Faclorise fully \( 125 \mathrm{~m}^{3}+15 \mathrm{~m} \)

To factorise the expression \( 125 \mathrm{~m}^{3}+15 \mathrm{~m} \), we first observe that both terms share a common factor of \( 5\mathrm{~m} \). We can start by factoring out \( 5\mathrm{~m} \): \[ 125 \mathrm{~m}^{3} + 15 \mathrm{~m} = 5\mathrm{~m}(25 \mathrm{~m}^{2} + 3) \] Now, the expression inside the brackets, \( 25 \mathrm{~m}^{2} + 3 \), doesn't factor further over the real numbers. So the fully factorised form is: \[ 5\mathrm{~m}(25 \mathrm{~m}^{2} + 3) \] You’ve just transformed a cubic and linear expression into a product, which is a handy skill for simplifying many algebraic problems! Speaking of simplifications, always look for the greatest common factor first; it’s the quickest route to factorising complex expressions. A common mistake is trying to factor in too many steps, forgetting to check for simple factors like single numbers or variables. It can be a real time-waster, so always glance for that easiest factor first!