OpenStudy (anonymous):

factor x^3+5x^2-9x-45

5 years ago
OpenStudy (anonymous):

... I don't think it can be factored. You can't take out an x because of the -45 and you can't pull out a 3 because of the x^3 not having a 3 before it.

5 years ago
OpenStudy (anonymous):

x(x^2 +5x -9) -45

5 years ago
OpenStudy (anonymous):

obv it can be facotred more, but it won't be that simple

5 years ago
OpenStudy (anonymous):

x^3(1+(5/x)-(9/x^2)-(45/x^3)

5 years ago
OpenStudy (anonymous):

x^3(1+(5/x)-(9/x^2)-(45/x^3))

5 years ago
OpenStudy (anonymous):

it could be one of these..... . (x+5)(x-3)(x+3) (x+5)(x²+9) (x-5)(x-3)(x+3) (x-5)(x²+9)

5 years ago
OpenStudy (anonymous):

Christine in that case, I would do the problem in reverse. You could multiply them together and find the answer. I think it's one of the two answers with three sections. Probably the first or third.

5 years ago
OpenStudy (anonymous):

ok thank you!! ill try

5 years ago
OpenStudy (zehanz):

I think 3 is a root, so x-3 is a factor. See: 3^3+5*3^2-9*3-45=27+45-27-45=0 You can write it as: (x-3)(some 2nd degree polynomial). How to find the 2nd degree polynomial? You could try more numbers, -3, 5 an -5 are good candidates, but if you want you can just do long division to find the other factor. Is this something you are supposed to know? I'd be glad to help you with it.

5 years ago
OpenStudy (anonymous):

thank you

5 years ago
OpenStudy (zehanz):

I have this tip for you: In the past, when seeing something like x^3+5x^2-9x-45, I didn't know how to begin, because of the x^3. It seemd to me that you would have to get a brilliant idea to magically see how such a thing can be factored. However, after a while it occured to me that it only can be done if there are "nice" numbers involved. Lik in this case: it it not very hard to find 3 or -5. Just try a few simple numbers, like 1, 2, 3, 4, 5 and -1, -2 and so on. In nine out of ten cases you'll find something! There is no magic, you just have to try a few numbers. The 2nd part is trickier: this is the part with the long division. Some have learnt how to do it, others haven't. Still it is a very handy way of "cracking" these things!

5 years ago
OpenStudy (anonymous):

thank you tons!!! :)

5 years ago
OpenStudy (zehanz):

YW ;)

5 years ago