suppose you can factor x^2+bx+c as (x+p)(x+q). If c<0, what could be possible values of p and q?
a. p=-3 ,q=7
b. p=11 ,q=4
c. p=-2 ,q=-5
d. p=1 ,q=10
you wrong
as before,
since c=pq, that means that p and q have opposite signs.
To find the values of p and q when factoring the quadratic expression x^2+bx+c as (x+p)(x+q), we need to consider the relationship between the coefficients b and c.
First, let's examine the coefficient c. Since c<0, it means that c is a negative number.
Next, let's analyze the values of p and q. We want to find two numbers whose sum is equal to the coefficient b, and whose product is equal to the coefficient c. In other words, p+q=b and p*q=c.
Since c is negative, we need to find two numbers whose sum is equal to b but whose product is negative.
Let's go through the given options:
a. p=-3, q=7:
The sum of -3 and 7 is 4 (p+q=4), and the product is -21 (p*q=-21). Since the product is negative, this could be a possible value.
b. p=11, q=4:
The sum of 11 and 4 is 15 (p+q=15), and the product is 44 (p*q=44). Since the product is positive, this is not a possible value.
c. p=-2, q=-5:
The sum of -2 and -5 is -7 (p+q=-7), and the product is 10 (p*q=10). Since the product is not negative, this is not a possible value.
d. p=1, q=10:
The sum of 1 and 10 is 11 (p+q=11), and the product is 10 (p*q=10). Since the product is not negative, this is not a possible value.
Therefore, the possible values of p and q when c<0 are p=-3 and q=7. Hence, the correct answer is option a.