Question 117827


Looking at {{{6x^2+7x+2}}} we can see that the first term is {{{6x^2}}} and the last term is {{{2}}} where the coefficients are 6 and 2 respectively.


Now multiply the first coefficient 6 and the last coefficient 2 to get 12. Now what two numbers multiply to 12 and add to the  middle coefficient 7? Let's list all of the factors of 12:




Factors of 12:

1,2,3,4,6,12


-1,-2,-3,-4,-6,-12 ...List the negative factors as well. 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)


note: remember two negative numbers multiplied together make a positive number



Now which of these pairs add to 7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 7


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



From this list we can see that 3 and 4 add up to 7 and multiply to 12



Now looking at the expression {{{6x^2+7x+2}}}, replace {{{7x}}} with {{{3x+4x}}} (notice {{{3x+4x}}} adds up to {{{7x}}}. So it is equivalent to {{{7x}}})


{{{6x^2+highlight(3x+4x)+2}}}



Now let's factor {{{6x^2+3x+4x+2}}} by grouping:



{{{(6x^2+3x)+(4x+2)}}} Group like terms



{{{3x(2x+1)+2(2x+1)}}} Factor out the GCF of {{{3x}}} out of the first group. Factor out the GCF of {{{2}}} out of the second group



{{{(3x+2)(2x+1)}}} Since we have a common term of {{{2x+1}}}, we can combine like terms


So {{{6x^2+3x+4x+2}}} factors to {{{(3x+2)(2x+1)}}}



So this also means that {{{6x^2+7x+2}}} factors to {{{(3x+2)(2x+1)}}} (since {{{6x^2+7x+2}}} is equivalent to {{{6x^2+3x+4x+2}}})


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Answer:


So {{{6x^2+7x+2}}} factors to {{{(3x+2)(2x+1)}}}