Question

1. \\(\frac{15n^3 - 6n^2 - 9n}{15n^2 + 12n}\\)

Explanation:

Step1: Factor numerator and denominator

Factor numerator: \(15n^3 - 6n^2 - 9n = 3n(5n^2 - 2n - 3)\). Further factor \(5n^2 - 2n - 3=(5n + 3)(n - 1)\), so numerator is \(3n(5n + 3)(n - 1)\).
Factor denominator: \(15n^2 + 12n = 3n(5n + 4)\)? Wait, no, \(15n^2 + 12n = 3n(5n + 4)\)? Wait, original denominator: \(15n^2 + 12n\), factor out \(3n\): \(3n(5n + 4)\)? Wait, no, let's re - check the numerator's quadratic. Wait, \(5n^2-2n - 3\): using quadratic formula, \(n=\frac{2\pm\sqrt{4 + 60}}{10}=\frac{2\pm8}{10}\), so \(n = 1\) or \(n=-\frac{3}{5}\). So \(5n^2-2n - 3=(n - 1)(5n+3)\). Then numerator: \(3n(n - 1)(5n + 3)\). Denominator: \(15n^2+12n=3n(5n + 4)\)? Wait, no, \(15n^2+12n = 3n(5n + 4)\)? Wait, \(15n^2+12n=3n(5n + 4)\)? Wait, no, \(15n^2+12n = 3n(5n + 4)\)? Wait, let's compute \(3n(5n + 4)=15n^2+12n\), yes. Wait, but numerator has \((5n + 3)\) and denominator has \((5n + 4)\), no common factors? Wait, maybe I made a mistake in factoring the denominator. Wait, the original denominator is \(15n^2 + 12n\), factor out \(3n\): \(3n(5n + 4)\). The numerator: \(15n^3-6n^2 - 9n=3n(5n^2-2n - 3)=3n(5n + 3)(n - 1)\) (since \(5n^2-2n - 3=(5n + 3)(n - 1)\) as \(5n\times n=5n^2\), \(5n\times(-1)+3\times n=-5n + 3n=-2n\), \(3\times(-1)=-3\)). Then denominator: \(15n^2+12n = 3n(5n + 4)\). Wait, but there is a mistake here? Wait, no, let's check the problem again. Wait, maybe the denominator is \(15n^2 + 12n\) or is it \(15n^2+21n\)? Wait, the user's problem: denominator is \(15n^2 + 12n\). Wait, maybe I misread the denominator. Wait, let's re - express the fraction:

\(\frac{15n^3-6n^2 - 9n}{15n^2+12n}=\frac{3n(5n^2-2n - 3)}{3n(5n + 4)}\) (after factoring out \(3n\) from numerator and denominator, \(n eq0\))

Cancel out the common factor \(3n\) (for \(n eq0\)):

\(\frac{5n^2-2n - 3}{5n + 4}\). But we can also perform polynomial long division or rewrite the numerator. Wait, \(5n^2-2n - 3=(5n + 4)(n)-6n - 3\). Wait, no, let's do polynomial long division of \(5n^2-2n - 3\) by \(5n + 4\).

Divide \(5n^2\) by \(5n\) to get \(n\). Multiply \(5n + 4\) by \(n\) to get \(5n^2+4n\). Subtract from \(5n^2-2n - 3\): \((5n^2-2n - 3)-(5n^2+4n)=-6n - 3\). Now divide \(-6n\) by \(5n\) to get \(-\frac{6}{5}\). Multiply \(5n + 4\) by \(-\frac{6}{5}\) to get \(-6n-\frac{24}{5}\). Subtract: \((-6n - 3)-(-6n-\frac{24}{5})=-3+\frac{24}{5}=\frac{9}{5}\). So \(\frac{5n^2-2n - 3}{5n + 4}=n-\frac{6}{5}+\frac{\frac{9}{5}}{5n + 4}\). But this seems complicated. Wait, maybe I made a mistake in factoring the denominator. Wait, the original denominator: \(15n^2 + 12n\), maybe it's \(15n^2+21n\)? Let's check the numerator and denominator again. Wait, if the denominator was \(15n^2 + 21n\), then denominator would be \(3n(5n + 7)\), no. Wait, maybe the user made a typo, but assuming the problem is as given. Wait, no, let's re - factor the numerator and denominator correctly.

Wait, numerator: \(15n^3-6n^2 - 9n = 3n(5n^2-2n - 3)=3n(5n + 3)(n - 1)\) (correct, as \((5n + 3)(n - 1)=5n^2-5n+3n - 3=5n^2-2n - 3\)). Denominator: \(15n^2+12n = 3n(5n + 4)\) (correct, as \(3n(5n + 4)=15n^2+12n\)). So there are no common factors between numerator and denominator except \(3n\) (when \(n eq0\)). Wait, but that can't be. Wait, maybe the denominator is \(15n^2+21n\)? Let's check: \(15n^2+21n = 3n(5n + 7)\), no. Wait, maybe the denominator is \(15n^2+18n\)? \(15n^2+18n=3n(5n + 6)\), no. Wait, maybe I made a mistake in the numerator's quadratic. Wait, \(5n^2-2n - 3\): discriminant \(D = 4 + 60 = 64\), square root of \(D = 8\), so roots are \(n=\frac{2\pm8}{10}\), so \(n = 1\) or \(n=-\frac{3}{5}\), so factoring is correct.

Wait, maybe the problem is \(\frac{15n^3-6n^2 - 9n}{15n^2+21n}\)? Let's try that. Denominator \(15n^2+21n=3n(5n + 7)\), no. Wait, maybe the denominator is \(15n^2+12n\) and numerator is \(15n^3-6n^2 - 9n\), then we can simplify by canceling \(3n\) (for \(n eq0\)):

\(\frac{15n^3-6n^2 - 9n}{15n^2+12n}=\frac{3n(5n^2-2n - 3)}{3n(5n + 4)}=\frac{5n^2-2n - 3}{5n + 4}\) (for \(n eq0\))

We can also perform polynomial long division on \(\frac{5n^2-2n - 3}{5n + 4}\):

\(5n^2-2n - 3=(5n + 4)(n)-6n - 3\)

\(-6n - 3=(5n + 4)(-\frac{6}{5})+\frac{24}{5}-3=(5n + 4)(-\frac{6}{5})+\frac{9}{5}\)

So \(\frac{5n^2-2n - 3}{5n + 4}=n-\frac{6}{5}+\frac{9}{5(5n + 4)}\)

But this is not a simple simplification. Wait, maybe I made a mistake in the problem's numbers. Alternatively, maybe the denominator is \(15n^2+21n\), but as per the given problem, we proceed with the given numbers.

Wait, another approach: factor numerator and denominator for common factors.

Numerator: \(15n^3-6n^2 - 9n = 3n(5n^2-2n - 3)=3n(5n + 3)(n - 1)\)

Denominator: \(15n^2+12n = 3n(5n + 4)\)

Since there are no common factors between \((5n + 3)(n - 1)\) and \((5n + 4)\), the simplified form (for \(n eq0\)) is \(\frac{(5n + 3)(n - 1)}{5n + 4}=\frac{5n^2-2n - 3}{5n + 4}\)

Or we can expand the numerator: \(\frac{5n^2-2n - 3}{5n + 4}\)

Wait, maybe the original problem has a typo, but based on the given problem, the simplification is \(\frac{5n^2-2n - 3}{5n + 4}\) (after canceling \(3n\) for \(n eq0\))

Answer:

\(\frac{5n^2 - 2n - 3}{5n + 4}\) (for \(n
eq0\))