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To factor the quadratic expression \(2n^2 + 3n - 9\), we'll look for two numbers that multiply to \(2(-9) = -18\) (the product of the coefficient of \(n^2\) and the constant term) and add to 3 (the coefficient of \(n\)). The numbers that meet these criteria are 6 and -3 since \(6 \times -3 = -18\) and \(6 + (-3) = 3\). Using these numbers, we can split the middle term \(3n\) into two terms \(6n\) and \(-3n\): \[2n^2 + 6n - 3n - 9\] Next, we group the terms into pairs and factor out the greatest common factor (GCF) from each pair: \[2n^2 + 6n - 3n - 9 = (2n^2 + 6n) + (-3n - 9)\] From the first group \((2n^2 + 6n)\), we can factor out \(2n\): \[2n(n + 3)\] From the second group \((-3n - 9)\), we can factor out \(-3\): \[-3(n + 3)\] Now we have: \[2n(n + 3) - 3(n + 3)\] We can see that \((n + 3)\) is a common factor: \[(2n - 3)(n + 3)\] So, the factored form of the quadratic expression \(2n^2 + 3n - 9\) is: \[(2n - 3)(n + 3)\]

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