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16 June, 07:07

Find the factors and zeros of 6x2 + 8x - 28 = 2x2 + 4?

A) 2 (x + 4) (x + 2); {-4, - 2}

B) 4 (x + 4) (x - 2); {-4, 2}

C) 4 (x - 4) (x + 2); {4, - 2}

D) 4 (x - 4) (x - 2); {4, 2}

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Answers (1)
  1. 16 June, 07:25
    0
    First solve the equation:

    6x^2 + 8x - 28 = 2x^2 + 4

    => 6x^2 - 2x^2 + 8x - 28 - 4 = 0

    => 4x^2 + 8x - 32 = 0

    Extract common factor 4:

    => 4[x^2 + 2x - 8] = 0

    Now factor the polynomial:

    4 (x + 4) (x - 2) = 0

    => the solutions are x + 4 = 0 = > x = - 4, and x - 2 = 0 = > x = 2.

    So the answer is the option B: 4 (x + 4) (x - 2); {-4, 2}
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