What is the solution set to the inequality 5(x – 2)(x + 4) > 0?

(-∞, -4) U (2,∞)

Step-by-step explanation:

5(x – 2)(x + 4) > 0

First we solve for x, we replace the inequality sign by = sign

5(x – 2)(x + 4) = 0

Divide both sides by 5

(x – 2)(x + 4) = 0

Now we set each factor =0  and solve for x

x-2 =0 , so x= 2

x+4 =0, so x= -4

Now we use number line and make three intervals

First interval -infinity to -4

second interval -4 to 2

third interval 2 to infinity

Now we check each interval with our inequality

First interval -infinity to -4, pick a number in this interval and check with our inequality. lets pick -5

5(-5 – 2)(-5 + 4) > 0

35>0 is true

second interval -4 to 2, pick a number in this interval and check with our inequality. lets pick 0

5(0– 2)(0 + 4) > 0

-40>0 is false

Third interval 2 to infinity, pick a number in this interval and check with our inequality. lets pick 3

5(3 – 2)(3 + 4) > 0

35>0 is true

solution set are the intervals that make the inequalities true

(-∞, -4) U (2,∞)

The answer is B

Step-by-step explanation:

For those who didn't understand like me

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

the domain of a function is the set of input values, x , for which a function is defined. the domain is shown in the left oval in the picture below. the function provides an output value, f(x) , for each member of the domain.