factor completely
80c^2+200cf+125f^2

Question

mertsj · Answer

Factor out 5:  5(16c^2 + 40cf + 25f^2)  = 5(4c+5f)^2

anonymous · Answer

huh

mertsj · Answer

Do you understand about factoring out the 5?

anonymous · Answer

a little but not a lot

mertsj · Answer

5 is a factor of 80, 200, and 125.  Therefore it must be factored out using the distributive property:  $$80c ^{2}+200cf +125f ^{2}  =5(16c ^{2}+40cf + 25f ^{2}$$

mertsj · Answer

Do you understand that?

anonymous · Answer

yeah

mertsj · Answer

Ok.  Then consider $$16c ^{2}+ 40cf + 25f ^{2} $$
That is a trinomial and came from multiplying two binomials.  It is our job to find the two binomials.  To get 16c^2 we could multiply 4c by 4c.  Let's try that.   (4c    )(4c    )
To get 25f^2 on the end we could have 5f times 5f.  Let's try that and see if we get the correct middle term of 40cf   (4c+5f)(4c+5f).  The outer is 20cf and the inner is 20 cf.  If we add those, we get 40 cf.  So that seems to work and the factored form is therefore 5(4c+5f)(4c+5f) of 5(4c+5f)^2

anonymous · Answer

so then i would be at the answer of 5(4c+5f)^2 right now where do i go from there

mertsj · Answer

That's it.  It's completely factored.

anonymous · Answer

so in my problem the answer was sort of there in a way right with the factoring of 5 right

mertsj · Answer

That was the first step.  That is always the first step in any factoring problem.  To look for a common factor.  Sometimes there will be a common factor but not always.  If there is a common factor, the first step is to factor it out using the distributive property.

mertsj · Answer

Then the second step is to examine what remains after factoring out the common factor. Sometimes it will be factorable but not always.

anonymous · Answer

you explained that really easy for me in a way,thank you