Factor completely.If polynomial is prime state so: 49y^2 - 49w^2
AND;
Again, factor completely.If polynomial is prime state so: 24x^2y^3 + 14xy^2 + 2y
jamjam,
it would be helpful if you could post some of your work. Post an attempt or something to help us know where you are at. This will improve our effectiveness and it will improve your learning.
I will work through the process of the first one but I would ask for you to then try the second one.
To start the first problem you must find out what needs to be factored. The final form is something like
\(\displaystyle (y_-^+aw)(y_-^+bw)\)
I would start by factoring the 49 out as follows:
\(\displaystyle 49[y^2-w^2]\)
Then you have factor it out to the general form
\(\displaystyle (y_-^+aw)(y_-^+bw)\) this comes from the general form \(\displaystyle y^2+(a+b)yw+w^2\) with any plus sign being a negative sign.
so all we have to do is determine out a , b, and the signs.
if you can see we do not have the middle term \(\displaystyle (a+b)yw\) in our equation so it must have canceled out some how.
the way that this term can cancel out is when a and b are equal and oppoite. In our case a=1 and b=-1.
so if a=1 and b=-1 then our final answer is as follows
\(\displaystyle 49[y^2-w^2]=49[(y+w)(y-w)]\)
you can always check to see if you did it right by foiling your solution back out. If you were right you will get what you started with.
is this clear?
