Question

Find a polynomial with integer coefficients that satisfies the given conditions.

P has degree 2 and zeros 5 + i and 5 − i.

Answer 1

To construct a polynomial with integer coefficients and zeros at 5 + i and 5 - i, the factors (x - (5 + i)) and (x - (5 - i)) are multiplied to yield the quadratic polynomial x^2 - 10x + 26.

Step-by-step explanation:

To find a polynomial with integer coefficients that has the given zeros of 5 + i and 5 - i, we should use the fact that complex roots of polynomials with real coefficients occur in conjugate pairs. This means that if a polynomial has a complex zero like 5 + i, it must also have a zero of 5 - i. As the polynomial is of degree 2 and has integer coefficients, we can start from the factors corresponding to its zeros.

So for the zeros 5 + i and 5 - i, we can write the factors of the polynomial as (x - (5 + i)) and (x - (5 - i)). Multiplying these two factors will give us our polynomial:

Expanding this, we get:

Therefore, the polynomial with integer coefficients and the given zeros is x2 - 10x + 26.

Answer 2

`\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b) \\\\\\ \textit{also recall that }~~~i^2=-1\\\\ -------------------------------\\\\ \begin{cases} x=5+i\implies &x-5-i=0\\ x=5-i\implies &x-5+i=0 \end{cases} \\\\\\ (x-5-i)(x-5+i)=\stackrel{\textit{original polynomial}}{0} \\\\\\\ [(x-5)~~-~~i]~[(x-5)~~+~~i]=0\implies [(x-5)^2~~-~~i^2]=0 \\\\\\ (x^2-10x+25)~~-~~(-1)=0\implies x^2-10x+25+1=y \\\\\\ x^2-10x+26=y`