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Given equation : f(x) =  3x³ + ax² + bx - 8 = 0

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The quadratic factor = two factors = x² - 4

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x² - 4 = ( x - 2)(x + 2)

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two factors = ( x - 2)(x + 2)

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(x + 2) is factor of f(x).So substitute x = -2.

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f(-2) =  3(-2)³ + a(-2)² + b(-2) - 8 = 0

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-24 + 4a - 2b - 8 = 0

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4a - 2b = 32

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2a - b = 16

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b =2a - 16

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( x - 2) is factor of f(x).So substitute x = 2.

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f(2) =  3(2³) + a(2²) + b2 - 8 = 0

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24 + 4a + 2b - 8 = 0

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4a + 2b = -16

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2a + b = -8

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Substitute b = 2a - 16

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2a + 2a - 16 = -8

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4a = 8

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a = 2

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b = 2a - 16 = 2(2) - 16 = 4 - 16

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b = -12

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Solution is a = 2 and b = -12

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