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All the real zeros of the given polynomial are integers. Find the zeros. (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities.) \[ P(x)=x^{4}+9 x^{3}+17 x^{2}-9 x-18 \] Write the polynomial in factored form. \( P(x)=\square \)

Ask by Lawson Brewer. in the United States
Feb 27,2025

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Zeros: 1, −1, −3, −6 \( P(x) = (x - 1)(x + 1)(x + 3)(x + 6) \)

Solución

We are given the polynomial   P(x) = x⁴ + 9x³ + 17x² − 9x − 18. Step 1. Find an Integer Zero Check x = 1:   P(1) = 1 + 9 + 17 − 9 − 18 = 0. So, x = 1 is a zero. Step 2. Factor by Synthetic Division Divide P(x) by (x − 1):   Coefficients: 1, 9, 17, −9, −18   Using synthetic division with 1:       1 | 1  9  17  −9 −18           1  10  27  18       --------------------------         1  10  27  18  0 The quotient is Q(x) = x³ + 10x² + 27x + 18. Step 3. Factor the Cubic Q(x) Search for an integer zero of Q(x). Try x = −1:   Q(−1) = (−1)³ + 10(−1)² + 27(−1) + 18 = −1 + 10 − 27 + 18 = 0. So, x = −1 is a zero. Perform synthetic division with −1 on Q(x):   Coefficients: 1, 10, 27, 18   Using synthetic division with −1:       −1 | 1  10  27  18            −1  −9  −18       --------------------------         1   9  18   0 The resulting quadratic is R(x) = x² + 9x + 18. Step 4. Factor the Quadratic R(x) Find two numbers that add to 9 and multiply to 18. They are 3 and 6:   x² + 9x + 18 = (x + 3)(x + 6). Step 5. Write the Complete Factorization Now, we have:   P(x) = (x − 1)(x + 1)(x + 3)(x + 6). The zeros (with multiplicity) are: 1, −1, −3, −6. Final Answers:   Zeros: 1, −1, −3, −6   P(x) in factored form: (x − 1)(x + 1)(x + 3)(x + 6).

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Beyond the Answer

To find the real zeros of the polynomial \( P(x) = x^{4}+9x^{3}+17x^{2}-9x-18 \), we can use the Rational Root Theorem. The potential integer zeros could be the factors of -18, which include ±1, ±2, ±3, ±6, ±9, and ±18. By testing these values, we find that \( x = -2 \) is a root. Using synthetic division with \( x + 2 \), we can factor the polynomial to find the other roots. After performing the division, we end up with a cubic polynomial which can be factored further to identify the remaining roots. In this particular case, the cubic polynomial can be verified to have the other zeros of \( x = 1 \) with a multiplicity of 2. So the complete set of zeros are \( -2, 1, 1 \), and the factored form of the polynomial is: \( P(x) = (x + 2)(x - 1)^{2} = (x + 2)(x^2 - 2x + 1) \). In full, it can be expressed as: \[ P(x) = (x + 2)(x^2 - 2x + 1) \] So, the real zeros of the polynomial are: -2, 1, 1 And the polynomial in factored form is: \( P(x) = (x + 2)(x - 1)^2 \)

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