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I need some help on three homework questions, thank you!

 

1. Let P(x) = 0 be the polynomial equation of least possible degree, with rational coefficients, having \(\sqrt[3]{7} + \sqrt[3]{49}\) as a root. Compute the product of all of the roots of P(x) = 0.

 

2. A polynomial with integer coefficients is of the form \(3x^3 + a_2 x^2 + a_1 x - 6 = 0\).Enter all the possible integer roots of this polynomial, separated by commas.

 

3. The polynomial f(x) is divided by the polynomial d(x) to give a quotient of q(x) and a remainder of r(x). If deg f = 9 and deg r = 3, what is the maximum possible value of deg q?

 Apr 18, 2019
 #1
avatar+7763 
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\(\quad (\sqrt[3]{7}+\sqrt[3]{49})^3\\ = 7 + 49 + 3\sqrt[3]{7\cdot49}(\sqrt[3]{7}+\sqrt[3]{49})\\ = 56 + 21(\sqrt[3]{7}+\sqrt[3]{49})\)

Therefore P(x) is x3 - 21x - 56.

Product of roots = \(-\dfrac{-56}{1} = 56\)

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 Apr 18, 2019
 #4
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+2

Thank you for your help!

Guest Apr 18, 2019
 #2
avatar+7763 
+2

\(P(x) = 3x^3+a_2x^2+a_1x-6\)

Product of roots = \(-\dfrac{-6}{3}\)= 2.

 

Possible roots = {1,-1,2,-2}.

 Apr 18, 2019
 #3
avatar+7763 
+2

\(f(x) = q(x)d(x)+r(x)\)

deg f = 9, deg r = 3, deg d > deg r.

For maximum deg q. deg d is minimum. => deg d = 4.

\(O(x^9) = q(x) O(x^4) + O(x^3)\)

q(x) = O(x5)

Therefore deg q = 5.

 Apr 18, 2019

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