How do you solve $(x-4)(x+1)>=0$ using a sign chart?

Question

1737 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T11:24:33

The answer is $x in ] -oo,-1 [ uu [4, +oo[$

Let $f(x)=(x-4)(x+1)$

Let's do the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-1$ $color(white)(aaaa)$ $4$ $color(white)(aaaa)$ $+oo$

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

$color(white)(aaaa)$ $x-4$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

So, $f(x)>=0$ when $x in ] -oo,-1 [ uu [4, +oo[$

graph{(x-4)(x+1) [-12.66, 12.65, -6.33, 6.33]}

How do you solve (x-4)(x+1)>=0(x-4)(x+1)>=0 using a sign chart?