How do you solve $(x - 1) (x + 2) (2 x - 5) < 0$ $(x-1)(x+2)(2x-5)<0$ using a sign chart?

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How do you solve $(x - 1) (x + 2) (2 x - 5) < 0$ $(x-1)(x+2)(2x-5)<0$ using a sign chart?

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Answer 1 · 2017-01-17T11:09:48

The answer is $x \in] - \infty, - 2 [\cup] 1, \frac{5}{2} [$ $x in ] -oo,-2 [ uu ] 1, 5/2[$

Explanation:

Let $f (x) = (x - 1) (x + 2) (2 x - 5)$ $f(x)=(x-1)(x+2)(2x-5)$

Now, we can do our sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a a$ $color(white)(aaaaaa)$ $- \infty$ $-oo$ $a a a$ $color(white)(aaa)$ $- 2$ $-2$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a$ $color(white)(aaaa)$ $\frac{5}{2}$ $5/2$ $a a a a a$ $color(white)(aaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 2$ $x+2$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 1$ $x-1$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $2 x - 5$ $2x-5$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (x) < 0$ $f(x)<0$ when $x \in] - \infty, - 2 [\cup] 1, \frac{5}{2} [$ $x in ] -oo,-2 [ uu ] 1, 5/2[$

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How do you solve $(x - 1) (x + 2) (2 x - 5) < 0$ $(x-1)(x+2)(2x-5)<0$ using a sign chart?

1 Answer

Explanation:

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How do you solve (x-1)(x+2)(2x-5)<0(x−1)(x+2)(2x−5)<0(x-1)(x+2)(2x-5)<0 using a sign chart?

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Explanation:

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How do you solve $(x - 1) (x + 2) (2 x - 5) < 0$ $(x-1)(x+2)(2x-5)<0$ using a sign chart?