Question 190752


{{{49xy^2-25x}}} Start with the given expression



{{{x(49y^2-25)}}} Factor out the GCF {{{x}}}



Now let's focus on the inner expression {{{49y^2-25}}}





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{{{49y^2-25}}} Start with the given expression.



{{{(7y)^2-25}}} Rewrite {{{49y^2}}} as {{{(7y)^2}}}.



{{{(7y)^2-(5)^2}}} Rewrite {{{25}}} as {{{(5)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=7y}}} and {{{B=5}}}.



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.



{{{(7y)^2-(5)^2=(7y+5)(7y-5)}}} Plug in {{{A=7y}}} and {{{B=5}}}.



So this shows us that {{{49y^2-25}}} factors to {{{(7y+5)(7y-5)}}}.



In other words {{{49y^2-25=(7y+5)(7y-5)}}}.


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Answer:

So {{{49xy^2-25x}}} completely factors to {{{x(7y+5)(7y-5)}}}