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Sagot :
The remaining root of f(x) will be 3 - [tex]\sqrt{11}[/tex]
To answer this question we have to find the conjugate of 3+[tex]\sqrt{11}[/tex],
For example, the conjugate of p+q will be p-q or vice versa.
Let us consider x= 2+[tex]\sqrt{11}[/tex]
To simplify the equation let us subtract 3 on both sides of the equation
x - 3 = 3+[tex]\sqrt{11}[/tex] - 3
x - 3 = [tex]\sqrt{11}[/tex]
Square both,
[tex](x - 3)^{2}[/tex] = 11
[tex](x - 3)^{2}[/tex] - 11 =0
Now we will use the principle of Square difference to factorize both sides-
[tex](d^{2} - e^{2}) = (d - e)(d + e)[/tex]
(x - 3) - 11 × (x - 3) + 11 = 0
on solving the equation we get,
(x - 3) - 1 = 0 let us consider this first equation
or (x - 3) + 11 = 0 let us consider this second equation
Add [tex]\sqrt{11}[/tex] on both sides of the first equation,
[tex]x - 3 = \sqrt{11}[/tex]
Subtract [tex]\sqrt{11}[/tex] on both sides of the second equation,
[tex]x - 3 = -\sqrt{11}[/tex]
Now by adding 3 on both sides of the first and second equation,
[tex]x = 3 + \sqrt{11}[/tex] or [tex]x = 3 - \sqrt{11}[/tex]
Therefore, the remaining root of f(x) will be = [tex]3 - \sqrt{11}[/tex]
To learn about Rational Coefficients,
https://brainly.com/question/19157484
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Complete Question - If a polynomial function f(x), with rational coefficients, has roots 0, 4, and 3 + [tex]\sqrt{11}[/tex], what must also be a root of f(x)?.
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