 2x2 – 10rx + 18 = 0
    Now, if 8 and d are its roots, then
    Sum of roots = 8 + d =
    => 8 + d = 5r ... (1)
    Product of roots = 8d =
    => 8d = 9 ... (2)
    From (2), we get d =

    Putting this value in (1), we get

    r =
  • Question 4
    1 / -0
    Find out the cubical polynomial with sum, sum of the product of its zeros taken two at a time, and product of its zeros as 11, -15 and -9, respectively.
    Solution
    Let the polynomial be ax3 + bx2 + cx + d and the zeros be p, q and r.
    It is given that:
    Sum of all zeros =
    Sum of the product of zeros =
    Product of its zeros =
    If a = 1, then b = -11, c = -15, d = 9
    Hence, the polynomial is x3 – 11x2 – 15x + 9.
  • Question 5
    1 / -0
    If s and r are the zeros of the polynomial f(x) = 4x2 – 8q(x + 1) + 8c, then is equal to
    Solution
    4x2 – 8q(x + 1) + 8c = 4x2 -8qx – 8q + 8c = 4x2 – 8qx + 8(c – q)
    Since s and r are the zeros;
    => s + r = = 2q and

    =
    =
    =
    = 2(2c – 2q – 2q + 1)
    = 4c – 8q + 2
  • Question 6
    1 / -0
    When 2x3 + 8x2 – 4x -12 is divided by x2 + 2x – 1, the quotient and the remainder respectively are:
    Solution
    By long division method, we get

    Therefore, the quotient will be (2x + 4) and the remainder will be (-10x – 8).
  • Question 7
    1 / -0
    What should be subtracted from f(x) = 9x3 – 12x2 + 40x – 72 so that f(x) is exactly divisible by 3x2 – 2x + 1?
    Solution
    By long division method, we get



    We must subtract the remainder so that f(x) is exactly divisible by 3x2 – 2x + 1.
    Hence, (33x – 70) is to be subtracted.
  • Question 8
    1 / -0
    If one zero of the polynomial f(x) = x2 – (k2 – 2)x + 8 is double of the other zero, then k is equal to
    Solution
    f(x) = x2 – (k2 – 2)x + 8
    According to the question,
    Let one zero be a and the other zero be 2a.

    Sum of zeros =
    =>
    => 3a = k2 – 2

    Product of zeros =
    =>
    =>

    Let a = 2
    3a = k2 – 2
    3(2) = k2 – 2
    8 = k2

    Let a = -2
    3a = k2 – 2
    3(-2) = k2 – 2
    -4 = k2, which is not possible.
    Hence,
  • Question 9
    1 / -0
    If f(x) = dx7 + px6 – gx5 + ax2 – fx + s, then which of the following is true?
    Solution
    Since the degree of the given polynomial is 7, it will have at most 7 zeros.
  • Question 10
    1 / -0
    What is the number of zeroes of the polynomial y = p(x) in the given graph?

    Solution


    Number of zeroes is the number of times the graph touches or cuts the x-axis.
    ∴ Number of zeroes is 4. [1]
  • Question 11
    1 / -0
    If a, b and c are roots of the equation 6x3+ 11x2 – 5 = 0, then the equation whose roots are (9a – 2b – 3c), (2a -3b – 9c) and (-9a + 7b + 14c) is
    Solution
    Given equation: 6x3 + 11x2 – 5 = 0
    Sum of zeros = … (1)
    To find the equation whose roots are (9a – 2b – 3c), (2a - 3b – 9c) and (-9a + 7b + 14c), we will solve as follows:

    Suppose the required equation is px3 + qx2 + rx + t= 0
    Sum of zeros = (9a – 2b – 3c) + (2a -3b – 9c) + (-9a + 7b + 14c) =
    => 2a + 2b + 2c =

    => 2(a + b + c) =

    From equation (1), we get



    Therefore, p = 3 and q = 11, which satisfy the equation given in option 1 only.
  • Question 12
    1 / -0
    For x3 + x2 – 3x + 6 to be a factor of x5 + 2x4 + ax2 – b + 2, the values of a and b should respectively be
    Solution
    For x3 + x2– 3x + 6 to be a factor of x5+ 2x4 + ax2– b + 2, the remainder should be zero.



