6х³-18х²-144х+96=0

To solve the cubic equation \(6x^3 - 18x^2 - 144x + 96 = 0\), we can start by simplifying it. First, let's divide the entire equation by 6: \[ x^3 - 3x^2 - 24x + 16 = 0 \] Next, we can use the Rational Root Theorem to test possible rational roots. The potential rational roots are the factors of the constant term (16) divided by the factors of the leading coefficient (1). The possible rational roots are: \[ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16 \] Let's test some of these values: 1. Testing \(x = 2\): \[ 2^3 - 3(2^2) - 24(2) + 16 = 8 - 12 - 48 + 16 = -36 \quad \text{(not a root)} \] 2. Testing \(x = 4\): \[ 4^3 - 3(4^2) - 24(4) + 16 = 64 - 48 - 96 + 16 = -64 \quad \text{(not a root)} \] 3. Testing \(x = -2\): \[ (-2)^3 - 3(-2)^2 - 24(-2) + 16 = -8 - 12 + 48 + 16 = 44 \quad \text{(not a root)} \] 4. Testing \(x = 1\): \[ 1^3 - 3(1^2) - 24(1) + 16 = 1 - 3 - 24 + 16 = -10 \quad \text{(not a root)} \] 5. Testing \(x = -1\): \[ (-1)^3 - 3(-1)^2 - 24(-1) + 16 = -1 - 3 + 24 + 16 = 36 \quad \text{(not a root)} \] 6. Testing \(x = -4\): \[ (-4)^3 - 3(-4)^2 - 24(-4) + 16 = -64 - 48 + 96 + 16 = 0 \quad \text{(found a root)} \] Thus, \(x = -4\) is a root. Now we can perform polynomial long division to factor the cubic polynomial. Dividing \(x^3 - 3x^2 - 24x + 16\) by \(x + 4\): 1. \(x^3 \div x = x^2\) 2. Multiply: \(x^2(x + 4) = x^3 + 4x^2\) 3. Subtract: \((x^3 - 3x^2) - (x^3 + 4x^2) = -7x^2\) 4. Bring down \(-24x\): \(-7x^2 - 24x\) 5. Divide: \(-7x^2 \div x = -7x\) 6. Multiply: \(-7x(x + 4) = -7x^2 - 28x\) 7. Subtract: \((-7x^2 - 24x) - (-7x^2 - 28x) = 4x\) 8. Bring down \(16\): \(4x + 16\) 9. Divide: \(4x \div x = 4\) 10. Multiply: \(4(x + 4) = 4x + 16\) 11. Subtract: \((4x + 16) - (4x + 16) = 0\) The polynomial factors as: \[ x^3 - 3x^2 - 24x + 16 = (x + 4)(x^2 - 7x + 4) \] Next, we need to solve \(x^2 - 7x + 4 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 - 16}}{2} = \frac{7 \pm \sqrt{33}}{2} \] Thus, the solutions to the original equation \(6x^3 - 18x^2 - 144x + 96 = 0\) are: \[ x = -4, \quad x = \frac{7 + \sqrt{33}}{2}, \quad x = \frac{7 - \sqrt{33}}{2} \]