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Polynomials Test - 7

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Polynomials Test - 7
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  • Question 1
    1 / -0
    If y - 2 and y - are the factors of py2 + 5y + r, then
    Solution
    y - 2 is a factor of py2 + 5y + r.
    So, p(2)2 + 5(2) + r = 0
    4p + r + 10 = 0 ... (1)

    y - is a factor of py2 + 5y + r.
    So, p+ 5+ r = 0
    + + r = 0 … (2)
    On solving (1) and (2), we get
    p = -2
    r = -2
    Hence, p = r
  • Question 2
    1 / -0
    If , are roots of ax2 + bx + b = 0, then
    (a and b are of the same sign)
    Solution
    + = - , = =
    = = -
    + = -
    + + = 0
  • Question 3
    1 / -0
    Which of the following is the value of the coefficient b in the polynomial y = ax2 + bx + c in the given graph? (Given: a ≠ 0)

    Solution

    The zeros are -2 and 2.
    Sum of zeros (S) = -2 + 2 = 0
    i.e. -b/a = 0
    So, b = 0
  • Question 4
    1 / -0
    If a, b and c are the roots of the cubic equation x3 - 7x2 + 14x - 8 = 0, then find the value of a2 + b2 + c2.
    Solution
    Sum of roots = a + b + c = = 7
    Sum of the roots taken two at a time = ab + bc + ca = = 14
    a2 + b2 + c2 = (a + b + c)2 - 2(ab + bc + ca)
    = 72 - 2(14) = 21
  • Question 5
    1 / -0
    The sum of zeros, the product of zeros, and the sum of the product of zeros taken two at a time of the polynomial x3 + bx2 + cx + d are 3, -24 and -10, respectively. What are the respective values of b, c and d?
    Solution
    Sum of zeros =
    3 =
    b = -3
    Product of zeros =
    -24 =
    d = 24
    Sum of product of zeros taken two at a time =
    -10 =
    c = -10
    Hence, (4) is the correct option.
  • Question 6
    1 / -0
    What is the product of sum of zeros and reciprocal of product of zeros of the polynomial 25x2 + 79x + 12?
    Solution
    Zeros, solution and roots, all mean the same thing.

    Let p(x) = 25 x2 + 79x + 12

    Sum of zeros =
    =
    Now, product of zeros =
    =
    Its reciprocal =

    Now, product of sum of zeros and reciprocal of product of zeros

    =

    =

    Hence, (2) is the correct option.
  • Question 7
    1 / -0
    Find the value of g(x), if on dividing 3x2 + 5x + 2 by g(x) the quotient and the remainder are and , respectively.
    Solution
    p(x) = 3x2 + 5x + 2

    g(x) = ?

    q(x) =

    r(x) =

    According to division algorithm,

    p(x) = g(x) q(x) + r(x)

    3x2 + 5x + 2 = g(x)

    Or 3x2 + 5x + 2 - = g(x)

    Or

    Or 12x2 + 20x + 7 = g(x)(6x + 7)

    Or g(x) =



    g(x) = 2x + 1

    Hence, (2) is the correct option.
  • Question 8
    1 / -0
    If kx + 1 is a factor of 2x3 - x2 - 5x - 2, find the value of k.
    Solution
    p(x) = 2x3 – x2 – 5x – 2
    In this, the constant is –2.
    The factors of 2 are 1 and2.
    Now, let x = 2
    p(2) = 2(2)3 – 22 – 5(2) – 2
    = 16 – 4 – 10 – 2
    = 0
    So, (x – 2) is a factor of p(x).
    Now, to find other factors, divide p(x) by (x – 2).



    Further factorising,
    2x2 + 3x + 1
    = 2x2 + 2x + 1x + 1
    = 2x(x + 1) + 1(x + 1)
    = (2x + 1) (x + 1)
    Factors of p(x) = (x – 2)(2x + 1)(x + 1)
    On comparing the factors with (kx + 1), we get k = 2 or k = 1.

    ALTERNATIVE METHOD

    In the given polynomial, the constant coefficient is -2.
    The factors of (2) are 1, 2, -1 and -2.
    By trial method, the polynomial P(x) = 2x3 - x2 - 5x - 2 vanishes at x = 2. Hence, (x - 2) is a factor of the polynomial.
    By synthetic division,



    On dividing P(x) by (x – 2), we get 2x2 + 3x + 1.
    On further factorising, we get (2x + 1)(x + 1).
    The polynomial can be re-written as: (x – 2)(2x + 1)(x + 1).
    On comparing the possible values of k, we get k = 2 or k = 1.
  • Question 9
    1 / -0
    What is the minimum degree of the polynomial for the plotted graph?