    Now, the remainder should be equal to zero.
    Therefore, a – 5 = 0 and -b – 10 = 0
    => a = 5, b = -10
  • Question 13
    1 / -0
    If a and b are roots of the given equation, then which of the following equations has its roots as 2a and 2b?

    f(x) = 10x2- 32x + 24 = 0
    Solution
    We have
    f(x) = 10x2 – 32x + 24
    Solving the above equation, we get
    10x2 – 12x – 20x + 24 = 0
    Or (2x - 4)(5x - 6) = 0
    Or x = 2 and
    a = 2 and b = 6/5
    New roots of the required equation = 2a and 2b = 4 and
    From this, we get
    (x – 4)(x –) = 0
    Or x2 –– 4x + = 0
    Solving further, we get
    5x2 – 32x + 48 = 0, which is the required equation.
    Therefore, option 2 is the right answer.
  • Question 14
    1 / -0
    Which of the following should be subtracted from the dividend x8 - 6x + 8 so that -2 + x2 as its factor?
    Solution
    We have
    f(x) = -2 + x2 = x2 - 2
    g(x) = x8 - 6x + 8
    For f(x) to be a factor of the polynomial g(x), the remainder should be zero. So, on dividing g(x) by f(x).
    we get


    So, the remainder is -6x + 24.
    So, this should be subtracted from the dividend to make -2 + x2 as its factor.
  • Question 15
    1 / -0
    If x3 + ax2 + bx + 12 leaves remainder 4 when divided by (x - 1), which of the following is the value of a + b?
    Solution
    Let p(x) be x3 + ax2 + bx + 12
    g(x) = x - 1
    On division of p(x) by g(x), remainder = 4
    Now, g(x) is linear.
    So, g(x) = 0
    x - 1 = 0
    x = 1
    Now, (+1)3 + a (+1)2 + b (+1) + 12 = 4
    Or, 1 + a + b + 12 = 4
    a + b + 13 = 4
    a + b = 4 - 13
    a + b = -9
  • Question 16
    1 / -0
    The area of a trapezoidal shaped wooden board is 36x4 + 6x3 - 69x2 - 8x + 28. The lengths of the parallel sides of the shape are 15x2 - 4x + 4 and 9x2 + 8x - 18. What will be the height of the wooden board?
    Solution
    Length of the first parallel side (a) = 15x2 - 4x + 4
    Length of the second parallel side (b) = 9x2 + 8x - 18
    Area of the wooden board = 36x4 + 6x3 - 69x2 - 8x +28
    We know,
    Area of the trapezoid = (a + b) × h
    36x4 + 6x3 - 69x2 - 8x + 28
    = (15x2 - 4x + 4 + 9x2 + 8x - 18) × h
    = (24x2 + 4x - 14) × h = (12x2 + 2x - 7) × h
    So, finding the height:


    So, height = 3x2 - 4
  • Question 17
    1 / -0
    The perimeter of a square-shaped plot is 16x2+ 8x + 16. What will be the area of a square-shaped garden whose area isof the area of the given plot?
    Solution
    Perimeter of the square-shaped plot = 16x2+ 8x + 16
    We know that P = 4a (Let a be the side of the square plot.)
    16x2+ 8x + 16 = 4a
    a =
    a = 4x2 + 2x + 4
    Now, area of a square = a2
    (4x2+ 2x + 4)2
    Using trinomial equation:
    (16x4 + 4x2 + 16 + 16x3 + 16x + 32x2)
    = (16x4 + 36x2 + 16 + 16x3 + 16x)
    = 4(4x4 + 4x3+ 9x2 +4x + 4); which is the area of the plot.
    Since,it is 4 times the area of the garden;
    Area of the garden = 4x4 + 4x3 + 9x2 + 4x + 4
  • Question 18
    1 / -0
    An automobile company manufactures 35x6 + 63x5 + 56x4 + 64x3 + 27x2 + 24x + 21 cars. If there are 7x3 + 3 showrooms in a city, then how many cars are there in one showroom?
    Solution
    Total number of cars manufactured = 35x6 + 63x5 + 56x4 + 64x3 + 27x2 + 24x + 21
    Number of showroom in a city = 7x3 + 3
    Solving, we get:

    5x3 + 9x2 + 8x + 7 is the number of cars in one showroom.
    Therefore, option 2 is the right answer.
  • Question 19
    1 / -0
    If the weights of two different quantities are (3x3 - 6x2 - 4x + 8) kg and (5x3 - 10x2 + 2x - 4) kg, then what is the biggest measure that can weigh both the quantities?
    Solution
    Weights of two different quantities:
    (3x3 - 6x2- 4x + 8) kg and (5x3 - 10x2 + 2x - 4) kg

    Weight of the first quantity = (3x3 - 6x2- 4x + 8) kg
    = 3x2(x - 2) - 4(x - 2)
    = (x - 2)(3x2- 4)

    Weight of the second quantity = (5x3- 10x2+ 2x - 4) kg
    = 5x2(x - 2) + 2(x - 2)
    = (x - 2)(5x2+ 2)
    Therefore, required measure is the H.C.F. of both the weights, i.e (x - 2) kg.
  • Question 20
    1 / -0
    The radius and height of a cylindrical pipe are (5x2 - 27x) m and (3x2- 20x) m, respectively. Find the surface area of the pipe which is open from both the sides when x = .
    Solution
    Let the radius of the pipe be r and the height be h.

    r = (5x2 - 27x) m

    h = (3x2- 20x) m

    When x == 7;

    r = (5(7)2- 27(7)) m = 245 - 189 = 56 m

    h = (3(7)2- 20(7)) m = 147 - 140 = 7 m

    Surface area (open cylindrical pipe) = 2πrh

    = 2 ×× 56 × 7 = 2 × 22 × 8 × 7 = 2464 m2
  • Question 21
    1 / -0
    Find the roots of, if the polynomial 2x5+ 4x3– 3x2- px + q is exactly divisible by x2– 3.
    Solution
    2x5 + 4x3 – 3x2 - px + q is exactly divisible by x2 – 3.
    => Remainder must be zero.



