Complicated But Fun!! Try It See What U Get!

[aisha]

(x^(2)+6x+5)/(x^(2)+7x+12) MULTIPLIED BY (x^(2)+2x-8)/(x^(2)-25)
__________________________________________________ __________
ALL DIVIDED BY
(x^(2)-x-2)/(x^(2)-2x-15)

OK SIMPLIFY THIS

THE ANSWER I GOT WAS a simple 1
IS THIS CORRECT? :eek:

[jisland85]

i get the same . . . 1

[aisha]

I FORGOT RESTRICTIONS!! :yuck: but I dont know what step to get the restrictions from, anyone have any ideas?

x cannot = -4,-3,-5,5,2,-1??
AHHHHHHHHHHHHHHHHHHHHH PLEASE HELP ME SOMEONE :eek:

[Sirus]

Careful. Restrictions here are due to the undefined nature of a number divided by zero. For the denominator to never be equal to zero, neither of the polynomials in the denominator can ever equal zero. Therefore, to find restrictions, set each polynomial equal to zero and solve the equation to find the x-intercepts.

[aisha]

Thats exactly what I did but in this question there are soo many denominators I dont know which one to take the restriction from so the numbers in my last reply were all the restrictions from all the denominators.... AHhhhhhhhhhhhhhh
SOMEONE please help me :uhh:

[dextercioby]

Thats exactly what I did but in this question there are soo many denominators I dont know which one to take the restriction from so the numbers in my last reply were all the restrictions from all the denominators.... AHhhhhhhhhhhhhhh
SOMEONE please help me :uhh:

Why would u need any help??U did it splendidly.Like a mathematician would. :wink:
Everytime u're dealing with denominators (that means u have to devide something through another),make sure they're never zero.Sometimes that's simple to do,sometimes not.Practically u have to solve and to find all the roots of the equation "denominator=0".That's the rule.In your case,it was simple as u were able to decompose the polynomial in the denominator in simple monoms whose roots could have easily been found.

Daniel.

[aisha]

so those 6 restrictions I stated are correct? :rofl:

[BobG]

Except for -1. -1 will set the numerator of the left fraction to zero, not the denominator.

[dextercioby]

Except for -1. -1 will set the numerator of the left fraction to zero, not the denominator.

Actually Bob,since the result is 1,ALL THE ROOTS OF THE DENOMINATOR'S POLYNOMIAL WILL ANULLATE THE NUMERATOR AS WELL.So you're saying there's no restriction on the entire fraction...?? :wink: :confused:

[aisha]

Hey guys do I keep the -1 in my restriction or not :tongue2:

[dextercioby]

Hey guys do I keep the -1 in my restriction or not :tongue2:

Yep,and the other numbers as well.For if u didn't,u'd be dividing something (incidentally it's 0) by 0.And that should no be okay.

Daniel.