Question

Illustrate a sign graph that shows the signs
of each factor.x²+x-12>0?​

Answer 1

Since the right-hand side (RHS) is already 0, we start by factoring the left-hand side (LHS):

x

2

−

x

−

12

>

0

⇒

(

x

−

4

)

(

x

+

3

)

>

0

In its factored form, this inequality tells us that the product of two numbers

(

x

−

4

)

and

(

x

+

3

)

is positive (greater than

0

).

In order for a product of two terms to be positive, either both terms must be positive or both terms must be negative. So, we require either

x

−

4

>

0

∩

x

+

3

>

0

or

x

−

4

<

0

∩

x

+

3

<

0

.

The former simplifies to

x

>

4

∩

x

>

-

3

,

which is only true when

x

>

4

. The latter simplifies to

x

<

4

∩

x

<

-

3

,

which is only true when

x

<

-

3

. Since either of these situations makes the inequality true, we combine these statements with the logical "or" (

∪

) to get

x

2

−

x

−

12

>

0

⇒

x

<

-

3

∪

x

>

4

.

Answer 2

Given:

The factor $x^{2}-x-12>0$.

To find:

The shows  the signs of each factor.

Step-by-step explanation:

$x^{2}-x-12=0$

$(x-4)(x+3)=0$

$x-4=0,x+3=0$

$x-4=0,x=4$

$x+3=0,x=-3$

$x=4,-3$

$-3$<$x$<$4$

Answer:

Therefore, that shows  the signs of each factor is $-3$<$x$<$4$.