#### Question

# Precalculus polynomial functions!~~?

Given 4(2x-1)(4x+5)^2(10x+2),

Show and determine the intervals on which f(x) <0 and f(x) >0

#### Answers

Well basically first find the "zeroes" where f(x)=0. That's at -0.5 from (2x-1), -0.8 from (4x+5), and -5 from (10x+2).

Now the ends of the graph are going to go to the same direction as x--> infinity and as x --> negative infinity (fancy way of saying "on the graph as the line goes left/right") f(x) will either go to positive or negative infinity (fancy way of saying "on the graph the line will either go up at both ends or down at both ends") (aka one side won't go negative and the other go positive, both sides will end up going towards the same direction, up or down). that's because the highest degree (exponent) is an even number (4) if you were to factor it all out.

Anyway, since it is a positive function, (not -4(2x-1)(blah)^2(blah)), it is going to be a normal parabola and start by going down, and end by going up. It starts at up, it ends at up (like i said in the last paragraph) so from negative infinity till the first x intercept (-5), it will be positive (aka f(x)>0), since, it is going down. But then it cross over the x axis, and becomes negative, so it changes to negative (aka f(x)<0), and will stay there till it decides to go up. When it does decide to go up, it will go up till -0.8. Here's something tricky now: when a quotient on a graph is multiplicity 2 (to the second power) it doesn't hop over the x axis, it BOUNCES. There's a calculus proof I could show you involving why that's a turning point as well put I don't feel like doing first derivative right now and I know you don't want to read all of it. So it bounces off and goes back down, and then goes back up and crosses over again at the next x intercept (0.5), and goes on till infinity being positive.

Giant wall of text aside, the answer it from negative infinity to -5, and from 1/2 to positive infinity, f(x)>0 and from -5 to 1/2 f(x)<0.

edit: an easier way to do this, since you're in precalc and they MIGHT allow graphing calculators (or if you're just doing homework) is put the function in and click graph and you can see a visual of it to make it easier. Then just find the zeroes (it's all factored out for you so that's like what, basic algebra?) and you can see where the f(x) is negative and positive.

72<x<528(pi)

Bam.