The second one can be written as this
Since there are 2 terms that have a 5 in them, you can factor out a 5
If you distribute the 5, you will get 5y-15

the 3rd problem can be written as
Where the 5 is factored out

The fourth is similar, but the common term is x. So factor out an x
If you distribute the x you get x^2+2x.
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Here are few problems that need to be factored:
8y-5= 8(y-(5/8))
This could also be factored as:
5((8/5)y-1)
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5y-15= 5(y-3)
This could also be factored as:
15((1/3)y-1)
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10y-5= 5(2y-1)
This could also be factored as:
10(y-(1/2))
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x^(2)+2x = x(x+2)
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Cheers,
Stan H.
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