Complex Math Problem Solver

Summary To solve this polynomial equation, we need to find the value of X which makes the equation true. One way to solve such equations is to use the Rational Root Theorem to find a list of possible rational roots, and then test them using synthetic division or long division. If we find a root, we can then use polynomial division to factor the equation and solve for the remaining roots. Prerequisites and Skills Required Knowledge of polynomial equations, roots, and factorization. Familiarity with synthetic division or long division of polynomials. Understanding of the Rational Root Theorem and its application. Solution Step 1: Identify the coefficients of the polynomial equation. Coefficient of X^3: 3 Coefficient of X^2: 1 Coefficient of X: 6 Constant term: -2 Step 2: Find the possible rational roots of the equation using the Rational Root Theorem. The possible rational roots are factors of the constant term (-2) divided by factors of the leading coefficient (3). Factors of -2: ±1, ±2 Factors of 3: ±1, ±3 Possible rational roots: ±1/3, ±2/3, ±1, ±2 Step 3: Test the possible rational roots using synthetic division or long division. We start by testing the root X = 1/3. markdown 1/3 | 3  1  6  -2     |___ 1  1  7       3  2  7  5 Since the remainder is not zero, X = 1/3 is not a root of the equation. We continue by testing the next possible rational root, X = -1/3. -1/3 | 3  1  6  -2      |___ -1  0 -2        3  0  6 -4 Since the remainder is not zero, X = -1/3 is also not a root of the equation. We continue testing with the next possible root, X = 1. 1 | 3  1  6  -2   |___ 3  4 10     3  4 10  8 Since the remainder is not zero, X = 1 is also not a root of the equation. We continue testing with the next possible root, X = -1. -1 | 3  1  6  -2    |___ -3  2 -8      3 -2  8 -10 Since the remainder is zero, X = -1 is a root of the equation. Step 4: Use polynomial division to factor the equation and solve for the remaining roots. We use synthetic division to factor out the root X = -1 and obtain a quadratic equation. (X + 1)(3X^2 - 2X + 2) = 0 Setting each factor equal to zero, we obtain: X + 1 = 0  -> X = -1  3X^2 - 2X + 2 = 0 We can solve the quadratic equation using the quadratic formula. X = (-b ± sqrt(b^2 - 4ac)) / 2a Plugging in the coefficients, we get Continuing from the last step of the previous solution, X = (-(-2) ± sqrt((-2)^2 - 4(3)(2))) / 2(3) X = (2 ± sqrt(4 - 24)) / 6 X = (2 ± sqrt(-20)) / 6 X = (2 ± 2i√5) / 6 X = (1 ± i√5) / 3 Therefore, the complete solution of the given polynomial equation is: 3X^3 + X^2 + 6X - 2 = 0  Roots: X = -1, (1 ± i√5) / 3 Conclusion Solving polynomial equations can be challenging, but the Rational Root Theorem provides a systematic way to find possible rational roots. Once we find a root, we can use polynomial division to factor the equation and solve for the remaining roots. This solution illustrates the importance of testing multiple possible roots to find all the roots of the equation.