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Let \(r_1,r_2,...,r_7\) be the roots of \(f(x) = x^7 + 5x^6 - 4x^4 + 5x^3 - 9x^2 + x + a\). Let \(g(x) = x3 + 3x − 3\). Then the product \(g(r_1)g(r_2)...g(r_7)\) can be expressed as a polynomial in \(a\). Find this polynomial.

 Jun 30, 2024
edited by asdasdasddwasdwasd  Jun 30, 2024
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By Vieta's formulas, the sum of the roots of f(x) is −5, so the sum of the roots of x3+3x−3=0 is −(r1​+r2​+⋯+r7​)=5.

 

By the Factor Theorem, x−5 is a factor of x3+3x−3. We can use polynomial division (or inspection) to find that x3+3x−3=(x−5)(x2+8x+3). Then

 

\begin{align*} g(r_1) g(r_2) \dotsm g(r_7) &= (r_1 - 5)(r_1^2 + 8r_1 + 3) (r_2 - 5)(r_2^2 + 8r_2 + 3) \dotsm (r_7 - 5)(r_7^2 + 8r_7 + 3) \ &= (r_1 - 5)(r_2 - 5) \dotsm (r_7 - 5) (r_1^2 + r_2^2 + \dots + r_7^2 + 8(r_1 + r_2 + \dots + r_7) + 21). \end{align*}

 

By Vieta's formulas again, r12​+r22​+⋯+r72​=49−2⋅4+2⋅5−9+1=43, and r1​+r2​+⋯+r7​=−5.

 

Hence, \begin{align*} g(r_1) g(r_2) \dotsm g(r_7) &= (-5)(-5) \dotsm (-5) (43 + 8 \cdot -5 + 21) \ &= \boxed{125a + 3125}. \end{align*}

 Jul 1, 2024

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