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How many Integral values of k are there for which x^2+kx+24 is factorable?

How many Integral values of k are there for which x^2+kx+24 is factorable?-example-1
User Lorick
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Answer:

there are 10 integral values of k for which x^2 + kx + 24 is factorable.

Explanation:

A quadratic expression of the form x^2 + kx + 24 can be factored if and only if its discriminant (b^2 - 4ac) is a perfect square, where a=1, b=k and c=24 are the coefficients of the quadratic.

The discriminant of the quadratic is k^2 - 96. To be a perfect square, k^2 - 96 must be equal to the square of some integer, say m^2. Then we have:

k^2 - 96 = m^2

k^2 - m^2 = 96

(k + m)(k - m) = 96

The factors on the left-hand side of this equation must be pairs of integers whose product is 96. We can list the pairs of factors of 96:

(1, 96), (2, 48), (3, 32), (4, 24), (6, 16), (8, 12)

For each pair of factors, we can solve the system of equations:

k + m = factor

k - m = factor

The solution to this system is k = (factor + factor)/2 = factor, which gives us the possible values of k that make x^2 + kx + 24 factorable.

Using this method, we find that there are 10 integral values of k that make x^2 + kx + 24 factorable:

k = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Therefore, there are 10 integral values of k for which x^2 + kx + 24 is factorable.

User Jowett
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