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Question: Let $P = 27+108x+90x^2-80x^3-60x^4+48x^5-8x^6$, find $\sqrt[3]{P}$.

Just wondering if there is a general way for dealing with this sort of question, I was able to figure how to find $\sqrt{P}$, but $\sqrt[3]{P}$ is beyond me. Long division is used in an example, unfortunately it lacks justification.

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I think a procedure is given in some 19th century algebra textbooks, but I don't have time now to look. Freely available .pdf files of the books I'm thinking of are linked in this recent math stackexchange post of mine, in the paragraph that begins with In older advanced "school level" algebra texts from the 1800s ... –  Dave L. Renfro Apr 25 '13 at 16:03
    
@Dave L. Renfro: yes, the above example is from a Hall/Knight algebra book; some really good explanations in your thread there. –  Orpheus Apr 30 '13 at 2:54

2 Answers 2

up vote 3 down vote accepted

The hope is that $\sqrt[3]{P}$ is also a polynomial. If $\sqrt[3]{P}$ were to be a polynomial, then $\sqrt[3]{P}$ should be of the form $$\sqrt[3]{P} = (-2x^2 + ax + 3)$$ since the coefficient of $x^6$ is $-8$ and the constant term is $27$. Can you take this further?

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Yes, $\sqrt[3]{P}$ is also a polynomial. Just can't figure out how it is done. –  Orpheus Apr 25 '13 at 3:39
    
@Orpheus What are your thoughts on how to proceed? –  user17762 Apr 25 '13 at 3:41
    
yep it's needed to compare coefficients, cheers. –  Orpheus Apr 25 '13 at 3:48

Let $(3+ax-2x^2)^3=27+108x+90x^2-80x^3-60x^4+48x^5-8x^6$

Now use this and compare the coefficients of the different powers of $x$ to find the unique value of $a$ such that all the equation formed by comparing the coefficients remain consistent.

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It should suffice to look at just, say, the $x^5$ terms, with possibly the linear terms as a check, since they require the fewest permutations in multiplying out the cube of the putative cube-root polynomial. The $x^5$ can only be formed from three sets of $(-2x^2)(-2x^2)(ax)$ and the linear terms from three sets of $3 \cdot 3 \cdot ax$. If those don't produce the same value of $a$, our polynomial won't serve as a cube root of $P$. –  RecklessReckoner Apr 25 '13 at 17:26

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