Question

if we know that (6 + i) is a root of a 2nd degree polynomial, find the remaining root. do not use any spaces in your answer.

question 6
1 pts

if we know that (4+sqrt{3}) is a root of a 2nd degree polynomial, find the remaining root. do not use any spaces in your answer. use \sqrt(... )\ to represent the square root symbol in your answer, such as \sqrt(5)\ to mean (sqrt{5}).

question 7
1 pts

if we know that (-7-sqrt{11}) and (-6 - 8i) are roots of a 4th degree polynomial, find the remaining roots. do not use any spaces in your answer. use \sqrt(... )\ to represent the square root symbol in your answer, such as \sqrt(5)\ to mean (sqrt{5}).

a. (-7-sqrt{11}) tells us that is also a root.

b. (-6 - 8i) tells us that is also a root.

Explanation:

Step1: Recall the conjugate root theorem

For a polynomial with real - coefficients, if \(a + bi\) is a root, then \(a - bi\) is also a root. If \(a+\sqrt{b}\) is a root (\(a,b\in\mathbb{Q}\), \(b>0\) and \(\sqrt{b}
otin\mathbb{Q}\)), then \(a - \sqrt{b}\) is also a root.

For the first polynomial (with root \(6 + i\)):

Step2: Find the conjugate of \(6 + i\)

The conjugate of a complex number \(z=a+bi\) is \(\overline{z}=a - bi\). For \(z = 6 + i\) (\(a = 6\), \(b = 1\)), the conjugate is \(6 - i\).

For the second polynomial (with root \(4+\sqrt{3}\)):

Step3: Find the conjugate of \(4+\sqrt{3}\)

The conjugate of \(a+\sqrt{b}\) (where \(a = 4\), \(b = 3\)) is \(a-\sqrt{b}\). So the conjugate is \(4-\sqrt{3}\).

For the third polynomial (with roots \(-7-\sqrt{11}\) and \(-6 - 8i\)):

Step4: Find the conjugate of \(-7-\sqrt{11}\)

The conjugate of \(a-\sqrt{b}\) (where \(a=-7\), \(b = 11\)) is \(a+\sqrt{b}\). So the conjugate of \(-7-\sqrt{11}\) is \(-7+\sqrt{11}\).

Step5: Find the conjugate of \(-6 - 8i\)

The conjugate of \(a+bi\) (where \(a=-6\), \(b=-8\)) is \(a - bi\). So the conjugate of \(-6 - 8i\) is \(-6+8i\).

Answer:

6 - i
4 - sqrt(3)
A. - 7 + sqrt(11)
B. - 6 + 8i