Question 131488
Factor:
{{{4x^2-49y^2}}} This is the difference of two squares:
{{{(2x)^2-(7y)^2}}}
The difference of two squares is factored like this:
{{{A^2-B^2 = (A+B)(A-B)}}} Apply this toyour problem:
{{{(2x)^2-(7y)^2 = (2x+7y)(2x-7y)}}}
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Factor:
{{{4x^2-28x+49}}} Since the first and last terms are perfect squares: {{{4x^2 = (2x)^2}}} and {{{49 = 7^2}}}, also, twice 2(7) = 28, the middle term of your equation, try:
{{{4x^2-28x+49 = (2x-7)(2x-7)}}} = {{{(2x-7)^2}}}