(x-1)(x+4)
x6 + 3x4 - x2 - 3 = 0(x6 + 3x4) - (x2 + 3) = 0x4(x2 + 3) - (x2 + 3) = 0(x2 + 3)(x4 - 1) = 0(x2 + 3)[(x2)2 - 12] = 0(x2 + 3)(x2 + 1)(x2 - 1) = 0(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0x2 + 3 = 0x2 = -3x = ±√-3 = ±i√3 ≈ ±1.7ix2 + 1 = 0x2 = -1x = ±√-1 = ±i√1 ≈ ±ix + 1 = 0x = -1x - 1 = 0x = 1The solutions are x = ±1, ±i, ±1.7i.
The four solution values of x of 3x4 + 7x3 + 4x2 = 0 are: x = -11/3, -1, 0 (repeated) 3x4 + 7x3 + 4x2 = 0 ⇒ x2(3x2 + 7x + 4) = 0 ⇒ x2(3x + 4)(x + 1) = 0 ⇒ x2 = 0 → x = 0 (repeated) or (3x + 4) = 0 → x = -4/3 = -11/3 or (x + 1) = 0 → x = -1
(x + 3)(x + 4)
You don't factor an equation. You factor an expression.So we'll forget about the "equals 0" for a second, while we factor "x2 plus 5x plus 6".x2 + 5x + 6 = (x + 3) (x + 2).Now, if you want this expression to be equal to zero, that can only happen when 'x' is -2 or -3.
(x - 3)(x - 2)
x2- 7x + 6 = 0 factor, (x-6)(x-1) = 0 x = 6 x = 1
x2 - 8x + 32 = 0 (assuming)you can factor or use the quadratic equation.I recommend the quadratic equation.
X2 + 4X - 5 = 0 what two factor of - 5 add up to 4? (X - 1)(X + 5) ----------------- ( so, X = 1 and X = - 5 )
x2+12x-13=0 The factor of x2+12x-13 is: (x+13)(x-1) So: x+13 = 0 or x-1 = 0 The answer is: x=-13 or x=1
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
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.
The discriminant is less than 0 so the equation has no real roots and therefore, no real factors.