To factor the trinomial \(15n^2 - 27n - 6\) completely, we can follow these steps:
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Identify a, b, and c from the standard quadratic form \(ax^2 + bx + c\): \[ a = 15, \quad b = -27, \quad c = -6 \]
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Multiply \(a\) and \(c\): \[ ac = 15 \times -6 = -90 \]
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Find two numbers that multiply to \(ac\) (-90) and add to \(b\) (-27). The numbers we are looking for are -30 and 3 because: \[ -30 \times 3 = -90 \quad \text{and} \quad -30 + 3 = -27 \]
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Rewrite the middle term using these two numbers: \[ 15n^2 - 30n + 3n - 6 \]
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Group the terms: \[ (15n^2 - 30n) + (3n - 6) \]
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Factor out the common factors in each group: \[ 15n(n - 2) + 3(n - 2) \]
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Factor out the common binomial factor: \[ (n - 2)(15n + 3) \]
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Finally, to express it in its completely factored form, we can also factor out the 3 from the second binomial: \[ (n - 2)(3(5n + 1)) = 3(n - 2)(5n + 1) \]
Thus, the complete factorization of the trinomial \(15n^2 - 27n - 6\) is: \[ \boxed{3(n - 2)(5n + 1)} \]
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