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Find all zeros of the function \( f(x)=8 x^{3}+2 x^{2}-7 x-3 \). Enter the zeros separated by commas. You may use technology to find ONE rational zero. You must use algebraic methods to find any additional rational solution(s), and/or irrational solution(s), showing all work, steps, and logic.

Ask by Stewart Little. in the United States
Feb 20,2025

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The zeros of the function are \( x=-\frac{3}{4}, x=-\frac{1}{2}, x=1 \).

Solución

Solve the equation \( 8x^{3}+2x^{2}-7x-3=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(8x^{3}+2x^{2}-7x-3=0\) - step1: Factor the expression: \(\left(x-1\right)\left(2x+1\right)\left(4x+3\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-1=0\\&2x+1=0\\&4x+3=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=1\\&x=-\frac{1}{2}\\&x=-\frac{3}{4}\end{align}\) - step4: Rewrite: \(x_{1}=-\frac{3}{4},x_{2}=-\frac{1}{2},x_{3}=1\) The zeros of the function \( f(x)=8x^{3}+2x^{2}-7x-3 \) are \( x=-\frac{3}{4}, x=-\frac{1}{2}, x=1 \).

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Mind Expander

To find the zeros of the function \( f(x) = 8x^3 + 2x^2 - 7x - 3 \), we can begin by applying the Rational Root Theorem. This theorem suggests that any rational solution is of the form \( \frac{p}{q} \) where \( p \) is a factor of the constant term (-3) and \( q \) is a factor of the leading coefficient (8). The factors of -3 are: \( \pm 1, \pm 3 \) The factors of 8 are: \( \pm 1, \pm 2, \pm 4, \pm 8 \) Thus, the potential rational zeros are: \[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{8} \] Testing \( x = 1 \) in \( f(x) \): \[ f(1) = 8(1)^3 + 2(1)^2 - 7(1) - 3 = 8 + 2 - 7 - 3 = 0 \] So, \( x = 1 \) is a zero. Now we will perform synthetic division of \( f(x) \) by \( x - 1 \): \[ \begin{array}{r|rrrr} 1 & 8 & 2 & -7 & -3 \\ & & 8 & 10 & 3 \\ \hline & 8 & 10 & 3 & 0 \\ \end{array} \] This gives us \( 8x^2 + 10x + 3 \). Now we need to find the zeros of \( 8x^2 + 10x + 3 \). We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 8 \), \( b = 10 \), and \( c = 3 \). Calculating the discriminant: \[ D = b^2 - 4ac = 10^2 - 4(8)(3) = 100 - 96 = 4 \] Now substituting back into the formula: \[ x = \frac{-10 \pm \sqrt{4}}{2 \cdot 8} = \frac{-10 \pm 2}{16} \] Calculating both solutions: 1. \( x = \frac{-10 + 2}{16} = \frac{-8}{16} = -\frac{1}{2} \) 2. \( x = \frac{-10 - 2}{16} = \frac{-12}{16} = -\frac{3}{4} \) Thus, the zeros of \( f(x) = 8x^3 + 2x^2 - 7x - 3 \) are: \[ \boxed{1, -\frac{1}{2}, -\frac{3}{4}} \]

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