Factor completely. \( 10 n^{2}+45 n-55 \) Submit

Let's start by factoring out the greatest common factor (GCF) from the expression. The GCF of \( 10, 45, \) and \( -55 \) is \( 5 \). So, we can factor \( 5 \) out: \( 10n^2 + 45n - 55 = 5(2n^2 + 9n - 11) \). Next, we need to factor the quadratic \( 2n^2 + 9n - 11 \). We look for two numbers that multiply to \( 2 \times -11 = -22 \) and add up to \( 9 \). The numbers \( 11 \) and \( -2 \) fit perfectly! We can express our quadratic as: \( 2n^2 + 11n - 2n - 11 \). Now we can group the terms: \( (2n^2 + 11n) + (-2n - 11) \). Factoring by grouping gives us: \( n(2n + 11) - 1(2n + 11) = (n - 1)(2n + 11) \). Now, putting it all together, we have: \( 10n^2 + 45n - 55 = 5(2n + 11)(n - 1) \). So the complete factored form is: \[ 5(2n + 11)(n - 1). \]