Question 190751


{{{75r^2-48}}} Start with the given expression



{{{3(25r^2-16)}}} Factor out the GCF {{{3}}}



Now let's focus on the inner expression {{{25r^2-16}}}





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{{{25r^2-16}}} Start with the given expression.



{{{(5r)^2-16}}} Rewrite {{{25r^2}}} as {{{(5r)^2}}}.



{{{(5r)^2-(4)^2}}} Rewrite {{{16}}} as {{{(4)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=5r}}} and {{{B=4}}}.



So let's use the difference of squares formula {{{A^2-B^2=(A+B)(A-B)}}} to factor the expression:



{{{A^2-B^2=(A+B)(A-B)}}} Start with the difference of squares formula.



{{{(5r)^2-(4)^2=(5r+4)(5r-4)}}} Plug in {{{A=5r}}} and {{{B=4}}}.



So this shows us that {{{25r^2-16}}} factors to {{{(5r+4)(5r-4)}}}.



In other words {{{25r^2-16=(5r+4)(5r-4)}}}.



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Answer:

So {{{75r^2-48}}} factors to {{{3(5r+4)(5r-4)}}}