Posted by Emily on Wednesday, December 17, 2008 at 8:49pm.
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

Algebra II  Damon, Wednesday, December 17, 2008 at 9:02pm
( 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)

Algebra II  Damon, Wednesday, December 17, 2008 at 9:07pm
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

Algebra II  Damon, Wednesday, December 17, 2008 at 9:17pm
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)

Algebra II  Damon, Wednesday, December 17, 2008 at 9:22pm
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

Algebra II  Damon, Wednesday, December 17, 2008 at 9:23pm
Just brute force think your way through these problems !!!

Algebra II  Emily, Thursday, December 18, 2008 at 12:30am
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!!!!!!

Algebra II  Mike, Friday, March 30, 2012 at 1:54am
6x^415x^3+10x^210x+4 divided by 3x^2+2
Please help!!!
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