Answer:
To show that the polynomial x² + 7 is irreducible over the real field (R), we need to demonstrate that it cannot be factored into a product of lower-degree polynomials with coefficients in R.
Assume, for contradiction, that x² + 7 is reducible over R. This would imply that it can be factored as:
x² + 7 = (ax + b)(cx + d)
Expanding the right side of the equation, we have:
x² + 7 = acx² + (ad + bc)x + bd
Comparing the coefficients on both sides, we get:
ac = 1 (1)
ad + bc = 0 (2)
bd = 7 (3)
From equation (1), we see that both a and c must be non-zero, as the coefficient of x² on the left side is 1. Similarly, from equation (3), b and d must also be non-zero.
Now, considering equation (2), we see that ad + bc = 0. Since a, b, c, and d are non-zero, this implies that ad and bc must have opposite signs.
However, when we look at equation (3), we see that bd = 7. As 7 is prime, it can only be factored as 7 * 1 or (-7) * (-1). In either case, bd has the same sign.
Therefore, we have a contradiction, as we cannot satisfy equations (2) and (3) simultaneously.
Since assuming the reducibility of x² + 7 leads to a contradiction, we can conclude that x² + 7 is irreducible over the real field (R).