    Solution


    B and C are the zeroes as the graph cuts x-axis at B and C.
    And point A represents one zero, but repeated; so, minimum degree is 2 + 1 + 1 = 4.
    Hence, (2) is the correct option.
  • Question 10
    1 / -0
    When the polynomial x3 + 3x2 – 6ax – 5 is divided by (x – 1), the remainder is P; and when the polynomial x3 + ax2 – 15x + 20 is divided by (x + 2), the remainder is Q. Find the value of 'a' if 2P + Q = 0.
    Solution
    Let f(x) = x3 + 3x2 – 6ax – 5
    A.T.Q.
    f(1) = P
    (1)3 + 3(1)2 – 6a(1) – 5 = P
    1 + 3 – 6a – 5 = P
    –1 – 6a = P … (1)
    Let p(x) = x3 + ax2 – 15x + 20
    p(–2) = Q
    (–2)3 + a(–2)2 – 15(–2) + 20 = Q
    –8 + 4a + 30 + 20 = Q
    4a + 42 = Q ... (2)
    Given, 2P + Q = 0
    2(–1 – 6a) + 4a + 42 = 0
    –2 – 12a + 4a + 42 = 0
    –8a + 40 = 0
    a = = 5
  • Question 11
    1 / -0
    If 2 and -2 are the zeros of the polynomial x4 - 5x2 + 4, what are the other possible zeros of the polynomial?
    Solution
    The two zeros are 2 and -2.
    ∴ The factors are (x - 2) and (x + 2).
    Then, (x - 2)(x + 2) = x2 - 4
    Now,


    q(x) = x2 - 1
    To find zeros, put q(x) = 0.
    x2 - 1 = 0
    x2 = 1
    Or x = 1
    The other zeros are -1 and 1.
  • Question 12
    1 / -0
    Which of the following is/are the factor(s) of ?
    Solution

    Putting 3x2 + 4x = a,
    6(a)2 + 14(a) + 8
    = 6a2 + 14a + 8
    = 6a2 + 6a + 8a + 8
    = 6a(a + 1) + 8(a + 1)
    = (a + 1)(6a + 8) … (1)
    Now, substituting the value of 'a' in equation (1), we get
    =
    = (3x2 + 3x + x + 1)(18x2 + 24x + 8)
    = (18x2 + 24x + 8)
    =
    = (3x + 1)(x + 1)(18x2 + 24x + 8)
    Option (4) is correct.
  • Question 13
    1 / -0
    If ab = q and a + b = p, then what are the factors of the algebraic expression x2 + px + q?
    Solution
    Expression: x2 + px + q
    Given: p = (a + b) and q = ab,
    So, x2 + (a + b)x + ab = x2 + ax + bx + ab
    = x(x + a) + b(x + a)
    = (x + a)(x + b)
  • Question 14
    1 / -0
    If one of the zeros of the polynomial x3 + 2x2 - 9x - 18 is -2, what are the other possible zeros?
    Solution
    Since -2 is a zero, (x + 2) is a factor.
    Now,



    x2 - 9
    To find other zeros, put x2 - 9 = 0.
    Or x2 = 9
    Square-rooting both sides,
    x = ±
    x =
    Hence, (2) is the correct option.
  • Question 15
    1 / -0
    If and are the roots of the equation x2 - p(x + 1) - q = 0, then what is the value of ?
    Solution
    The given equation is:
    x2 - p(x + 1) - q = 0
    Or, x2 - px - (p + q) = 0

    Therefore, sum of roots = + = p
    And product of roots = = -(p + q) ... Eq. (1)
    Therefore, we have

    = [Substituting the value of 'q' from Eq. (1)]

    =

    = = 1

    Thus, the required value is 1.
  • Question 16
    1 / -0
    The area of a parallelogram is (x5 + 4x3 + 45x2+ 32x - 38) m and the height of the parallelogram is (x2+ x - 1) m. Find the base of the parallelogram.
    Solution
    Area of the parallelogram = (x5 + 4x3 + 45x2 + 32x - 38) m
    Height of the parallelogram = (x2 + x - 1) m
    Area = Base × Height
    (x5 + 4x3 + 45x2 + 32x - 38) = Base × (x2 + x - 1)

