Question 186951
Let's factor {{{3x^2+8x+4}}}





Looking at the expression {{{3x^2+8x+4}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{8}}}, and the last term is {{{4}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{4}}} to get {{{(3)(4)=12}}}.



Now the question is: what two whole numbers multiply to {{{12}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{8}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{12}}} (the previous product).



Factors of {{{12}}}:

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12



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



These factors pair up and multiply to {{{12}}}.

1*12
2*6
3*4
(-1)*(-12)
(-2)*(-6)
(-3)*(-4)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{8}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>1+12=13</font></td></tr><tr><td  align="center"><font color=red>2</font></td><td  align="center"><font color=red>6</font></td><td  align="center"><font color=red>2+6=8</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>3+4=7</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>-1+(-12)=-13</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-2+(-6)=-8</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-3+(-4)=-7</font></td></tr></table>



From the table, we can see that the two numbers {{{2}}} and {{{6}}} add to {{{8}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{6}}} both multiply to {{{12}}} <font size=4><b>and</b></font> add to {{{8}}}



Now replace the middle term {{{8x}}} with {{{2x+6x}}}. Remember, {{{2}}} and {{{6}}} add to {{{8}}}. So this shows us that {{{2x+6x=8x}}}.



{{{3x^2+highlight(2x+6x)+4}}} Replace the second term {{{8x}}} with {{{2x+6x}}}.



{{{(3x^2+2x)+(6x+4)}}} Group the terms into two pairs.



{{{x(3x+2)+(6x+4)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(3x+2)+2(3x+2)}}} 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)(3x+2)}}} Combine like terms. Or factor out the common term {{{3x+2}}}



So {{{3x^2+8x+4}}} factors to {{{(x+2)(3x+2)}}}.



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So {{{3x^2+8x+4=0}}} then transforms to {{{(x+2)(3x+2)=0}}}




Now set each factor equal to zero:


{{{x+2=0}}} or  {{{3x+2=0}}} 


{{{x=-2}}} or  {{{x=-2/3}}}    Solve for x in each case



So our answers are 


 {{{x=-2}}} or  {{{x=-2/3}}}