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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

Solución

**Question 18** We are given \[ f(x)=x^4-625 \] and we need to find its zeros. 1. Recognize that \(625=25^2\) and write the expression as a difference of squares: \[ x^4-625 = \left(x^2\right)^2 - 25^2. \] 2. Factor using the difference of squares formula, \(a^2-b^2=(a-b)(a+b)\): \[ x^4-625 = (x^2-25)(x^2+25). \] 3. Factor the first quadratic further since \(25=5^2\): \[ x^2-25=(x-5)(x+5). \] 4. Set the factors equal to zero: - For \(x-5=0\), we have \(x=5\). - For \(x+5=0\), we have \(x=-5\). - For \(x^2+25=0\), we solve \[ x^2=-25 \quad\Longrightarrow\quad x=\pm\sqrt{-25}=\pm5i. \] The zeros of \(f(x)\) are \[ x=5,\quad -5,\quad 5i,\quad -5i. \] Thus, the correct answer is option (c). --- **Question 19** We are given \[ f(x)=x^3-2x^2-11x+52 \] and we need to find its zeros. 1. Use the Rational Root Theorem to test possible rational roots. Test \(x=-4\): \[ (-4)^3-2(-4)^2-11(-4)+52 = -64-2(16)+44+52 = -64-32+44+52 = 0. \] Since \(f(-4)=0\), \(x=-4\) is a root. 2. Factor \(x+4\) from the polynomial using synthetic division. - Write the coefficients: \(1, -2, -11, 52\). - Perform synthetic division with \(x=-4\): \[ \begin{array}{r|rrrr} -4 & 1 & -2 & -11 & 52\\[5mm] & & -4 & 24 & -52\\ \hline & 1 & -6 & 13 & 0\\ \end{array} \] The quotient is \(x^2-6x+13\). 3. Solve the quadratic equation \[ x^2-6x+13=0. \] Use the quadratic formula: \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, \] where \(a=1\), \(b=-6\), and \(c=13\). Calculate the discriminant: \[ b^2-4ac = (-6)^2-4(1)(13)=36-52=-16. \] Since the discriminant is negative, the solutions are complex: \[ x=\frac{6\pm\sqrt{-16}}{2}=\frac{6\pm4i}{2}=3\pm2i. \] The zeros of the function \(f(x)\) are \[ x=-4,\quad 3+2i,\quad 3-2i. \] Thus, the correct answer is option (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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