algebra 2 help?

Question

x^2 + 11x + 24 > 0

A -8 < x < -3
B x< -8 or x> -3
C -8 < x < 3
D x = +- 8
E x < -3 or x > 8
F for all x

Answer 1

B

Answer 2

x^2 + 11x + 24 > 0
Look at the left side of the inequality
x² + 11x + 24 factors into
(x+3)(x+8)
For that product to be > 0, both factors have to be less than 0, or they both have to be greater

Let’s look at them when the product = 0
For that to be, x= -3, or x= -8

There are three ranges we need to consider
x < -8, -8 -3
Pick x = -10 < -8
(-10 + 3)(-10 + 8) = (-7)(-2) = 14 > 0 Ok x< -8 works
Pick x= -5, -8 < -5 < -3
(-5 +3)(-5 +8) = (-2)(3) = -6 < 0 Ok -8 Pick x = 0 > -3
(0+3)(0+8)=(3)(8) = 24>0 Ok so x> -3 works

So we’re looking at x< -8 or x> -3... looks like B

Answer 3

Easiest way to see answer is to sketch a quick graph. You know the two roots are -8 and -4 because (x+8)(x+3)=0. You also know the parabola is concave up (like a U) because x^2 term is positive so f(x) is positive for all x <-8 and all x > -3.

x is negative for -8
The answer is B

Answer 4

(x+3)(x+8)>0
x<-8 and x>-3
B