How do you solve $x^{3} + 7 x^{2} - x - 7 < 0$ $x^3+7x^2-x-7<0$ ?

Question

How do you solve $x^{3} + 7 x^{2} - x - 7 < 0$ $x^3+7x^2-x-7<0$ ?

1 Answer

Impact of this question

4443 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-29T07:38:20

The answer is $x \in] - \infty, - 7 [\cup] - 1, 1 [$ $x in ]-oo, -7 [uu]-1, 1[$

Explanation:

Let $f (x) = x^{3} + 7 x^{2} - x - 7$ $f(x)=x^3+7x^2-x-7$

$f (1) = 1 + 7 - 1 - 7 = 0$ $f(1)=1+7-1-7=0$

so, $(x - 1)$ $(x-1)$ is a factor of $f (x)$ $f(x)$

To find the other factors, we do a long division

$a a a a$ $color(white)(aaaa)$ $x^{3} + 7 x^{2} - x - 7$ $x^3+7x^2-x-7$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $x - 1$ $color(blue)(x-1)$

$a a a a$ $color(white)(aaaa)$ $x^{3} - x^{2}$ $x^3-x^2$ $a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaa)$ $∣$ $|$ $x^{2} + 8 x + 7$ $color(red)(x^2+8x+7)$

$a a a a a$ $color(white)(aaaaa)$ $0 + 8 x^{2} - x$ $0+8x^2-x$

$a a a a a a a$ $color(white)(aaaaaaa)$ $+ 8 x^{2} - 8 x$ $+8x^2-8x$

$a a a a a a a a a$ $color(white)(aaaaaaaaa)$ $+ 0 + 7 x - 7$ $+0+7x-7$

$a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaa)$ $+ 7 x - 7$ $+7x-7$

$a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaa)$ $+ 0 - 0$ $+0-0$

Therefore,

$f (x) = (x - 1) (x^{2} + 8 x + 7) = (x - 1) (x + 1) (x + 7)$ $f(x)=(x-1)(x^2+8x+7)=(x-1)(x+1)(x+7)$

Now, we can build the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 7$ $-7$ $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a a$ $color(white)(aaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 7$ $x+7$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x + 1$ $x+1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x + 1$ $x+1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (x) < 0$ $f(x)<0$ when $x \in] - \infty, - 7 [\cup] - 1, 1 [$ $x in ]-oo, -7 [uu]-1, 1[$

Science

Math

Humanities

... and beyond

How do you solve $x^{3} + 7 x^{2} - x - 7 < 0$ $x^3+7x^2-x-7<0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve x^3+7x^2-x-7<0x3+7x2−x−7<0x^3+7x^2-x-7<0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $x^{3} + 7 x^{2} - x - 7 < 0$ $x^3+7x^2-x-7<0$ ?