Question 190671

{{{(3x+2)(3x-2)}}} Start with the given expression.



Now let's FOIL the expression.



Remember, when you FOIL an expression, you follow this procedure:



{{{(highlight(3x)+2)(highlight(3x)-2)}}} Multiply the <font color="red">F</font>irst terms:{{{(3*x)*(3*x)=9*x^2}}}.



{{{(highlight(3x)+2)(3x+highlight(-2))}}} Multiply the <font color="red">O</font>uter terms:{{{(3*x)*(-2)=-6*x}}}.



{{{(3x+highlight(2))(highlight(3x)-2)}}} Multiply the <font color="red">I</font>nner terms:{{{(2)*(3*x)=6*x}}}.



{{{(3x+highlight(2))(3x+highlight(-2))}}} Multiply the <font color="red">L</font>ast terms:{{{(2)*(-2)=-4}}}.



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So we have the terms: {{{9*x^2}}}, {{{-6*x}}}, {{{6*x}}}, {{{-4}}} 



{{{9*x^2-6*x+6*x-4}}} Now add every term listed above to make a single expression.



{{{9*x^2-4}}} Now combine like terms.



So {{{(3x+2)(3x-2)}}} FOILs to {{{9*x^2-4}}}.



In other words, {{{(3x+2)(3x-2)=9x^2-4}}}.


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Now, let's factor the answer to check our work:



{{{9x^2-4}}} Start with the given expression.



{{{(3x)^2-4}}} Rewrite {{{9x^2}}} as {{{(3x)^2}}}.



{{{(3x)^2-(2)^2}}} Rewrite {{{4}}} as {{{(2)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=3x}}} and {{{B=2}}}.



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.



{{{(3x)^2-(2)^2=(3x+2)(3x-2)}}} Plug in {{{A=3x}}} and {{{B=2}}}.



So this shows us that {{{9x^2-4}}} factors to {{{(3x+2)(3x-2)}}}.



In other words {{{9x^2-4=(3x+2)(3x-2)}}}.



So this confirms our answer