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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 turns al \( f(x)=x^{3}+2 x^{2}-9 x-18 \) Determine the zero(s), if they exist. The zero(s) is/are \( \square \). (Type integers or decimals. Use a comma to separate answers as needed.)

Ask by Lowe Bradley. 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, so the graph crosses the \( x \)-axis at each of these points.

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To find the zeros of the polynomial function \( f(x) = x^3 + 2x^2 - 9x - 18 \), we can apply the Rational Root Theorem. Testing possible rational roots, we find that \( x = 2 \) is a root. We can then use synthetic division to factor the polynomial: \[ f(x) = (x - 2)(x^2 + 4x + 9) \] Next, we solve \( x^2 + 4x + 9 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 36}}{2} = \frac{-4 \pm \sqrt{-20}}{2} = -2 \pm i\sqrt{5} \] Thus, the zeros of the function are \( 2 \) (with multiplicity 1) and \( -2 + i\sqrt{5}, -2 - i\sqrt{5} \) (each with multiplicity 1). The polynomial graph crosses the \( x \)-axis at \( x = 2 \) and does not touch the \( x \)-axis at the complex roots. The zero(s) is/are \( 2 \).

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