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Factor the polynomial completely: [tex]$x^2 - x + 7$[/tex]

Select one:
A. [tex]$(x+7)(x-1)$[/tex]
B. [tex][tex]$(x+1)(x+7)$[/tex][/tex]
C. Cannot be factored
D. [tex]$(x-7)(x+1)$[/tex]

Sagot :

To tackle the problem of factoring the polynomial [tex]\(x^2 - x + 7\)[/tex], let's examine it in detail.

Consider the quadratic polynomial [tex]\(x^2 - x + 7\)[/tex]. To factor this polynomial, we typically look for two binomials [tex]\((x - a)(x - b)\)[/tex] whose product equals the given polynomial. To do this, the coefficients and constants will need to match the original polynomial once expanded.

1. Identify Possible Factor Pairs: Let's assume it can be factored into [tex]\((x - a)(x - b)\)[/tex].

2. Expand the Binomials: By expanding [tex]\((x - a)(x - b)\)[/tex], we get:
[tex]\[ (x - a)(x - b) = x^2 - (a + b)x + ab \][/tex]
In this case, we need:
- The sum of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] to equal -1 (since the coefficient of [tex]\(x\)[/tex] is -1).
- The product of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] to equal 7, the constant term.

3. Solve the System of Equations:
- We need [tex]\(a + b = 1\)[/tex]
- We need [tex]\(ab = 7\)[/tex]

Let's find pairs [tex]\((a, b)\)[/tex] that satisfy these conditions adequately:
- The pairs [tex]\((a, b)\)[/tex] that multiply to 7 could be [tex]\((1, 7)\)[/tex], [tex]\((-1, -7)\)[/tex], [tex]\((\sqrt{7}, -\sqrt{7})\)[/tex], etc.
- None of these pairs will simultaneously sum up to -1.

4. Conclusion:
After analyzing the polynomial, it becomes apparent that there are no pairs of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] consisting of real numbers that satisfy both conditions [tex]\(a + b = 1\)[/tex] and [tex]\(ab = 7\)[/tex]. As a result:

Therefore, the polynomial [tex]\(x^2 - x + 7\)[/tex] cannot be factored into simpler binomials with real coefficients. Hence, the correct answer is:
- Cannot be factored