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Factoring the quadratic expression \(2n^2 + 3n - 9\) involves finding two binomials that multiply to give the original expression. Here's a step-by-step guide to help you factor it: 1. **Multiply the leading coefficient and the constant term**: The leading coefficient is \(2\) (from \(2n^2\)), and the constant term is \(-9\). Multiply these together: \[ 2 \times (-9) = -18 \] 2. **Find two numbers that multiply to \(-18\) and add to the middle coefficient, \(3\)**: We need two numbers that multiply to \(-18\) and add to \(3\). These numbers are \(6\) and \(-3\) because: \[ 6 \times (-3) = -18 \] \[ 6 - 3 = 3 \] 3. **Rewrite the middle term using the two numbers found**: Rewrite \(3n\) as \(6n - 3n\) in the original quadratic expression: \[ 2n^2 + 6n - 3n - 9 \] 4. **Group the terms**: Group the first two terms together and the last two terms together: \[ (2n^2 + 6n) + (-3n - 9) \] 5. **Factor out the greatest common factor (GCF) from each group**: - From the first group, \(2n^2 + 6n\), factor out \(2n\): \[ 2n(n + 3) \] - From the second group, \(-3n - 9\), factor out \(-3\): \[ -3(n + 3) \] 6. **Factor out the common binomial**: Both groups now contain the common binomial \((n + 3)\): \[ 2n(n + 3) - 3(n + 3) \] 7. **Factor out \((n + 3)\)**: \[ (n + 3)(2n - 3) \] So, the factored form of \(2n^2 + 3n - 9\) is: \[ (n + 3)(2n - 3) \] I hope this helps! Let me know if you have any further questions.

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