    Base =


    Base = (x3 - x2 + 6x + 38) m
  • Question 17
    1 / -0
    A triangular park with length of the first side (3x3 + 7x2 - 8) m and length of the second side (5x3 - 9x2 - 2) m has perimeter (12x3 - 8x2 - 10) m. Find the length of the third side of the park.
    Solution
    Length of the first side = (3x3 + 7x2 – 8) m
    Length of the second side = (5x3 – 9x2 – 2) m
    Perimeter = (12x3 – 8x2 – 10) m
    Now,
    Perimeter = Length of the first side + Length of the second side + Length of the third side
    (12x3 – 8x2 – 10) = (3x3 + 7x2 – 8) + (5x3 – 9x2 – 2) + Length of the third side
    (12x3 – 8x2 – 10) = (8x3 – 2x2 – 10) + Length of the third side
    Length of the third side = (12x3 – 8x2 – 10) - (8x3 – 2x2 – 10)
    = 12x3 – 8x2 – 10 - 8x3 + 2x2 + 10
    = (4x3 - 6x2) m

  • Question 18
    1 / -0
    In the first week, Priya spent Rs. (10x3 + 5x2 + 2) for buying pencils. In the second week, she bought (9x2 + 10) pencils and the price of each pencil was Rs. (2x + 4). How much did she spend altogether in two weeks?
    Solution
    Money spent in the first week = Rs. (10x3 + 5x2 + 2)
    For the second week:
    Number of pencils she bought = (9x2 + 10)
    Cost of one pencil = Rs. (2x + 4)
    Amount spent by her = Rs. (2x + 4)(9x2 + 10) = Rs. (18x3 + 36x2 + 20x + 40)
    Money spent in two weeks altogether = Rs. {(10x3 + 5x2 + 2) + (18x3 + 36x2 + 20x + 40)}
    Money spent in two weeks altogether = Rs. (28x3 + 41x2 + 20x + 42)
  • Question 19
    1 / -0
    If the weight of one basket is (2x3 – 8x2 + x – 4) kg and the weight of another basket is (4x3 + 12x2 + 2x + 6)kg, then what will be the biggest measure that can measure both quantities exactly?
    Solution
    Weight of one basket = (2x3 – 8x2 + x – 4) kg = 2x2(x – 4) + 1(x – 4) = (x – 4)(2x2 + 1)
    Weight of another basket = (4x3 + 12x2 + 2x + 6) kg = 2(2x3 + 6x2 + x + 3) = 2(2x2(x + 3) + 1(x + 3)) = 2((x + 3)(2x2 + 1))
    Therefore, the required measure is the H.C.F. of both the measures, i.e. (2x2 + 1) kg.
  • Question 20
    1 / -0
    The radius and slant height of a cone are (6x2 + 5x) m and (3x2 + 4x + 1) m, respectively. Find the lateral surface area of the right cone when.
    Solution
    Radius of the cone = (6x2 + 5x) m
    Slant height of the cone = (3x2 + 4x + 1)m



    Radius = (6x2 + 5x) m = 6(2)2 + 5(2) = 34 m
    Slant height = (3x2 + 4x + 1)m = 3(2)2 + 4(2) + 1 = 12 + 8 + 1 = 21 m
    Lateral surface area of the right cone = πrl == 2244 m2
  • Question 21
    1 / -0
    On dividing x3 + 2x2 - 4x + 7 by x + 2, the quotient and the remainder are x2 - a and 15, respectively. What is the value of a?
    Solution
    p(x) = x3 + 2x2 - 4x + 7
    g(x) = x + 2
    By division algorithm,
    p(x) = g(x)q(x) + r(x)
    x3 + 2x2 - 4x + 7 = (x + 2)(x2 - a) + 15
    Or, x3 + 2x2 - 4x + 7 - 15 = (x + 2)(x2 - a)
    Or, x3 + 2x2 - 4x - 8 = (x + 2)(x2 - a)
    Or, x2(x + 2) - 4(x + 2) = (x + 2)(x2 - a)
    Or, (x2 - 4)(x + 2) = (x + 2)(x2 - a)
    Or, (x2 - 4) = (x2 - a)
    Or, x2 - x2 = -a + 4
    Or, a = 4
    Hence, (2) is the correct option.
  • Question 22
    1 / -0
    Which of the following options holds?

