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Algebraic Expressions and Identities Test - 13

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Algebraic Expressions and Identities Test - 13
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  • Question 1
    1 / -0

    What must be subtracted from 3a2 − 6ab − 3b2 − 1 to get 4a2 − 7ab − 4b2 + 1 ?

    Solution

    Let subtrahend be 'a? Then (3a2 − 6ab − 3b2 − 1) − A =(4a2 − 7ab − 4b2 + 1) ⇒A = [3a2 − 6ab − 3b2 − 1) − (4a2 − 7ab − 4b2 + 1) and proceed.

  • Question 2
    1 / -0

    Find the sum of the given expression −5x2 + 3x − 8,4x + 7 − 2x2 and  6 − 2x + 2x2

  • Question 3
    1 / -0

    Simplify :  (4x + 5y)(4x − 5y)

    Solution

    use (a + b)(a − b) = a2 − b2.

  • Question 4
    1 / -0

    Solve (4x2 + 5)(4x2 + 5)

    Solution

    apply  (a + b)2 = a2 + b2 + 2ab.

  • Question 5
    1 / -0

    Multiply (3x + 2y)(3x + 2y)

  • Question 6
    1 / -0

    The product of (3x2 + y2) and (2x2 + 3y2) is

    Solution

    (3x2 + y2) × (2x2 + 3y2) =3x2(2x2 + 3y2) + y2(2x2 + 3y2) =6x4 + 9x2y2 + 2x2yy + 3y4 = 6x4 + 11x2y2 + 3y4

  • Question 7
    1 / -0

    The product of  (−3x2y) + (4x2y − 3xy2 + 4x − 5y) is

  • Question 8
    1 / -0

    (4x − 7y)(4x − 7y) equal 10

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