Question

write a polynomial function of least degree with rational coefficients so that p(x)=0 has the given root. 4 - 7i p(x)=x² - □x + □

Explanation:

Step1: Identify the conjugate root

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

Step2: Form the polynomial from roots

If \(r_1\) and \(r_2\) are roots of a quadratic polynomial, the polynomial can be written as \(P(x)=(x - r_1)(x - r_2)\). Substituting \(r_1 = 4 - 7i\) and \(r_2 = 4 + 7i\), we get:
\[

$$\begin{align*} P(x)&=(x-(4 - 7i))(x-(4 + 7i))\\ &=(x - 4 + 7i)(x - 4 - 7i) \end{align*}$$

\]

Step3: Expand the product

Using the formula \((a + b)(a - b)=a^2 - b^2\), where \(a=(x - 4)\) and \(b = 7i\):
\[

$$\begin{align*} P(x)&=(x - 4)^2-(7i)^2\\ &=x^2-8x + 16-(49i^2) \end{align*}$$

\]
Since \(i^2=-1\), we substitute:
\[

$$\begin{align*} P(x)&=x^2-8x + 16+49\\ &=x^2-8x + 65 \end{align*}$$

\]

Answer:

The polynomial is \(P(x)=x^2 - 8x + 65\), so the first box (coefficient of \(x\)) is \(8\) and the second box (constant term) is \(65\).