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
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 - 2)(x - 2)(x2 + 2x + 4)
(x - 4)(x - 2)
(x + 7)(x + 5)
x-14
X((x+1)^2)
(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)
x2 + 6x + 8 = (x + 4)(x + 2)
x2 + 6x + 8 =(x + 2)(x +4)
-x2 + 6x + 16 = -(x2 - 6x - 16) = -(x - 8)(x + 2) = -(8 - x)(x + 2)
16 + 6x - x2 = 16 + 8x - 2x - x2 = 8*(2 + x) - x*(2 + x) = (8 - x)*(2 + x)
4x+8
(X + 2)(X + 4) Factored
x2 + 2x - 48 = x2 + 8x - 6x - 48 = x(x + 8) - 6(x + 8) = (x - 6)(x + 8)
The quadratic expression x2+6x+8 when factorised equals (x+2)(x+4)
(x - 2)(x - 4)
x2 - 6x - 16 = (x + 2)(x - 8)
(-3,-1)
x3 + 3x2 - 6x - 8 = (x - 2)(x2 + 5x + 4) = (x - 2)(x + 1)(x + 4)