To determine which polynomial divides evenly by \( x + 3 \), we can use synthetic division. A polynomial \( f(x) \) will be divisible by \( x + 3 \) if the remainder is 0 when \( f(-3) = 0 \). Let's evaluate each polynomial at \( x = -3 \).
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For \( 5x^2 + 7x - 12 \): \[ f(-3) = 5(-3)^2 + 7(-3) - 12 = 5(9) - 21 - 12 = 45 - 21 - 12 = 12 \quad (\text{not divisible}) \]
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For \( x^3 - 3x^2 - 13x + 15 \): \[ f(-3) = (-3)^3 - 3(-3)^2 - 13(-3) + 15 = -27 - 27 + 39 + 15 = -27 - 27 + 54 = 0 \quad (\text{divisible}) \]
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For \( x^3 - 5x^2 + 10x - 15 \): \[ f(-3) = (-3)^3 - 5(-3)^2 + 10(-3) - 15 = -27 - 45 - 30 - 15 = -117 \quad (\text{not divisible}) \]
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For \( 3x^2 - 6x + 9 \): \[ f(-3) = 3(-3)^2 - 6(-3) + 9 = 3(9) + 18 + 9 = 27 + 18 + 9 = 54 \quad (\text{not divisible}) \]
After performing the evaluations:
- \( 5x^2 + 7x - 12 \): Not divisible
- \( x^3 - 3x^2 - 13x + 15 \): Divisible
- \( x^3 - 5x^2 + 10x - 15 \): Not divisible
- \( 3x^2 - 6x + 9 \): Not divisible
Thus, the polynomial that divides evenly by \( x + 3 \) is: \[ \boxed{x^3 - 3x^2 - 13x + 15} \]