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Polynomials and identities — Practice MCQs

Algebraic identities, remainder and factor theorem

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    Practise 12 questions on this topic
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    1. Terms

    2. Remainder and factor theorems

    (ASO 2015) x⁴ + x³ − 2x² + x + 1 ÷ (x − 1): p(1) = 2.
    (ASO 2015) (x − 1) a factor of 4x³ + 3x² − 4x + k → 3 + k = 0 → k = −3.

    3. Zeros and coefficients

    PolynomialRelations
    ax² + bx + cα + β = −b/a, αβ = c/a
    ax³ + bx² + cx + dα + β + γ = −b/a; αβ + βγ + γα = c/a; αβγ = −d/a
    (ASO 2015) Sum of squares of zeros of x² − 8x + k is 40 → 64 − 2k = 40 → k = 12.

    4. Identities (most used)

    5. Solved examples

    a + b = 7, ab = 10 → a² + b² = 49 − 20 = 29; a³ + b³ = 343 − 210 = 133.
    x + 1/x = 4 → x² + 1/x² = 14; x³ + 1/x³ = 64 − 12 = 52.
    (−12)³ + 7³ + 5³ = 3 × (−12) × 7 × 5 = −1260 (since −12 + 7 + 5 = 0).
    Dividing a degree-5 polynomial by a degree-2 polynomial gives a quotient of degree 3 (n − m).

    Solved examples by question type

    Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.

    Basic level

    Example 1. The remainder when x^4 + x^3 − 2x^2 + x + 1 is divided by (x + 1) is:
    (a) 0   (b) -2   (c) -4   (d) -1
    Answer: (b) -2
    Solution: Remainder theorem: put x = -1. Value = -2.
    Example 2. If p(x) = 2x³ − 5x − 1, then p(-1) = ?
    (a) 2   (b) 6   (c) -2   (d) 4
    Answer: (a) 2
    Solution: p(-1) = 2(-1)³ + -5(-1) + -1 = 2.
    Example 3. If p(x) = 3x³ + 6x − 6, then p(-1) = ?
    (a) 15   (b) -9   (c) -15   (d) -13
    Answer: (c) -15
    Solution: p(-1) = 3(-1)³ + 6(-1) + -6 = -15.

    ASO exam level

    Example 4. The sum of the squares of the zeroes of x² − 6x + k is 52. The value of k is:
    (a) -8   (b) 8   (c) -6   (d) -16
    Answer: (a) -8
    Solution: α² + β² = (α+β)² − 2αβ = 36 − 2k = 52 → k = -8.
    Example 5. The sum of the squares of the zeroes of x² − 6x + k is 52. The value of k is:
    (a) -6   (b) -16   (c) -8   (d) 8
    Answer: (c) -8
    Solution: α² + β² = (α+β)² − 2αβ = 36 − 2k = 52 → k = -8.
    Example 6. If (x − 1) is a factor of 1x³ + 2x² − 5x + k, then k = ?
    (a) 2   (b) -2   (c) 3   (d) 0
    Answer: (a) 2
    Solution: Put x = 1: 1 + 2 + -5 + k = 0 → k = 2.
    Example 7. The remainder when x³ − 1x² + 4x − 5 is divided by (x − 4) is:
    (a) 60   (b) 59   (c) -5   (d) -101
    Answer: (b) 59
    Solution: Remainder theorem: put x = 4: 59.
    Example 8. The remainder when x³ − 1x² − 4x + 4 is divided by (x − 1) is:
    (a) 1   (b) 6   (c) 0   (d) 4
    Answer: (c) 0
    Solution: Remainder theorem: put x = 1: 0.
    Example 9. If a + b = 12 and ab = 20, then a³ + b³ = ?
    (a) 1,488   (b) 2,448   (c) 1,008   (d) 1,728
    Answer: (c) 1,008
    Solution: a³ + b³ = (a + b)³ − 3ab(a + b) = 1728 − 720 = 1008.

    Advanced level

    Example 10. If α and β are zeroes of x² − 7x + 10, then 1/α + 1/β = ?
    (a) 70   (b) 10/7   (c) 17   (d) 7/10
    Answer: (d) 7/10
    Solution: 1/α + 1/β = (α + β)/αβ = 7/10.
    Example 11. If α and β are zeroes of x² − 3x + 2, then 1/α + 1/β = ?
    (a) 2/3   (b) 6   (c) 3/2   (d) 5
    Answer: (c) 3/2
    Solution: 1/α + 1/β = (α + β)/αβ = 3/2.
    Example 12. If (x − 4) is a factor of x³ + 4x² + 2x + k, then k = ?
    (a) -138   (b) -136   (c) 136   (d) -135
    Answer: (b) -136
    Solution: p(4) = 0 → 64 + 64 + 8 + k = 0 → k = -136.
    Example 13. If (x + 3) is a factor of x³ − 1x² − 2x + k, then k = ?
    (a) -30   (b) 31   (c) 30   (d) 28
    Answer: (c) 30
    Solution: p(-3) = 0 → -27 + -9 + 6 + k = 0 → k = 30.

    Topic-wise practice for Polynomials and identities (Quant): Algebraic identities, remainder and factor theorem. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.

    Question types covered

    Sourced from the Odisha Public Service Commission where applicable. opsc.gov.in. EduSaint is not affiliated with OPSC.