Algebra II
posted by Emily .
I honestly forgot how to do these type of problems. Can someone help me! Thanks!!

24. ( 9x^2  4 )/( 3x^2  10x  8 )
31. 2x^2 + 5 = 12x
39. Find the maximum value of g(x) = 4x^2 + 16x  15
42. Graph y9 = 2(x+3)^2

( 9x^2  4 )/( 3x^2  10x  8 )
well the numerator is difference of two squares so factor
(3x2)(3x+2)
now I bet one of those is a factor of the denominator
(3x  2)(x+4) no
(3x+2)(x4) yes
so we really have
(3x+2)(3x2) / (3x+2)(x4)
which is
(3x2) / (x4) 
2 x^2  12 x + 5 = 0
I can not factor that, sure it is not 11 x?
If not a typo, use quadratic equation
x = [ b +/ sqrt(b^24ac)] / 2a 
g(x) = 4x^2 + 16x  15
parabola
g is  for bi + or  x so upside down (sheds water)
Find the vertex.
4 x^2  16 x + 15 = g
x^2  4 x = (g15)/4
x^2  4 x + 4 = g/4  15/4 + 16/4
(x2)^2 = (1/4) (g +1)
AT VERTEX x = 2, g = +1
so max at (2,1) 
y9 = 2(x+3)^2
pretty much the same as the last one
same both sides of x = 3 so that is the vertical axis of symmetry
vertex at (3,9)
when x gets big, y gets small, so sheds water
when y = 0
9/2 = (x+3)^2
x + 3 = +/ 3 /sqrt 2
x = 3 +/ 3/sqrt 2 is where he zeros are 
Just brute force think your way through these problems !!!

OMG you are FREAKIN AWESOME for helping me out with these problems!!!!!! With your thorough explanations it all came back to me I thank you SOOOOOO MUCH!!!!!!

6x^415x^3+10x^210x+4 divided by 3x^2+2
Please help!!!
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