Question 28854: Im sorry if this is in the wrong place, knew it went in either Polynomials or quadratic. It's factoring equations so...
Ok we just got into this stuff in Alg. II and I can't figure out for the life of me how to do these, we worked on them in groups at school so I had help on the other ones but these last ones are the ones I didn't get done, I have looked on the internet how to do them, my parents are oblivious to this stuff, etc.
So if anyone can help me with these I would really appreciate it.
Also if you know the answers could you show me how you did them in a step by step format as well as the answer, then I could write them down in my notebook to study later, I really want to understand this stuff.
Factoring:
1. 3a2-24a+48
2. 3x2+xy-6x-2y
3. 52-5st-14t2
4. 6x2+13x-5
Setting to Zero
1.y3+5y2+4y=0
2. (d-3)(d+5)=-12
Also my teacher said there is a possibility that any of these may be prime.
Thnx a bunch.
If you want what I got to the problems let me know although I didn't get past like the 2nd line on my paper for each problem.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! #1. Common factor of "3".
3(a^2-8a+16)
3(a-4)(a-4)=a(a-4)^2
#2Factor and 1st two then the 2nd two terms as follows:
x(3x+y)-2(3x+y)
Common factor of (3x=y)
(3x+y)(x-2)
#3 prime
#4 Use the AC method.
ac=6(-5)=-30
b=13
Find two numbers whose product is ac=-30
and whose sum is b=13
The numbers are 15 and -2
So, rewrite the problem using 15x-2x in place of the 13x, as follows:
Rewrite as 6x^2+15x-2x-5
Factor the 1st two then the last two terms, as follows:
3x(2x+5)-(2x+5)
Common factor of (2x+5)
(2x+5)(3x-1)
Cheers,
Stan H.
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