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Can I factor this by grouping? 25b^2+10b+1?
Prime factors of 25 are 5 and 5 so we'll have (5x + a)(5x+b). Multiply that out and you get \[(5x+a)(5x+b) = 25x^2 + 5ax + 5bx + ab\] To make that look like your equation, a and b have to = 1.
25b^2+10b+1 multiply 25 with 1 you will get 25. now write 10 as the sum of two numbers such that the addition of those numbers result in 10 and the product of those numbers result in 25
I got b(5b+1) 1(b+1) but that does not look correct?
No, (5b+1)(5b+1) is what you want.
(5b+1) and (5b+1) you correct it once again
t(4t+1)=0? I got t=0, -1/4
Alright cool! Thanks
Well, no one really did it the correct way.
25b^2+10b+1 25b^2 +5b+5b+1 5b(5b+1)+1(5b+1) (5b+1)(5b+1) (5b+1)^2
No, some of us did it in a way that shows why grouping works :-)
25b25b+5b+1=0 5(5b+1)+1(5b+1)=0 (5b+1)2=0 5b+1=0 5b=-1 b=-1/5
Yes, but you forgot to show the process, which is also important as well @whpalmer4. The question was "How do you factor by grouping?". Showing why it works is fine. Showing the process is equally important.