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simplify 7m^3+4m(3m-1)=5
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expand\[\large 4m(3m-1) = 12m^2-4m\]\[\large 7m^3+12m^2-4m=5\] factor out an m. \[\large m(7m^2+12m-4)=5\]\[\large \color{red}7m^2+12m-\color{red}4=5\]\[\large m^2+12m-28\] find two numbers hat add to get 12 and multiply to get 28. \[\large (m+14)(m-2)\] use the 7 we originally used in multiplying the constant in dividing these two factors. \[\large (m+\frac{14}{7})(m-\frac{2}{7})\] reduce what you can. \[\large (m+7)(m-\frac{2}{7})\] Nwo multiply the "7" to the m to simplify the factor. \[\large (m+7)(7m-2)\] put it all together. \[\large 4m(m+7)(7m-2)=5\] now multiply and solve
thank you so much(:
no problem.
wow you really took your time i appreciate that hehe.
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