If 3−√7 is a polynomial root, name another root of the polynomial. SHOW ALL SETPS USED TO GET THE ANSWER secondly- explain how you know it must also be a root.
Given that 3−√7 is a root of the polynomial, let's denote this root as x1.
We know that if 3−√7 is a root, then its conjugate 3+√7 must also be a root. This is because complex roots of polynomials with real coefficients come in conjugate pairs.
Therefore, the other root of the polynomial is x2 = 3+√7.
To show that x2 = 3+√7 is a root, we can set up the polynomial equation and plug in x2:
Let the polynomial be P(x) = (x - x1)(x - x2)(x - x3)...(x - xn)
Substitute x = 3+√7 into P(x):
P(3+√7) = (3+√7 - (3-√7))(3+√7 - (3+√7))(x - x3)...(x - xn)
= (2√7)(0)(x - x3)...(x - xn)
= 0
Therefore, x2 = 3+√7 is also a root of the polynomial.