Suppose you can factor as . If c > 0, what could be possible values of p and q? (1 point) Responses p = 5, q = –8 p = 5, q = –8 p = –2, q = 6 p = –2, q = 6 p = –4, q = –7 p = –4, q = –7 p = –13, q = 1

Answer 1

In the given statement, p and q are the coefficients in the quadratic factors.

Since the factored form is (x + p)(x + q), we can see that the values of p and q are the roots of the quadratic equation.

Given that c > 0, the roots of the quadratic equation must be real. This means that the discriminant of the quadratic equation, which is b^2 - 4ac, must be greater than or equal to 0.

Therefore, we have:

(p + q)^2 - 4ac ≥ 0
(p + q)^2 - 4(-4)(c) ≥ 0
(p + q)^2 + 16c ≥ 0
(5 - 8)^2 + 16c ≥ 0
(5)^2 - 2(5)(8) + 8^2 + 16c ≥ 0
25 - 80 + 64 + 16c ≥ 0
(63 + 16c) ≥ 0

Since c > 0, it is not possible for p = 5 and q = -8 to hold since this statement will not be true.

To find possible values of p and q, we need to find a combination of p and q that would satisfy the inequality:

(p + q)^2 + 16c ≥ 0
(p + q)^2 -16c ≤ 0

From the given options, p = -4 and q = -7 result in (p + q)^2 -16c = ( -4 -7) ^2 - 16(c) = 11^2 - 16(c) = 121 - 16c which is greater than 0 and would satisfy the condition.

Therefore, p = -4, q = -7 could be possible values.