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To factor the quadratic expression \( n^2 + 4n - 12 \), you'll need to find two numbers that multiply to give the constant term (-12) and add to give the coefficient of the middle term (4). Here is a step-by-step process: 1. **Identify a, b, and c in the quadratic expression \( an^2 + bn + c \)**: \[ a = 1, \quad b = 4, \quad c = -12 \] 2. **Find two numbers that multiply to \( a \cdot c = 1 \cdot (-12) = -12 \) and add to \( b = 4 \)**: - The pair of numbers that work here are 6 and -2 because: \[ 6 \times (-2) = -12 \quad \text{and} \quad 6 + (-2) = 4 \] 3. **Rewrite the middle term (4n) using the two numbers found**: \[ n^2 + 6n - 2n - 12 \] 4. **Group the terms into two pairs**: \[ (n^2 + 6n) + (-2n - 12) \] 5. **Factor out the greatest common factor (GCF) from each pair**: - From \( n^2 + 6n \), factor out \( n \): \[ n(n + 6) \] - From \( -2n - 12 \), factor out \(-2\): \[ -2(n + 6) \] 6. **Factor out the common binomial factor \((n + 6)\)**: \[ n(n + 6) - 2(n + 6) = (n - 2)(n + 6) \] So, the factored form of \( n^2 + 4n - 12 \) is: \[ (n - 2)(n + 6) \] You can verify this by expanding the factored form: \[ (n - 2)(n + 6) = n^2 + 6n - 2n - 12 = n^2 + 4n - 12 \] This confirms that the factorization is correct.

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