Find all integers b so that the trinomial 3x^2 +bx + 2 can be factored.
The discriminant must be positive for there to be real roots, that is, factors
so, b^2 - 24 > 0
b must be an integer such that |b| >= 5
3x^2-5x+2 = (3x-2)(x-1)
3x^2+5x+2 = (3x+2)(x+1)
3x^2-7x+2 = (3x-1)(x-2)
3x^2+7x+2 = (3x+1)(x+2)
Many others are possible, not all of which factor into rational roots. Any integer b > 5 will provide real roots, however.
To find all integers b so that the trinomial 3x^2 + bx + 2 can be factored, we can use the fact that for a trinomial to be factorable, the discriminant of the quadratic equation must be a perfect square.
The discriminant of the quadratic equation ax^2 + bx + c is given by D = b^2 - 4ac. In this case, the given quadratic equation is 3x^2 + bx + 2, so the discriminant is D = b^2 - 4(3)(2) = b^2 - 24.
For the trinomial to be factorable, the discriminant D = b^2 - 24 must be a perfect square. Therefore, we need to find all integers b for which b^2 - 24 is a perfect square.
To do this, we can systematically check the values of b. Start by setting b^2 - 24 = 0 and solve for b. Taking the square root of both sides, we find b = ±√24 ≈ ±4.899.
Now, check the values of b around ±4.899. Since we are looking for integers, we can check the nearby integers:
For b = -5: (-5)^2 - 24 = -1, which is not a perfect square.
For b = -4: (-4)^2 - 24 = 0, which is a perfect square (0^2).
For b = -3: (-3)^2 - 24 = 15, which is not a perfect square.
For b = -2: (-2)^2 - 24 = 20, which is not a perfect square.
For b = -1: (-1)^2 - 24 = 23, which is not a perfect square.
For b = 0: (0)^2 - 24 = -24, which is not a perfect square.
For b = 1: (1)^2 - 24 = -23, which is not a perfect square.
For b = 2: (2)^2 - 24 = -20, which is not a perfect square.
For b = 3: (3)^2 - 24 = -15, which is not a perfect square.
For b = 4: (4)^2 - 24 = -8, which is not a perfect square.
For b = 5: (5)^2 - 24 = 1, which is a perfect square (1^2).
So, the possible values of b that make the trinomial 3x^2 + bx + 2 factorable are b = -4 and b = 5.