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Find the zeros for the polynomial function and give the multiplicity for each zero. State whether the graph crosses the \( x \)-axis or touches the \( x \)-axis and turn \( f(x)=x^{3}+2 x^{2}-9 x-18 \) Determine the zero(s), if they exist. The zero(s) is/are (Type integers or decimals. Use a comma to separate answers as needed.)

Ask by Matthews Vega. in the United States
Mar 19,2025

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The zeros of the polynomial \( f(x) = x^{3} + 2x^{2} - 9x - 18 \) are \( x = -3 \), \( x = -2 \), and \( x = 3 \). Each zero has a multiplicity of 1, meaning the graph crosses the \( x \)-axis at each of these points.

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The Deep Dive

To find the zeros of the polynomial function \( f(x) = x^{3} + 2x^{2} - 9x - 18 \), we can use the Rational Root Theorem to test potential rational roots. By testing \( x = -3 \), we find: \[ f(-3) = (-3)^{3} + 2(-3)^{2} - 9(-3) - 18 = -27 + 18 + 27 - 18 = 0 \] So, \( x = -3 \) is a zero. Next, we can perform synthetic division on \( f(x) \) using \( x + 3 \): \[ \begin{array}{r|rrrr} -3 & 1 & 2 & -9 & -18 \\ & & -3 & 3 & 18 \\ \hline & 1 & -1 & -6 & 0 \\ \end{array} \] The quotient is \( x^{2} - x - 6 \). We can factor this further: \[ x^{2} - x - 6 = (x - 3)(x + 2) \] This gives us three zeros: \( x = -3 \), \( x = 3 \), and \( x = -2 \). The multiplicities for each zero are: - \( x = -3 \): multiplicity 1 (crosses \( x \)-axis) - \( x = 3 \): multiplicity 1 (crosses \( x \)-axis) - \( x = -2 \): multiplicity 1 (crosses \( x \)-axis) Thus, the zeros are: \(-3, 3, -2\).

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