Question 72547
The first one cannot be factored further, since there are no common factors amongst the terms. 
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The second one can be written as this
{{{5y-15=5y-5*3}}}Since there are 2 terms that have a 5 in them, you can factor out a 5
{{{5(y-3)}}}If you distribute the 5, you will get 5y-15
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the 3rd problem can be written as
{{{5*2y-5=5(2y-1)}}}Where the 5 is factored out
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The fourth is similar, but the common term is x. So factor out an x
{{{x^2+2x=x(x+2)}}}If you distribute the x you get x^2+2x.