algebra??/?

Question

steps of how to solve these problems!
16. (4x+3)(x-1)(3x-9)=0
33. (4x)(4x)+38+34
37. (x)(x)+27
43. (16x)(16x)-34x-15=0

Answer 1

16) 4x+3=0 or x1=-3/4
x2=1
x3=3

33)
no real x value

37)
no real x value

4)
17+ or - sqrt(4129)

Answer 2

16. Since the right side of the equation equals 0, then at least one of the factors on the left side of the equation must be zero. Therefore, 4x+3=0, or x-1=0, or 3x-9=0. So your solution set is {-.75, 1, 3}

Answer 3

16.) -3/4, 1, 3
The function will be zero whenever any of the terms in parenthesis is zero ( 0 * anything * anything = 0).

33. 16x^2 + 72 = 2(2x^2 + 9) = 0
This will evaulate to zero when 2x^2 +9 = 0 so
-9/2 = x^2
sqrt(-9/2), aka no real solutions

37) x^2 + 27 = 0
sqrt (-27) no real solution again

43)(16x)(16x)-34x-15 = 256x^2 - 34x - 15
Hmmm.... Quadratic formula I suppose
[-b +/- sqrt (b^2 -4ac)] / 2a

Subsitute in you get [34 +/- sqrt(16516)] / 512