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Question: n²+4n-12 factor

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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.

May 24, 2024, 5:56 a.m.

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