L luiz_322 New member Joined Apr 13, 2006 Messages 7 Apr 13, 2006 #1 i need help in this problem 4m^3-m^2+12m-3 and i tried solvin like this 3m+12m-3=15m-3 but i got it wrong and i dont know where i went wrong can you please help me in this problem thank you
i need help in this problem 4m^3-m^2+12m-3 and i tried solvin like this 3m+12m-3=15m-3 but i got it wrong and i dont know where i went wrong can you please help me in this problem thank you
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Apr 13, 2006 #2 Hello, luiz_322 ! Factor: \(\displaystyle \,4m^3\,-\,m^2\,+\,12m\,-\,3\) Click to expand... Usually, with four or more terms, try factoring "by grouping". Factor them "in pairs": \(\displaystyle \:m^2(\underbrace{4m\,-\,1})\,+\,3(\underbrace{4m\,-\,1})\) . . . . . . . . . . . . . . . . . . . . . See the common factor? Factor it out: \(\displaystyle \4m\,-\,1)(m^2\,+\,3)\)
Hello, luiz_322 ! Factor: \(\displaystyle \,4m^3\,-\,m^2\,+\,12m\,-\,3\) Click to expand... Usually, with four or more terms, try factoring "by grouping". Factor them "in pairs": \(\displaystyle \:m^2(\underbrace{4m\,-\,1})\,+\,3(\underbrace{4m\,-\,1})\) . . . . . . . . . . . . . . . . . . . . . See the common factor? Factor it out: \(\displaystyle \4m\,-\,1)(m^2\,+\,3)\)
L luiz_322 New member Joined Apr 13, 2006 Messages 7 Apr 14, 2006 #3 well thank you so much that really helped