    Statement I: The polynomial 14x3 - 35x2 - 16x + 40 has (2x - 5) as one of its factors.
    Statement II: When the polynomial 5x5 + 7x4 - 3x3 + 6x2 - 10 is divided by (x2 + 3x - 1), then the value of constant term in remainder is -90.
    Solution
    Statement I:
    14x3 - 35x2 - 16x + 40
    = 7x2(2x – 5) - 8(2x – 5)
    = (2x - 5)(7x2 - 8)
    Therefore, (2x - 5) is a factor of 14x3 - 35x2 - 16x + 40.
    Hence, statement I is true.

    Statement II:



    Remainder = 266x - 90
    Therefore, the value of the constant term is -90.
    Hence, statement II is true.
    Therefore, option 1 holds.
  • Question 23
    1 / -0
    Let R1 and R2 be the remainders when the polynomials x3 + 2x2 - 5ax - 7 and x3 + ax2 - 12x + 6 are divided by x + 1 and x - 2, respectively.

    Which of the following is the value of a if 2R1 + R2 = 6?
    Solution
    Let p(x) = x3 + 2x2 - 5ax - 7.
    Let R1 be the remainder when p(x) is divided by x + 1.
    R1 = p(-1) = (-1)3 + 2(-1)2 - 5(a)(-1) -7 = -1 + 2 + 5a - 7 = 5a - 6
    Let q(x) = x3 + ax2 - 12x + 6.
    Let R2 be the remainder when q(x) is divided by x - 2.
    R2 = q(2) = (2)3 + a(2)2 -12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10
    Now, 2R1 + R2 = 2 (5a - 6) + 4a - 10 = 6
    Or 10a - 12 + 4a - 10 = 6
    Or 14a - 22 = 6, 14a = 28
    Or a = 28/14 = 2
  • Question 24
    1 / -0
    Match the following columns

    (P) If one of the zeros of the polynomial f(x) = x2 + 24x + 242 is 22 and the other zero is half of the first zero, then q is equal to (i) -40
    (Q) If the sum of the zeros of the polynomial f(x) = 8x3+ 5kx2+ 6x + 10 is 25, then k is equal to (ii) -(1 + r)
    (R) If the polynomial f(x) = qx4+ rx2– sx + p is exactly divisible by g(x) = x2+ sx – p, then q(p + s2)is equal to (iii) 9

    Solution
    (P) f(x) = x2 + 24x + 242
    Let one zero be 22 and the other be 11.
    Product of zeros = 22 × 11 = 242

    Therefore,

    242 × ( – 2) = 242

     – 2 = 1

     = 3

    q = 9

    (Q) f(x) = 8x3 + 5kx2 + 6x + 10
    Sum of zeros =
    25 =

    k = -40

    (R) f(x) is exactly divisible by g(x), i.e. when f(x) is divided by g(x), the remainder must be zero.


    Therefore, -(s + rs + qs3 + 2psq)x + (p + pr + qp2 + qps2) = 0
    (p + pr + qp2 + qps2) = 0
    p(1 + r + qp + qs2 ) = 0
    1 + r + qp + qs2 = 0
    1 + r + q(p + s2) = 0
    q(p + s2) = -(1 + r)
  • Question 25
    1 / -0
    If 2 and -2 are zeros of the polynomial Qx5 – Px4 + Zx3 + Rx2 + M, find the values of the following:

    (i) 8R + 2M
    (ii) 4Q + Z

    Solution
    2 and -2 are zeros of Qx5 – Px4 + Zx3 + Rx2 + M.
    Put x = 2
    Q(2)5 – P(2)4 + Z(2)3 + R(2)2 + M = 0
    32Q – 16P + 8Z + 4R + M = 0 --- (1)

    Now, put x = -2
    Q(-2)5 – P(-2)4 + Z(-2)3 + R(-2)2 + M = 0
    -32Q – 16P - 8Z + 4R + M = 0 --- (2)

    For (i)
    Adding (1) and (2),
    -32P + 8R + 2M = 0
    8R + 2M = 32P

    For (ii)
    Subtracting (1) from (2),
    64Q + 16Z = 0
    4Q + Z = 0


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