p(x)=x^3 + 3ax^2 + (b-1)x + 24 , q(x)=x^3 + ax^2 - (b+1)x - 60 , h(x)=x + 2 , find a and b ?

Answer 1

Since there is no relationship between p(x) and q(x); h(x) does not help the question on hand at all. I have found that by taking the derivative of your original functions, p'(x) and q'(x), we can determine your maximum/minimum points on the original function. When you set the derivatives equal to each other, meaning that you set the zeros of both the functions, you find the relationship of the where the maximum/minimum of the function occurs on the x axis. From it we can extract the relationship of a to b:
x=-2(b+1)/(4a)=.5 (b+1)/a

However, since you do not have any further information (ie. the maximum/minimum value or point) we cannot investigate this question any further.

Answer 2

42

Answer 3

b(y)=y^2 + 2ay^2 + (b-1)y + 2 , x(y^)=y^2 + by^2 - (a+1)y - 4 , j(y)=y + 1 /x(y^)=y^2 y^2 + 2ay^2 -(a+1)y - 4+(b-1)y + 2 -p(x)=x^3+h(x)=x + 2*x^3 + ax^2 = a

(a+1)y - 4+(b-1)y+ 2*x^3-y^2 + by^2*p(x)/x^3+h+ (b-1)y + 2-y^2 + by^2+x(y^)+y^2 y^2 = b

Answer 4

uum I had to work my @$$ off to get the grade Igot in math this year. I did so much extra credit its not even funny. So your on your own buddy.

Answer 5

You need a relation between the equations.
Does
h(x) = p(x)/q(x)?

Answer 6

A AND B BEFORE C

Answer 7

Shouldn't this be under 'homework help?'

Answer 8

I not doing your homework for you, do it your self.

Answer 9

Why should I find A and B for you, when you do not even say please?

Answer 10

e=mc2