4) \( 2 x^{2}-5 x+2 \)

1. We are given the quadratic expression \[ 2x^2 - 5x + 2. \] 2. To factor it, we look for two numbers that multiply to \(2 \times 2 = 4\) (the product of the coefficient of \(x^2\) and the constant term) and add up to \(-5\) (the coefficient of \(x\)). The two numbers that work are \(-1\) and \(-4\) because \[ (-1)(-4) = 4 \quad \text{and} \quad (-1) + (-4) = -5. \] 3. Rewrite the middle term \(-5x\) using \(-x\) and \(-4x\): \[ 2x^2 - 5x + 2 = 2x^2 - x - 4x + 2. \] 4. Group the terms: \[ (2x^2 - x) + (-4x + 2). \] 5. Factor out the common factors in each group: - From \(2x^2 - x\), factor out \(x\): \[ x(2x - 1). \] - From \(-4x + 2\), factor out \(-2\): \[ -2(2x - 1). \] 6. Now the expression becomes: \[ x(2x - 1) - 2(2x - 1). \] 7. Notice that \((2x - 1)\) is a common factor: \[ (2x - 1)(x - 2). \] 8. Therefore, the factorization is: \[ 2x^2 - 5x + 2 = (2x - 1)(x - 2). \] 9. If we set the expression equal to zero to solve the quadratic equation: \[ (2x - 1)(x - 2) = 0, \] then by the zero product property, we have: \[ 2x - 1 = 0 \quad \text{or} \quad x - 2 = 0. \] 10. Solve each equation: - From \(2x - 1 = 0\): \[ 2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}. \] - From \(x - 2 = 0\): \[ x = 2. \] Thus, the factored form is \[ (2x-1)(x-2) \] and the solutions to the equation \(2x^2 - 5x + 2 = 0\) are \[ x = \frac{1}{2} \quad \text{and} \quad x = 2. \]