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Factur the polynomial and use the factored form to find the zerus. [Enter your answers as a corna-segarated list, Entar all answers using the apgropriate multiplicities.) \[ P(x)=x^{3}+2 x^{2}-4 x-8 \] Sketch the graph.

Ask by Lynch Estrada. in the United States
Feb 27,2025

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Answer

Zeros: 2, –2, –2

Solution

We start with the polynomial   P(x) = x³ + 2x² – 4x – 8. Step 1. Find a root by testing possible candidates. Trying x = 2 gives:   P(2) = 2³ + 2·2² – 4·2 – 8 = 8 + 8 – 8 – 8 = 0. Since x = 2 is a zero, (x – 2) is a factor. Step 2. Divide P(x) by (x – 2) to find the remaining quadratic factor. Using synthetic division with 2:   Coefficients: 1  2  –4  –8   Bring down the 1.   Multiply:   1×2 = 2; add to 2 to get 4.   Multiply:   4×2 = 8; add to –4 to get 4.   Multiply:   4×2 = 8; add to –8 to get 0. So the quotient is x² + 4x + 4. Step 3. Factor the quadratic:   x² + 4x + 4 = (x + 2)². Thus the completely factored form of P(x) is:   P(x) = (x – 2)(x + 2)². Step 4. Identify the zeros from the factors:   • From (x – 2) = 0, we get x = 2 (simple zero).   • From (x + 2)² = 0, we get x = –2 with multiplicity 2. Enter these as a comma-separated list with multiplicities:   2, –2, –2 Sketching the Graph: • x-intercepts: The graph crosses or touches the x-axis at x = 2 and x = –2. Notice that at x = –2 (a double zero) the graph will "touch" the axis and turn around rather than crossing it. • End Behavior: Since P(x) is a cubic polynomial with positive leading coefficient:   – as x → –∞, P(x) → –∞.   – as x → ∞, P(x) → ∞. • y-intercept: Substitute x = 0:   P(0) = 0 + 0 – 0 – 8 = –8.   So the point (0, –8) is on the graph. • Additional features: The graph will have an inflection point typical of a cubic function and will flatten out at x = –2 (because of the double root). In summary:  Factored Form: P(x) = (x – 2)(x + 2)²  Zeros (with multiplicities): 2, –2, –2

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Beyond the Answer

To factor the polynomial \( P(x) = x^3 + 2x^2 - 4x - 8 \), we can use synthetic division by testing possible rational roots. In this case, we find that \( x = -2 \) is a root. Using synthetic division on \( P(x) \), we obtain a quadratic \( x^2 + 2 \) that factors to \( (x + 2)(x^2 + 2) \) after the first division step. The roots of \( P(x) \) are then \( x = -2 \) (with multiplicity 1) and \( x^2 + 2 = 0 \), which gives complex roots \( x = i\sqrt{2} \) and \( x = -i\sqrt{2} \). Now, the factored form is: \[ P(x) = (x + 2)(x - i\sqrt{2})(x + i\sqrt{2}) \] For the zeros, we summarize them as: - \( x = -2 \) (multiplicity 1) - \( x = i\sqrt{2} \) (multiplicity 1) - \( x = -i\sqrt{2} \) (multiplicity 1) Using graphing tools, you can sketch the graph, which would show a real root at \( x = -2 \) crossing the x-axis and indicating that there are no other real x-intercepts due to the complex roots. The graph approaches infinity as \( x \) moves to positive or negative infinity. To ensure you’ve tackled polynomial problems effectively, remember to check for synthetic division accuracy and try out the Rational Root Theorem as a first step. Many students forget to check for all possible rational roots, so it’s essential to thoroughly test several values before concluding. Always verify your roots by substituting them back into the original polynomial!

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