What can be added to \( f(x)=2 x^{2}-3 x f(x)=2 x^{2}-3 x \) so that \( (2 x-7)(2 x-7) \) and \( (x+5)(x+5) \) are factors of \( f(x) f(x) \) ?

To determine what needs to be added to the function \( f(x) = 2x^2 - 3x \) so that \( (2x-7)^2 \) and \( (x+5)^2 \) are factors of \( [f(x)]^2 \), follow these steps: 1. **Understand the Requirement:** - For \( [f(x)]^2 \) to have \( (2x-7)^2 \) and \( (x+5)^2 \) as factors, the original function \( f(x) \) must contain \( (2x-7) \) and \( (x+5) \) as factors. This is because squaring \( f(x) \) would naturally square its factors. 2. **Express \( f(x) \) with the Required Factors:** - Assume \( f(x) \) can be written as: \[ f(x) = A(2x - 7)(x + 5) \] where \( A \) is a constant to be determined. 3. **Expand the Factored Form:** - Expand \( (2x - 7)(x + 5) \): \[ (2x - 7)(x + 5) = 2x^2 + 10x - 7x - 35 = 2x^2 + 3x - 35 \] - Therefore: \[ f(x) = A(2x^2 + 3x - 35) \] 4. **Compare with the Given \( f(x) \):** - The given \( f(x) \) is: \[ f(x) = 2x^2 - 3x \] - Set the two expressions for \( f(x) \) equal to each other: \[ A(2x^2 + 3x - 35) = 2x^2 - 3x \] - Equate the coefficients: \[ 2A = 2 \quad \Rightarrow \quad A = 1 \] \[ 3A = -3 \quad \Rightarrow \quad A = -1 \] - There seems to be a contradiction, indicating that simply scaling won't work. Instead, we need to adjust by adding a linear term. 5. **Determine the Correction Term:** - Let’s set: \[ g(x) = h(x) - f(x) \] where \( h(x) = (2x - 7)(x + 5) = 2x^2 + 3x - 35 \). - Then: \[ g(x) = (2x^2 + 3x - 35) - (2x^2 - 3x) = 6x - 35 \] 6. **Verification:** - Adding \( g(x) \) to \( f(x) \): \[ f(x) + g(x) = 2x^2 - 3x + 6x - 35 = 2x^2 + 3x - 35 = (2x - 7)(x + 5) \] - Squaring this expression: \[ [f(x) + g(x)]^2 = (2x - 7)^2 (x + 5)^2 \] which satisfies the requirement. **Conclusion:** You need to add the linear polynomial \( 6x - 35 \) to \( f(x) \). **Answer:** After simplifying, the required polynomial is six x minus thirty-five. In symbols, 6 x − 35