    (30 – p)x + (q – 9) = 0
    30 – p = 0
    => p = 30 and q – 9 = 0
    => q = 9
    Now, becomes x2 + 9x – 10.
    x2 + 9x – 10 = 0
    x2 + 10x - x – 10 = 0
    x(x + 10) – 1(x + 10) = 0
    (x + 10)(x – 1) = 0
    x = 1, -10
  • Question 22
    1 / -0
    Consider the following statements:

    Statement-I: One of the factors of the polynomial 3x3– 12x2+ 4x – 16 is (2x + 4).
    Statement-II: When 8x6+ 7x5– 4x4- x3+ 2x2+ 6 is divided by 2x3+ x2– 4, the coefficient of the highest degree variable in remainder is .

    Which of the following options holds?
    Solution
    Statement-I:
    3x3 – 12x2 + 4x – 16
    = 3x2(x – 4) + 4(x – 4)
    = (x – 4)(3x2 + 4)
    Hence, (2x + 4) is not a factor of 3x3 – 12x2 + 4x – 16.
    Therefore, Statement-I is false.

    Statement-II:


    Remainder =
    Therefore, the coefficient of the highest degree variable is -7/8.
    Hence, Statement-II is false.
  • Question 23
    1 / -0
    What will be the value of if m, n and o are the zeros of the polynomial px3 – qx2 + 5x - 2?
    Solution
    As m, n and o are the zeros of the polynomial px3 - qx2 + 5x - 2;

    (m + n + o) = (As sum of roots in ax3 + bx2 + cx + d = 0 is )
    Squaring,
    (m + n + o)2 = –- (1)

    And mno = (As product of roots in ax3 + bx2 + cx + d = 0 is)
    Squaring,
    (m2n2o2) = --- (2)

    Therefore, = = (By putting the values from equations (1) and (2))
  • Question 24
    1 / -0
    Match the following:

    Column I Column II
    (L) If p(x) = x2 - k and x = 7 is a zero of p(x), then the value of k is (I)
    (M) In the polynomial equation (-8p + 10)e2 - 11e - 6 = 0, if one of the roots is the negative reciprocal of the other, then the value of p is (II) 35
    (N) In the polynomial equation x3 + 3 + 13x + d = 0, if the product of the roots is -7, then the value of d is (III) 49
    Solution
    (L) Zero of a polynomial p(x) is the value c such that p(c) = 0.
    Here, p(x) = x2 - k and x = 7 is the zero of p(x).
    p(7) = 0
    72 - k = 0
    k = 49

    (M) (-8p + 10)e2 - 11e - 6 = 0
    Let one of the roots be b.
    Then, as per the question, the other root is .
    Now, product of roots = b × = -1 --- (1)
    And product of roots as per the given polynomial = --- (2) (As product of roots in ax2 + bx + c = 0 is )
    As per equations (1) and (2),

    -1 =

    8p - 10 = -6
    8p = -6 + 10
    8p = 4
    p =

    (N) x3 + 3 + 13x + d = 0
    From the given equation,
    Product of roots = --- (1) (As product of roots in ax3 + bx2 + cx + d = 0 is)
    As product of roots = - 7 ..........(2)(Given)

    Therefore, from equations (1) and (2),

     = -7

    d = 35
  • Question 25
    1 / -0
    Find the values of

    (i) 1296A + 36C + D
    (ii) -2592A + B - 72C - 2D
    (iii) B

    if 6 and -6 are the zeros of Ax5 + B + Cx3 + Dx.
    Solution
    Since 6 and -6 are the zeros of Ax5 + B + Cx3 + Dx;
    7776A + B + 216C + 6D = 0 --- (1)
    And -7776A + B – 216C – 6D = 0 --- (2)

    (i) Subtract (2) from (1):
    7776A + B + 216C + 6D + 7776A - B + 216C + 6D = 0
    15552A + 432C + 12D = 0
    1296A + 36C + D = 0 --- (3)

    (ii) Equation (1) + 2(equation (2)):
    7776A + B + 216C + 6D – 15552A + 2B – 432C – 12D = 0
    -7776A + 3B – 216C – 6D = 0
    -2592A + B – 72C – 2D = 0 –- (4)

    (iii) Now, equation (1) + equation (2):
    7776A + B + 216C + 6D - 7776A + B – 216C – 6D = 0
    B = 0
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