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

To ensure that both \( (2x - 7) \) and \( (x + 5) \) are factors of \( f(x) = 2x^2 - 3x + k \), we can use the property that if \( r \) is a root of the polynomial, then the polynomial evaluated at \( r \) is zero. First, we find the roots of each factor: 1. For \( 2x - 7 = 0 \) ⇒ \( x = \frac{7}{2} \) 2. For \( x + 5 = 0 \) ⇒ \( x = -5 \) Now, we substitute these values into \( f(x) \): 1. Substituting \( x = \frac{7}{2} \): \[ f\left(\frac{7}{2}\right) = 2\left(\frac{7}{2}\right)^2 - 3\left(\frac{7}{2}\right) + k = 0 \] Simplifying this: \[ f\left(\frac{7}{2}\right) = 2 \cdot \frac{49}{4} - \frac{21}{2} + k = 0 \] \[ = \frac{49}{2} - \frac{21}{2} + k = 0 \] \[ \frac{28}{2} + k = 0 \] \[ k = -14 \] 2. Substituting \( x = -5 \): \[ f(-5) = 2(-5)^2 - 3(-5) + k = 0 \] Simplifying this: \[ f(-5) = 2 \cdot 25 + 15 + k = 0 \] \[ 50 + 15 + k = 0 \] \[ k = -65 \] To satisfy both conditions, we need a different approach since \( k = -14 \) and \( k = -65 \) are contradictory. Therefore, it is impossible for both \( (2x - 7) \) and \( (x + 5) \) to be factors of the polynomial \( f(x) = 2x^2 - 3x \) without changing its form. Thus, you need to modify \( f(x) \) accordingly, or select only one of the factors to proceed.