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Nick_P
Can someone help me!!!
(2a^2-4a+2) / (3a^2-3) 2(a^2-2a+1) / 3(a^2-1) 2(a-1)(a-1) / 3(a+1)(a-1) 2(a-1) / 3(a+1) You can leave it at that. Or write it out as (2a-2) / (3a+3). I would just leave it factored though
im not to sure on these but @mathstudent55 do you think you could help him?
7. is correct. Good job @dusty
but im not for sure on 8 & 9
8. (2s^2 - 5s - 12) / (2s^2 - 9s + 4) The basic idea is the same as for 7. First factor the numerator and denominator.
You need to factor 2s^2 - 5s - 12 This is a trinomial of the form ax^2 + bx + c To factor do the following: 1. First try to factor a common factor from all terms, if there is one. In this case there isn't a common factor.
2. Multiply ac together. In this case, a = 2, and c = -12, so ac = 2(-12) = -24
3. Come up with two dfactors of ac that add up to b. In this case, we need two factors of -24 that add up to -5. Since 8*3 = 24, -8*3 = -24, and -8 + 3 = -5, so the factors are -8 and 3.
4. Now break up the middle term of the trinomial, bx, into a sum of the factors from step 3. In this case, -5s becomes -8s + 3s. 5. Rewrite the trinomial using the new broken-up middle term. In this case, 2s^2 - 8s + 3s - 12
6. Now factor by parts. That means, factor a common factor out of the first two terms, and factor a common factor out of the last two terms. In this case, 2s^2 - 8s + 3s - 12 = 2s(s - 4) + 3(s - 4) 7. Pull out the common factor to complete the factoring. In this case, 2s(s - 4) + 3(s - 4) = (s - 4)(2s + 3)
Now you need to do the same to teh denominator. You need to factor 2s^2 - 9s + 4 If you follow the steps above, youi'll get: (2s - 1)(s - 4)
Now that both the numerator and the denominator are factored, you have: (2s^2 - 5s - 12) / (2s^2 - 9s + 4) = [ (s - 4)(2s + 3) ] / [ (2s - 1)(s - 4) ] Notice that s - 4 is in both the numerator and denominator, so you can divide both the numerator and denominator by s - 4: (2s + 3) / (2s - 1)
That's the final answer of 8. For 9, you once again have to factor the numerator and denominator if possible. This happens to be a very simple problem of this type because the factoring involved is very easy.