For what values of b is the expression factorable?
x^2 + bx + 12
Did we forget 2,6 and -2,-6?
Thank you Damon and Steve. Thank you both.
(x-p)(x-q) = x^2 + b x + 12
p q = 12
b = +/-(p+q)
p and q must be the same sign and product 12
so
1) p = 1, q = 12
2) p = 3, q = 4
3) p = -1 , q = -12
4) p = -3, q = -4
so
b = 13
b = -13
b = 7
b = -7
try those
x^2 + 13 x + 12 = (x+12)(x+1)
x^2 - 13 x + 12 = (x-12)(x-1)
x^2 + 7 x + 12 = (x+3)(x+4)
x^2 - 7 x + 12 = (x-3)(x-4)
b = -13
13,8,7
To determine the values of b for which the expression x^2 + bx + 12 is factorable, we need to determine if there are two binomial factors that can multiply together to give us the original expression. Factorable expressions can be factored into the product of two binomials, whereas non-factorable expressions are either prime or can only be factored using complex numbers.
The general form of a quadratic expression is ax^2 + bx + c. In this case, a = 1, b = b, and c = 12. To factor the expression x^2 + bx + 12, we need to find two binomial factors of the form (x + m)(x + n), where m and n are constants.
Multiplying out (x + m)(x + n) gives us x^2 + (m + n)x + mn. Comparing this to our original expression, we can see that the constant term mn must be equal to 12, and the coefficient of x, (m + n), must be equal to b.
Now, we need to find the pairs of integers mn that multiply to give 12. The possible pairs are:
1, 12
-1, -12
2, 6
-2, -6
3, 4
-3, -4
For each pair of mn, we can sum them to find (m + n) and see if it matches with the coefficient b. For example, if b = 5, we need to find a pair of mn such that (m + n) = 5.
Trying out the pairs, we find that (m + n) = 3 + 4 = 7, which does not match with b = 5. None of the pairs of mn for 12 satisfy the equation (m + n) = b, which suggests that there are no values of b for which the expression x^2 + bx + 12 is factorable.
Therefore, the expression x^2 + bx + 12 is not factorable for any value of b.