Question

Select the two binomials that are factors of this trinomial

x^2-3x-28

x+7
x-7
x-4
x+4

Answer 1

To factor the trinomial \( x^2 - 3x - 28 \), we need to find two numbers that multiply to \(-28\) (the constant term) and add to \(-3\) (the coefficient of the linear term).

Looking at the provided options:

- \( x + 7 \)
- \( x - 7 \)
- \( x - 4 \)
- \( x + 4 \)

Let's find the pairs of factors of \(-28\):

1. \((-1, 28)\) -> Sums to 27
2. \((1, -28)\) -> Sums to -27
3. \((-2, 14)\) -> Sums to 12
4. \((2, -14)\) -> Sums to -12
5. \((-4, 7)\) -> Sums to 3
6. \((4, -7)\) -> Sums to -3

The correct pair that satisfies both conditions (multiplying to \(-28\) and adding to \(-3\)) is \((4, -7)\).

Thus, we can factor the trinomial as follows:

\[
x^2 - 3x - 28 = (x + 4)(x - 7)
\]

The two binomials that are factors of the trinomial \(x^2 - 3x - 28\) are:

1. \(x + 4\)
2. \(x - 7\)

So, the answer is \(x + 4\) and \(x - 7\).