Find all possible positive values of \( c \) so that the trinomial is factorable \[ d^{2}-3 d+c \]

To determine the positive values of \( c \) that make the trinomial \( d^{2} - 3d + c \) factorable, we can utilize the concept of the discriminant from the quadratic formula. A quadratic expression \( ax^2 + bx + c \) is factorable over the real numbers if the discriminant \( D = b^2 - 4ac \) is a perfect square. For our trinomial: - \( a = 1 \) - \( b = -3 \) - \( c = c \) The discriminant \( D \) can be calculated as follows: \[ D = (-3)^2 - 4 \cdot 1 \cdot c = 9 - 4c \] We need this discriminant to be a non-negative perfect square: \[ 9 - 4c \geq 0 \] Rearranging the inequality gives: \[ 9 \geq 4c \] \[ c \leq \frac{9}{4} = 2.25 \] Additionally, since we are interested in positive values of \( c \), we have: \[ 0 < c \leq 2.25 \] Next, we want \( 9 - 4c \) to also be a perfect square: Let \( k^2 = 9 - 4c \) for some integer \( k \): \[ 4c = 9 - k^2 \] This implies: \[ c = \frac{9 - k^2}{4} \] For \( c \) to be positive, we need \( 9 - k^2 > 0 \) or \( k^2 < 9 \). Hence, \( k \) can take values from \(-2\) to \(2\): - If \( k = 0 \), \( c = \frac{9 - 0}{4} = \frac{9}{4} = 2.25 \). - If \( k = 1 \), \( c = \frac{9 - 1}{4} = \frac{8}{4} = 2 \). - If \( k = 2 \), \( c = \frac{9 - 4}{4} = \frac{5}{4} = 1.25 \). - If \( k = -1 \), \( c = \frac{9 - 1}{4} = 2 \) (same as \( k = 1 \)). - If \( k = -2 \), \( c = \frac{9 - 4}{4} = 1.25 \) (same as \( k = 2 \)). Thus, the possible positive values of \( c \) such that the trinomial \( d^{2} - 3d + c \) is factorable are: \[ c = 2.25, \; 2, \; 1.25 \] In interval notation, this can be summarized as: \[ (0, 2.25] \] To ensure that all inequalities hold true, the positive values of \( c \) for which the trinomial is factorable are \( \{ 2.25, 2, 1.25 \} \), respecting that \( 0 < c \leq 2.25 \).