Question

Find a positive value of b so that the trinomial is factorable.
3x^2+bx-2

Answer 1

To determine a positive value of \( b \) such that the trinomial \( 3x^2 + bx - 2 \) is factorable, we'll need to ensure that its discriminant is a perfect square. The quadratic can be expressed in the form \( ax^2 + bx + c \), where \( a = 3 \), \( b = b \), and \( c = -2 \).

The discriminant \( D \) of a quadratic \( ax^2 + bx + c \) is given by:

\[
D = b^2 - 4ac
\]

Substituting in our values, we have:

\[
D = b^2 - 4(3)(-2)
\]
\[
D = b^2 + 24
\]

For the trinomial to be factorable, the discriminant must be a perfect square. This means there exists some integer \( k \) such that:

\[
b^2 + 24 = k^2
\]

Rearranging gives:

\[
k^2 - b^2 = 24
\]

Factoring the left-hand side:

\[
(k - b)(k + b) = 24
\]

Now, we can find pairs of integers that multiply to \( 24 \):

- \( (1, 24) \)
- \( (2, 12) \)
- \( (3, 8) \)
- \( (4, 6) \)
- \( (-1, -24) \)
- \( (-2, -12) \)
- \( (-3, -8) \)
- \( (-4, -6) \)

We will consider only the positive pairs for \( k - b \) and \( k + b \):

### Case 1: \( k - b = 1 \) and \( k + b = 24 \)
Adding these equations:

\[
2k = 25 \implies k = 12.5 \quad \text{(not an integer)}
\]

### Case 2: \( k - b = 2 \) and \( k + b = 12 \)
Adding these equations:

\[
2k = 14 \implies k = 7
\]
Subtracting:

\[
2b = 10 \implies b = 5
\]

### Case 3: \( k - b = 3 \) and \( k + b = 8 \)
Adding these equations:

\[
2k = 11 \implies k = 5.5 \quad \text{(not an integer)}
\]

### Case 4: \( k - b = 4 \) and \( k + b = 6 \)
Adding these equations:

\[
2k = 10 \implies k = 5
\]
Subtracting:

\[
2b = 2 \implies b = 1
\]

Thus, we have two potential positive values for \( b \): \( 5 \) and \( 1 \).

So, the possible positive values of \( b \) that make \( 3x^2 + bx - 2 \) factorable are:

\[
\boxed{1} \quad \text{and} \quad \boxed{5}
\]