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Please help me with the steps to solve the following: Factor completely (f+4)(2f-3)+(f+4)(f+4)
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Each term has a f+4 in common, so factor that out to get (f+4)(2f-3)+(f+4)(f+4) (f+4)[(2f-3)+(f+4)] (f+4)(2f-3+f+4) (f+4)(3f+1)
Oh ok so is (f+4)(3f+1) the final result or do i need to go further such as 3f+12f+f4=15f+4 to complete?
(f+4)(3f+1) is as factored as it gets
what you have isn't correct (it looks like you're expanding it out again, but that's the opposite of factoring)
Oh I get it now...I thought I had to keep going. So to do another problem to factor completely such as 121p(square)-121p would be 121(p-1)p?
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that's one way to do it, you can also write it as 121p(p-1)
Great! Thank you so much for your help.
yw
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