Question 369932


{{{6x^2+33x+15}}} Start with the given expression.



{{{3(2x^2+11x+5)}}} Factor out the GCF {{{3}}}.



Now let's try to factor the inner expression {{{2x^2+11x+5}}}



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Looking at the expression {{{2x^2+11x+5}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{11}}}, and the last term is {{{5}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{5}}} to get {{{(2)(5)=10}}}.



Now the question is: what two whole numbers multiply to {{{10}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{11}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{10}}} (the previous product).



Factors of {{{10}}}:

1,2,5,10

-1,-2,-5,-10



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{10}}}.

1*10 = 10
2*5 = 10
(-1)*(-10) = 10
(-2)*(-5) = 10


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{11}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=red>1</font></td><td  align="center"><font color=red>10</font></td><td  align="center"><font color=red>1+10=11</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>2+5=7</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>-1+(-10)=-11</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>-2+(-5)=-7</font></td></tr></table>



From the table, we can see that the two numbers {{{1}}} and {{{10}}} add to {{{11}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{10}}} both multiply to {{{10}}} <font size=4><b>and</b></font> add to {{{11}}}



Now replace the middle term {{{11x}}} with {{{x+10x}}}. Remember, {{{1}}} and {{{10}}} add to {{{11}}}. So this shows us that {{{x+10x=11x}}}.



{{{2x^2+highlight(x+10x)+5}}} Replace the second term {{{11x}}} with {{{x+10x}}}.



{{{(2x^2+x)+(10x+5)}}} Group the terms into two pairs.



{{{x(2x+1)+(10x+5)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x+1)+5(2x+1)}}} Factor out {{{5}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(x+5)(2x+1)}}} Combine like terms. Or factor out the common term {{{2x+1}}}



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So {{{3(2x^2+11x+5)}}} then factors further to {{{3(x+5)(2x+1)}}}



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Answer:



So {{{6x^2+33x+15}}} completely factors to {{{3(x+5)(2x+1)}}}.



In other words, {{{6x^2+33x+15=3(x+5)(2x+1)}}}.



Note: you can check the answer by expanding {{{3(x+5)(2x+1)}}} to get {{{6x^2+33x+15}}} or by graphing the original expression and the answer (the two graphs should be identical).



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Jim