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

1 answer:

8 0

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

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Solve for x in the equation x squared + 2 x + 1 = 17.

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Answer:

$x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\$

Step-by-step explanation:

given equation

$x^2 +2x +1 = 17$

subtracting 17 from both sides

$x^2 +2x +1 = 17\\x^2 +2x +1 -17= 17-17\\x^2 +2x - 16 = 0\\$

the solution for quadratic equation

$ax^2 + bx + c = 0$ is given by

x = $x = -b + \sqrt{b^2 - 4ac} /2a \\\\and \ \\-b - \sqrt{b^2 - 4ac} /2a$

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in our problem

a = 1

b = 2

c = -16

$x =( -2 + \sqrt{2^2 - 4*1*-16}) /2*1 \\x =( -2 + \sqrt{4 + 64}) /2\\x =( -2 + \sqrt{68} )/2\\x = ( -2 + \sqrt{4*17} )/2\\x = ( -2 + 2\sqrt{17} )/2\\x = - 1 + \sqrt{17}\\and\\\\x = - 1 - \sqrt{17}\\$