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Question 18 (1 point) Find all zeros of \( f(x)=x^{4}-625 \). a \( \quad x=5,5 i \) b \( \quad x=-5,5 \) c \( \dot{x}=5,-5,5 i,-5 i \) d \( x=5,5 i,-5 i \) Question 19 (1 point) Find the zeros of the function \( f(x)=x^{3}-2 x^{2}-11 x+52 \) a \( \quad-4,3+2 i, 3-2 i \) b \( \quad 4,3+2 i, 3-2 i \) c \( \quad-4,-3+2 i,-3-2 i \) d \( \quad 4,-3+2 i,-3-2 i \) Review Answers

Ask by Munoz French. in the United States
Mar 17,2025

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**Question 18** The zeros of \( f(x) = x^4 - 625 \) are \( x = 5, -5, 5i, -5i \). **Answer:** c **Question 19** The zeros of \( f(x) = x^3 - 2x^2 - 11x + 52 \) are \( x = -4, 3 + 2i, 3 - 2i \). **Answer:** a

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To solve \( f(x) = x^{4} - 625 \), we can factor it as a difference of squares, yielding \( f(x) = (x^2 - 25)(x^2 + 25) \). This gives us the zeros \( x = 5, -5 \) from \( x^2 - 25 = 0 \) and \( x = 5i, -5i \) from \( x^2 + 25 = 0 \). Thus, the complete set of zeros is \( \{5, -5, 5i, -5i\} \). For the cubic function \( f(x) = x^{3} - 2x^{2} - 11x + 52 \), one can apply synthetic division or the Rational Root Theorem to find potential rational roots. After testing various values, we can find that \( x = 4 \) is a root. By further factorization, we find the quadratic \( x^2 + 2x - 13 \), yielding additional roots \( 3+2i \) and \( 3-2i \).

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