(5) \( 2 x^{2}-7 x+5 \)

It appears you have the quadratic expression: \[ 2x^{2} - 7x + 5 \] Assuming you'd like to **factor** this quadratic expression or **solve** the equation \(2x^{2} - 7x + 5 = 0\), here's how you can proceed: ### **1. Factoring the Quadratic Expression** To factor the quadratic expression \(2x^{2} - 7x + 5\), we look for two numbers that: - Multiply to \(2 \times 5 = 10\) (the product of the coefficient of \(x^{2}\) and the constant term). - Add up to \(-7\) (the coefficient of the \(x\) term). The numbers \(-2\) and \(-5\) satisfy these conditions because: - \((-2) \times (-5) = 10\) - \((-2) + (-5) = -7\) Using these numbers, we can rewrite the middle term and factor by grouping: \[ 2x^{2} - 2x - 5x + 5 \] Group the terms: \[ (2x^{2} - 2x) - (5x - 5) \] Factor out the common factors in each group: \[ 2x(x - 1) - 5(x - 1) \] Now, factor out the common binomial \((x - 1)\): \[ (2x - 5)(x - 1) \] So, the factored form of the quadratic expression is: \[ 2x^{2} - 7x + 5 = (2x - 5)(x - 1) \] ### **2. Solving the Quadratic Equation** If you're looking to solve the equation \(2x^{2} - 7x + 5 = 0\), you can use the factored form to find the solutions: \[ (2x - 5)(x - 1) = 0 \] Set each factor equal to zero and solve for \(x\): 1. **First Factor:** \[ 2x - 5 = 0 \\ 2x = 5 \\ x = \frac{5}{2} \] 2. **Second Factor:** \[ x - 1 = 0 \\ x = 1 \] **Solutions:** \[ x = 1 \quad \text{and} \quad x = \frac{5}{2} \] ### **Alternative Method: Quadratic Formula** If factoring is challenging, you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] For the equation \(2x^{2} - 7x + 5 = 0\): - \(a = 2\) - \(b = -7\) - \(c = 5\) Plugging the values into the formula: \[ x = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4 \times 2 \times 5}}{2 \times 2} \\ x = \frac{7 \pm \sqrt{49 - 40}}{4} \\ x = \frac{7 \pm \sqrt{9}}{4} \\ x = \frac{7 \pm 3}{4} \] This gives two solutions: 1. \[ x = \frac{7 + 3}{4} = \frac{10}{4} = \frac{5}{2} \] 2. \[ x = \frac{7 - 3}{4} = \frac{4}{4} = 1 \] **Thus, the solutions are \(x = 1\) and \(x = \frac{5}{2}\).**