Tuesday, July 16, 2013

I need help please. I am having terrible difficulty in solving this problem. Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros of f. Degree 3, zeros: 2, -6 - i

the remaining zero is -6+i (since all coefficients are real complex root occur in conjugate pairs) 

:)
2|x+1/3|<9 
means 
-9 < 2(x+1/3) < 9 

-9/2 < x+1/3 < 9/2 

-9/2-1/3 < x < 9/2 - 1/3 
-4 5/6 < x < 4 1/6 

Interval:
(-4 5/6, 4 1/6) 

determine the x and y intercepts of the following linear equation 2x+3y=9

2x+3y=9 
x=0 -> 3y=9, y=3 ...so (0,3) 
y=0 -> 2x=9, x=9/2 ...so (9/2,0) 

:)

The operation * is defined on the set of real numbers by a*b=a+b+ab/2. Is the operation associative

we check that (a*b)*c = a*(b*c) 

(a*b)*c = (a+b+ab/2)*c = 
a+b+ab/2 +c + ((a+b+ab/2)c)/2= 



a*(b*c) = a*(b+c+bc/2) = 
a+(b+c+bc/2)+(a((b+c+bc/2))/2 


comparing... 
(a+b+ab/2 +c + ((a+b+ab/2)c)/2) - 
(a+(b+c+bc/2)+(a((b+c+bc/2))/2) 
which is 0 

the operation is associative 

:)