To factor the expression \( 20n^2 - 60n - 35 \) completely, follow these steps:
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Factor out the greatest common factor (GCF): The coefficients \(20\), \(-60\), and \(-35\) have a GCF of \(5\). So we can factor out \(5\):
\[ 20n^2 - 60n - 35 = 5(4n^2 - 12n - 7) \]
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Factor the quadratic \(4n^2 - 12n - 7\). We will use the factoring method known as the AC method or trial and error. First, we need to find two numbers that multiply to \(4(-7) = -28\) and add up to \(-12\).
The factor pairs of \(-28\) are:
- \(1\) and \(-28\)
- \(-1\) and \(28\)
- \(2\) and \(-14\)
- \(-2\) and \(14\)
- \(4\) and \(-7\)
- \(-4\) and \(7\)
The pair \(-14\) and \(2\) adds to \(-12\).
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Rewrite the middle term using the chosen pair:
\[ 4n^2 - 14n + 2n - 7 \]
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Group the terms:
\[ (4n^2 - 14n) + (2n - 7) \]
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Factor out the common factors in each group:
\[ 2n(2n - 7) + 1(2n - 7) \]
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Factor out the common binomial:
\[ (2n - 7)(2n + 1) \]
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Combine everything together:
Now, we will write the complete factorization including the GCF we factored out initially:
\[ 5(2n - 7)(2n + 1) \]
Thus, the completely factored form of the expression \(20n^2 - 60n - 35\) is:
\[ \boxed{5(2n - 7)(2n + 1)} \]