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16 (a) Find the rational zeros and then the other zeros of the polynomial function \( f(x)=2 x^{3}+3 x^{2}+18 x+27 \), that is, solve \( f(x)=0 \). (b) Factor \( f(x) \) into linear factors. (a) Select the correct choice below and fill in any answer box(es) within your choice. (Type an exact answer, using radicals and \( i \) as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) A. There are only rational zeros and they are B. There is only one rational zero, \( \square \), and the other zeros are C. There are no rational zeros. The other zeros are

Ask by Vaughan Griffin. in the United States
Mar 13,2025

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(a) There is only one rational zero, \(-\frac{3}{2}\), and the other zeros are \(3i\) and \(-3i\). (b) The factorization into linear factors is: \[ (2x+3)(x-3i)(x+3i). \]

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To tackle the polynomial \( f(x) = 2x^3 + 3x^2 + 18x + 27 \), let's begin by applying the Rational Root Theorem. By testing potential rational zeros, we find that \( x = -3 \) is a root. Now we can use synthetic division or polynomial long division to factor the polynomial as \( f(x) = (x + 3)(2x^2 - 3) \). Next, we need to find the zeros of \( 2x^2 - 3 = 0 \), which gives us the other zeros: \( x = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2} \). Therefore, the complete solution is: A. There are only rational zeros and they are -3. Now, for the factorization, we get: \( f(x) = (x + 3)(2x - \sqrt{6})(2x + \sqrt{6}) \). Enjoy the journey down polynomial lane! Each zero you find is like a treasure unearthed, revealing more about the relationship between numbers. And don’t forget to celebrate those factoring skills; they’re your trusty compass for solving polynomial mysteries!

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