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That doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: -8 plus or minus i times the square root of 11.

x = -8 + 3.3166247903554i

x = -8 - 3.3166247903554i

where i is the imaginary square root of -1

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9y ago

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More answers

This doesn't factor neatly. Applying the quadratic equation, we find two imaginary solutions: Zero plus or minus 4i times the square root of 1

x = 4i, -4i

where i is the square root of -1.

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Wiki User

12y ago
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x2 - 8x + 12

(x - 6)(x - 2)

x2 - 2x - 6x + 12

x2 - 8x + 12

x - 6 = 0

x = 6

x - 2 = 0

x = 2

Solution set: {2, 6}

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Wiki User

15y ago
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It is not possible - you can't have two numbers that multiply to get one but add up to four

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Wiki User

14y ago
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There is no factorisation with real roots. The complex roots are 8 +or- i*sqrt(11) where i is the imaginary square root of -1

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Wiki User

15y ago
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(x - 2)(x - 2)(x2 + 2x + 4)

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12y ago
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(x - 4)(x - 2)

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Wiki User

9y ago
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(x + 7)(x + 5)

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Wiki User

12y ago
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x-14

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Wiki User

13y ago
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X((x+1)^2)

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Wiki User

15y ago
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Q: Factor x2 - 6x plus 8?
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