(X² + 6x + 16) Since the 6 is positive, you want to find two number that multiply to get 16 and add to get 6. There are none.
You need the quadratic formula: (-b±√(b²-4ac))/2a
(-6±√(36-48))/2
(-6+√-12)/2 (-6-√-12)/2
-3+6i and -3-6i
Both are imaginary numbers, so there are no real solutions. ■
-x2 + 6x + 16 = -(x2 - 6x - 16) = -(x - 8)(x + 2) = -(8 - x)(x + 2)
(X + 5)(X + 1) FOIL this. X2 + 1X + 5X + 5 Gather terms. X2 + 6X + 5 ----------------------------those were the factors
16 + 6x - x2 = 16 + 8x - 2x - x2 = 8*(2 + x) - x*(2 + x) = (8 - x)*(2 + x)
x2 - 6x - 16 = (x - 8)(x + 2)
X2 - 6X + 27 = 0 what are the factors of 27 that add to - 6? None! This polynomial is unreal and does not intersect the X axis.
-x2 + 6x + 16 = -(x2 - 6x - 16) = -(x - 8)(x + 2) = -(8 - x)(x + 2)
-((x + 2)(x - 8))
x2 + 6x = 16=> x2 + 6x - 16 = 0=> x2 + 8x -2x - 16 = 0=> (x+8)(x-2) = 0=> x = -8 or x = 2So, the solutions of the quadratic equation x2 + 6x = 16 are -8 and 2.
(X + 5)(X + 1) FOIL this. X2 + 1X + 5X + 5 Gather terms. X2 + 6X + 5 ----------------------------those were the factors
16 + 6x - x2 = 16 + 8x - 2x - x2 = 8*(2 + x) - x*(2 + x) = (8 - x)*(2 + x)
x2 - 6x - 16 = (x - 8)(x + 2)
(x - 4)(x - 2)
X2 - 6X + 27 = 0 what are the factors of 27 that add to - 6? None! This polynomial is unreal and does not intersect the X axis.
x2 + 6x = x*(x + 6)
x2 + 6x = 7 ⇒ x2 + 6x + 9 = 7 + 9 ⇒ (x + 3)2 = 16 ⇒ x + 3 = ±4 ⇒ x = -7 or 1
That factors to (x + 3)(x + 3)
x2 + 6x + 8 = (x + 4)(x + 2)