Simplify this problem. |3r−15| if r<5

We have the problem:

|3r−15| if r<5

First we see the equality, if r = 5 we have:

I3r - 15I = I3*5 - 15I = I0I = 0.

Then the only restriction that we have is:

I3r - 15I > 0.

now, we could simplify it a bit further:

if r < 5, then the thing inside the absolute value will always be negative:

Then we can write:

I3*r - 15I = -(3*r -15) > 0

multiplying by -1 in both sides

(3r - 15) < 0.

if we keep simplifying this, we will get our initial restriction:

3r - 15 < 0

3r < 15

r < 15/3 = 5

r < 5

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)