Find all integers \( k \) such that the trinomial \( x^2 - kx + 12 \) can be factored over the integers.
Answer
By checking integers, we can find that the integers that satisfy this condition are \( k = -8, -6, 6, 8 \), since \( (-8)^2 = 64 \), \( (-6)^2 = 36 \), \( 6^2 = 36 \), and \( 8^2 = 64 \), all of which are 48 more than a perfect square.
Step 1 ：The trinomial can be factored over the integers if and only if the discriminant \( \Delta = b^2 - 4ac \) is a perfect square, where \( a, b, c \) are the coefficients of the quadratic equation.
Step 2 ：In this case, \( a = 1 \), \( b = -k \), and \( c = 12 \). Thus, \( \Delta = (-k)^2 - 4\cdot1\cdot12 = k^2 - 48 \).
Step 3 ：For \( \Delta \) to be a perfect square, \( k^2 - 48 \) must be a perfect square. This happens when \( k \) is an integer such that \( k^2 \) is 48 more than a perfect square.
Step 4 ：By checking integers, we can find that the integers that satisfy this condition are \( k = -8, -6, 6, 8 \), since \( (-8)^2 = 64 \), \( (-6)^2 = 36 \), \( 6^2 = 36 \), and \( 8^2 = 64 \), all of which are 48 more than a perfect square.
