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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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.
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}
It is not possible - you can't have two numbers that multiply to get one but add up to four
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
(x + 4)(x + 8)
x2 + 15x + 54 = (x + 6)(x + 9) or (x + 9)(x + 6) since 6*9 = 54 and 6 + 9 = 15However, commutativity is implicit in deriving this factorisation so that the two factorisations are the same.or factor by groupingx2 + 15x + 54= x2 + 6x + 9x + 54= (x2 + 6x) + (9x + 54)= x(x + 6) + 9(x + 6)= (x + 6)(x + 9)
(x + 3)(4x + 5)
x2(x - 8)
(x+2)(x2+2x+4)