To factor the quadratic expression [tex]\(x^2 - 7x + 8\)[/tex], we need to find two binomials whose product gives us the original expression. This involves finding two numbers that multiply to the constant term (8 in this case) and add up to the coefficient of the linear term (-7 in this case).
Let's break this down step-by-step:
1. Identify the constant term and the coefficient:
- Constant term: [tex]\(8\)[/tex]
- Coefficient of [tex]\(x\)[/tex]: [tex]\(-7\)[/tex]
2. Find two numbers that multiply to [tex]\(8\)[/tex] and add up to [tex]\(-7\)[/tex]:
- The pairs of factors for [tex]\(8\)[/tex] are:
- [tex]\(1 \cdot 8\)[/tex]
- [tex]\(2 \cdot 4\)[/tex]
- We need to account for the signs since the coefficient is negative. Let's consider negative pairs:
- [tex]\((-1) \cdot (-8)\)[/tex]
- [tex]\((-2) \cdot (-4)\)[/tex]
- Now we check if any of these pairs add up to [tex]\(-7\)[/tex]:
- [tex]\[(-1) + (-8) = -9\][/tex]
- [tex]\[(-2) + (-4) = -6\][/tex]
- Clearly, none of these pairs exactly add up to [tex]\(-7\)[/tex]. Thus, no simple pair fitting standard factoring rules exists.
As a result of our search, it becomes clear that no factorization into real, integer-coefficient binomials exists for this quadratic. Therefore, the provided options for factorization:
- [tex]\((x + 8)(x - 1)\)[/tex]
- [tex]\((x - 8)(x - 1)\)[/tex]
- [tex]\((x + 8)(x + 1)\)[/tex]
do not factorize the polynomial properly without checking that fit all terms both in multiplication and collecting them back to the original quadratic.
So, our factorization answer would be Prime - since no further factorization can simplify the quadratic [tex]\(x^2 - 7x + 8\)[/tex].
Thus, the correct answer is:
```
Prime
```
