To determine if a polynomial divides evenly by \( x + 3 \), we can use synthetic division or direct evaluation with the Remainder Theorem. The Remainder Theorem states that if you substitute \( -3 \) (the root of \( x + 3 = 0 \)) into the polynomial, and the result equals zero, then the polynomial is divisible by \( x + 3 \).
Let's evaluate each polynomial for \( x = -3 \):
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For \( 3x^2 - 6x + 9 \): \[ 3(-3)^2 - 6(-3) + 9 = 3(9) + 18 + 9 = 27 + 18 + 9 = 54 \quad \text{(not zero)} \]
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For \( x^3 - 3x^2 - 13x + 15 \): \[ (-3)^3 - 3(-3)^2 - 13(-3) + 15 = -27 - 27 + 39 + 15 = -27 - 27 + 39 + 15 = 0 \quad \text{(zero)} \]
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For \( x^3 - 5x^2 + 10x - 15 \): \[ (-3)^3 - 5(-3)^2 + 10(-3) - 15 = -27 - 45 - 30 - 15 = -117 \quad \text{(not zero)} \]
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For \( 5x^2 + 7x - 12 \): \[ 5(-3)^2 + 7(-3) - 12 = 5(9) - 21 - 12 = 45 - 21 - 12 = 12 \quad \text{(not zero)} \]
From these evaluations, only the polynomial \( x^3 - 3x^2 - 13x + 15 \) gives a result of zero when evaluated at \( x = -3 \).
Conclusion: The polynomial that divides evenly by \( x + 3 \) is \( x^3 - 3x^2 - 13x + 15 \).
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