View Single Post
(#5 (permalink))
Old
Silver_Wolf_Kitty's Avatar
Silver_Wolf_Kitty (Offline)
Super Moderator
 
Posts: 2,208
Join Date: Dec 2003
Zodiac Sign: Leo
Rating: 36 Votes / 3.83 Average
Default 09-22-2009, 12:12 PM

iluvdolphin, you had the first one correct You are just supposed to distribute it throughout the parentheses. Same applies to the others with it.
As for the second, you made a minor mistake. If 18 is in front of the x, it is a multiple of a number. Notice how the last was 6, which is a multiple of 18. You divided each number inside by 6, but you kept the 18 for some reason even though its a multiple So, while you had the inside correct, the 18 should be 6. To check this, you would need to try to redistribute it and see if it equals the original problem. When doing problems like this, you should always see if the number without the letter could be divided into the one in front of the letter. If it can, you can reduce the numbers in it.

antimaie, do you remember how those formulas are factored? In the end, you are supposed to get something like (#x+/-#)(#x+/-#). I'll take your first problem and show you how.
y^2 - 6y - 55
So, looking at this problem, you know absolutely that the first letters of each parenthese does not have a number in front of it because there are none proceeding the ys, so you have part of your problem (y+/-#)(y+/-#).
Next, we look at the signs. Notice that they are -s, which means that there is a + in one and a - in the other parenthese because when you distribute the equation, a positive number times a negative number produces negative numbers. So, we know that it must now be (y+#)(y-#).
We are left now figuring out what numbers not only multiply to equal -55, but also will equal -6 when added together. You'll want to list out all of the multiples.
-11 x 5
11 x -5
-1 x 55
1 x -55
Those are all you can get with whole numbers. Next, you determine if any will equal -6. So, if you look below, I have added them together now.
-11 + 5 = -6
11 + -5 = 6
-1 + 55 = 54
1 + -55 = -54
So, the first pair is the only option, so your factored equation would be (y+5)(y-11).
This would change a bit will numbers in front of the "y" value. In your second problem, 6t^2 - t - 5, you begin differently, first figuring out what values will multiply together to get 6. That is 6 and 1, 2 and 3, the best option would be the 6 and 1 because you need to determine what numbers will add/subtract to equal -1(-6+5=-1). So, you begin with (6t+/-#)(1t+/-#). Next, you'll want to figure out what'll equal 5, which is simply 1 x 5. The tough part is then figuring out which one each number belongs in. If 6 is multiplied by 5, you will end up with -29, so you know now that 6 and 5 have to be in the same parenthese so they aren't multiplied together so your factoring is now at (6t+/-5)(1t+/-1). The final step is determining the signs. To get a -1, the 6 would have to be multipled with a negative number since it is larger, so you now have (6t+5)(1t-1). Just redistribute to check.

Grim, first off, with exponents(which is what I'm assuming you mean by "to the power of 4"), it is not the lower number times the exponent(what you did was 6 x 4 = 24), it is the lower number multipled by itself the exponent number's amount of times(in this case, 4, so it'd be 6 x 6 x 6 x6 = 1296). So, your equation would then be
1296 + 3(t-6) = 3t -7
not what you listed. Your second issue is when you distribute a number through parentheses(in this case 3(t-6)), you need to multiply everything incide the parentheses by the outside number, in this case positive three, so your problem would then be:
1296+3t-18=3t-7
Strangely enough, I'm looking at that equation, and it wouldn't work because the ts would cancel each other out. To explain, you don't subtract the 3 from the t since they are being multipled together, so you would divide each side by three. Once that is done, you would want to isolate the t, but since they are equal to each other, they would cancel each other out(t-t=0). Are you sure that was written down right? That doesn't make sense and there is something seriously wrong with the equation since my only other thought is you are supposed to factor the left, but there is no variable with the exponent part.


Av & Sig Credit: Me

Last edited by Silver_Wolf_Kitty : 09-22-2009 at 12:15 PM.
Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!