Factors with integer and positivity constraints — Topic practice

A: The relation k = n + 12/n has minimum greater than 6.5 even over positive real n.

B — correct: For n = 3 or 4, the factorization (x − 3)(x − 4) gives k = 7.

C: A real root can satisfy n² − 7.5n + 12 = 0, but no positive integer n does. For n at least 8, n + 12/n exceeds 8; direct testing of 1 through 7 excludes 7.5.

D: For positive n, both n and 12/n are positive, so their sum cannot be zero.

A: Take r = 4 and s = −5: b = 3s + r = −11, so b/r = −11/4 is not integral.

B: The same allowed example gives b/3 = −11/3.

C: 20/3 is not an integer.

D — correct: The constant terms give rs = −20, so r is nonzero and 20/r = −s, an integer.

A: The root sum is −5/3, so a = 1 would give noninteger b = 5/3.

B: a = 2 would give noninteger b = 10/3.

C — correct: Expansion gives b = 5a/3 and c = −2a/3. Both are integers exactly when a is a multiple of 3; a = 3 is the smallest positive choice.

D: This works but is not smallest; a = 3 already gives integer coefficients.

A: Matching coefficients gives t = b + 4 = 0, which fails strict positivity.

B: This gives t = 0.5, positive but not an integer.

C: This gives t = −1, an integer but not positive.

D — correct: This gives t = 3, a positive integer, and the product is x² − x − 12.

The root sum is −1/3 and the product is −2/3, so b = a/3 and c = −2a/3. Both are integers only when a is a multiple of 3. With a > 4, the smallest allowed a is 6, giving b = 2.

6x² + 2x − 4 = (3x − 2)(2x + 2).

The factorization proves attainability. Choosing a = 5 violates integrality of b and c; choosing a = 3 violates the strict lower bound.

A: Dividing by t without checking zero loses solutions such as t = 0, k = 1.5.

B — correct: For nonzero t, divide by t. For t = 0, the original equation is 0 = 0 for every real k.

C: Substitution of zero makes both sides zero.

D: The equation at t = 0 imposes no integrality requirement on k.