Question 974957
The polynomial of degree 3 , P(x) , has a root of multiplicity 2 at x=4 and a root of multiplicity 1 at x=-4 . The y -intercept is y=-12.8 .
Find a formula for P(x) 
what I have tried.. -12.8=a(x-4)^2(x+4)
I get confused and start adding steps I shouldn't.
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<pre>
You were on the right track with

 a(x-4)^2(x+4), 

but leave P(x) equal to that:

P(x) = a(x-4)^2(x+4) 

Now use this given fact:
</pre>
The y -intercept is y=&#8722;12.8
<pre>
Think about what y-intercept means. Remember that y = P(x). The y-intercept
being -12.8 means that when you substitute x=0, you get y = P(0) = -12.8

So substitute x=0 in

P(x) = a(x-4)^2(x+4) 
P(0) = a(0-4)^2(0+4)
12.8 = a(-4)^2(4)
12.8 = a(16)(4)
12.8 = a(64)
{{{12.8/64)}}} = a
0.2 = a

Now substitute 0.2 for a in 

P(x) = a(x-4)^2(x+4)

Answer:

P(x) = 0.2(x-4)^2(x+4) 

That's the answer.  Or you can multiply it out:

P(x) = 0.2x^3-0.8x^2-3.2x+12.8

Edwin</pre>