Problem 43

Question

Solve the equation \(2 x^{3}-5 x^{2}+x+2=0\) given that 2 is a zero of \(f(x)=2 x^{3}-5 x^{2}+x+2\)

Step-by-Step Solution

Verified
Answer
The solutions to the equation are \(x = 2, \frac{1+\sqrt{3}}{2}, \frac{1-\sqrt{3}}{2}\).
1Step 1: Substitution
Since 2 is a root of the equation, substitute \(x = 2\) in the equation and simplify to confirm that it leads to zero. The equation is: \(2(2)^3-5(2)^2+2+2\). Simplify this to confirm that the result is zero.
2Step 2: Factorization
Since 2 is a root of the equation, it means that \((x-2)\) is a factor of the cubic equation. Use polynomial division or factor by grouping to factorize the cubic equation. \(f(x)\) should factor into \((x-2)(2x^2- x - 1)\) or equivalent.
3Step 3: Solving for the other roots
After factorization, solve the quadratic equation \(2x^2 - x - 1 = 0\). Use the quadratic formula \(x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\) where \(a\) is the coefficient of \(x^2\), \(b\) is the coefficient of \(x\), and \(c\) is the constant term. Calculate the roots from this quadratic equation. This gives the other two solutions of the original cubic equation.