How do you solve $\frac{n^{3} - 2 n^{2} - n + 2}{n^{3} + 3 n^{2} + 4 n + 12} < 0$ $(n^3-2n^2-n+2)/(n^3+3n^2+4n+12)<0$ ?

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How do you solve $\frac{n^{3} - 2 n^{2} - n + 2}{n^{3} + 3 n^{2} + 4 n + 12} < 0$ $(n^3-2n^2-n+2)/(n^3+3n^2+4n+12)<0$ ?

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Answer 1 · 2017-11-11T13:31:05

The solution is $n \in (- 3, - 1) \cup 1, 2)$ $n in (-3,-1) uu1,2)$

Explanation:

Factorise the numerator and the denominator

Start with the numerator

$n^{3} - 2 n^{2} - n + 2 = n^{3} - n - 2 n^{2} + 2$ $n^3-2n^2-n+2=n^3-n-2n^2+2$

$= n (n^{2} - 1) - 2 (n^{2} - 1)$ $=n(n^2-1)-2(n^2-1)$

$= (n^{2} - 1) (n - 2)$ $=(n^2-1)(n-2)$

$= (n + 1) (n - 1) (n - 2)$ $=(n+1)(n-1)(n-2)$

Proceed with the denominator

$n^{3} + 3 n^{2} + + 4 n + 12 = n^{3} + 4 n + 3 n^{2} + 12$ $n^3+3n^2++4n+12=n^3+4n+3n^2+12$

$= n (n^{2} + 4) + 3 (n^{2} + 4)$ $=n(n^2+4)+3(n^2+4)$

$= (n^{2} + 4) (n + 3)$ $=(n^2+4)(n+3)$

Let $f (n) = \frac{n^{3} - 2 n^{2} - n + 2}{n^{3} + 3 n^{2} + + 4 n + 12} = \frac{(n + 1) (n - 1) (n - 2)}{(n^{2} + 4) (n + 3)}$ $f(n)=(n^3-2n^2-n+2)/(n^3+3n^2++4n+12)=((n+1)(n-1)(n-2))/((n^2+4)(n+3))$

Build a sign chart

$a a a a$ $color(white)(aaaa)$ $n$ $n$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a a$ $color(white)(aaaaa)$ $- 3$ $-3$ $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a a$ $color(white)(aaaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $n + 3$ $n+3$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $n + 1$ $n+1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $n - 1$ $n-1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $n - 2$ $n-2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (n)$ $f(n)$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$

Therefore,

$f (n) < 0$ $f(n)<0$ when $n \in (- 3, - 1) \cup 1, 2)$ $n in (-3,-1) uu1,2)$

graph{(x^3-2x^2-x+2)/(x^3+4x+3x^2+12) [-12.66, 12.65, -6.33, 6.33]}

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How do you solve $\frac{n^{3} - 2 n^{2} - n + 2}{n^{3} + 3 n^{2} + 4 n + 12} < 0$ $(n^3-2n^2-n+2)/(n^3+3n^2+4n+12)<0$ ?

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How do you solve $\frac{n^{3} - 2 n^{2} - n + 2}{n^{3} + 3 n^{2} + 4 n + 12} < 0$ $(n^3-2n^2-n+2)/(n^3+3n^2+4n+12)<0$ ?