Question

write a polynomial function of least degree with rational coefficients so that ( p(x) = 0 ) has the given root. ( 4 - 7i ) ( p(x) = x^2 - square x + square )

Explanation:

Step1: Recall Conjugate Root Theorem

For a polynomial with rational coefficients, if a complex number \(a - bi\) is a root, then its conjugate \(a + bi\) is also a root. Given root is \(4 - 7i\), so the other root is \(4 + 7i\).

Step2: Form the polynomial

If \(r_1\) and \(r_2\) are roots of a quadratic polynomial, the polynomial is \(x^2-(r_1 + r_2)x + r_1r_2\).

First, find \(r_1 + r_2\):
\(r_1 = 4 - 7i\), \(r_2 = 4 + 7i\)
\(r_1 + r_2=(4 - 7i)+(4 + 7i)=8\)

Then, find \(r_1r_2\):
\(r_1r_2=(4 - 7i)(4 + 7i)=4^2-(7i)^2=16 - (-49)=16 + 49 = 65\)

So the polynomial is \(x^2 - 8x + 65\).

Answer:

\(P(x)=x^2 - 8x + 65\)