Question 682271
I'm assuming you want to factor this




{{{2x^2+6x+4}}} Start with the given expression.



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



Now let's try to factor the inner expression {{{x^2+3x+2}}}



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Looking at the expression {{{x^2+3x+2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{3}}}, and the last term is {{{2}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{2}}} to get {{{(1)(2)=2}}}.



Now the question is: what two whole numbers multiply to {{{2}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{3}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{2}}} (the previous product).



Factors of {{{2}}}:

1,2

-1,-2



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



These factors pair up and multiply to {{{2}}}.

1*2 = 2
(-1)*(-2) = 2


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{3}}}:



<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>2</font></td><td  align="center"><font color=red>1+2=3</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-1+(-2)=-3</font></td></tr></table>



From the table, we can see that the two numbers {{{1}}} and {{{2}}} add to {{{3}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{2}}} both multiply to {{{2}}} <font size=4><b>and</b></font> add to {{{3}}}



Now replace the middle term {{{3x}}} with {{{x+2x}}}. Remember, {{{1}}} and {{{2}}} add to {{{3}}}. So this shows us that {{{x+2x=3x}}}.



{{{x^2+highlight(x+2x)+2}}} Replace the second term {{{3x}}} with {{{x+2x}}}.



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



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



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



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



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Answer:



So {{{2x^2+6x+4}}} completely factors to {{{2(x+2)(x+1)}}}.



In other words, {{{2x^2+6x+4=2(x+2)(x+1)}}}.



Note: you can check the answer by expanding {{{2(x+2)(x+1)}}} to get {{{2x^2+6x+4}}} or by graphing the original expression and the answer (the two graphs should be identical).