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5th degree polynomial

Given the equation \(x^5+5(2x^3+x+2)=5x^2(x^2+2)\) has a real root in the form of \(a-\sqrt[5]{b}\) where \(a\) and \(b\) are real positive integers. Find \((a,b)\)

Note by William Isoroku
2 years, 1 month ago

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\[x^5+5(2x^3+x+2)=5x^2(x^2+2)\\ x^5+10x^3+5x+10 = 5x^4+10x^2 \\ x^5 -5x^4+10x^3-10x^2+5x+10 = 0 \\ x^5 -5x^4+10x^3-10x^2+5x-1+1 +10 = 0 \\ (x-1)^5 = -11 \\ x-1 = - \sqrt [5] {11} \\ x = 1 - \sqrt [5] {11} \\ \Rightarrow \boxed{a = 1 \quad b = 11} \] Chew-Seong Cheong · 2 years, 1 month ago

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@Chew-Seong Cheong Oh, so we have to know the \((x-1)^5\) factorization here right? William Isoroku · 2 years, 1 month ago

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@William Isoroku Yes, it is best that you remember first few rows of the Pascal triangle. They are actual the binomial expansion coefficients, which is given by \(\begin{pmatrix} n \\ r \end{pmatrix}\). For example, when \(n=5\), we have:

\[\begin{equation} \begin{split} (x-1)^5 & = \sum_{r=0}^5 {(-1)^r\begin{pmatrix} 5 \\ r \end{pmatrix}x^{5-r}} \\ & = \begin{pmatrix} 5 \\ 0 \end{pmatrix}x^5 - \begin{pmatrix} 5 \\ 1 \end{pmatrix}x^4 + \begin{pmatrix} 5 \\ 2 \end{pmatrix}x^3 - \begin{pmatrix} 5 \\ 3 \end{pmatrix}x^2 + \begin{pmatrix} 5 \\ 4 \end{pmatrix}x^1 - \begin{pmatrix} 5 \\ 5 \end{pmatrix}x^0 \\ & = x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1 \end{split} \end{equation} \] Chew-Seong Cheong · 2 years, 1 month ago

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@Chew-Seong Cheong How you thought it will be (x-1)^5? Sir. Ayush Verma · 2 years, 1 month ago

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@Ayush Verma Pascal triangle:

\[1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1\\ 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \] Chew-Seong Cheong · 2 years, 1 month ago

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@Chew-Seong Cheong Thanks Ayush Verma · 2 years, 1 month ago

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