Question 146141


Looking at {{{12y^2-11y+2}}} we can see that the first term is {{{12y^2}}} and the last term is {{{2}}} where the coefficients are 12 and 2 respectively.


Now multiply the first coefficient 12 and the last coefficient 2 to get 24. Now what two numbers multiply to 24 and add to the  middle coefficient -11? Let's list all of the factors of 24:




Factors of 24:

1,2,3,4,6,8,12,24


-1,-2,-3,-4,-6,-8,-12,-24 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 24

1*24

2*12

3*8

4*6

(-1)*(-24)

(-2)*(-12)

(-3)*(-8)

(-4)*(-6)


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



Now which of these pairs add to -11? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -11


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">24</td><td>1+24=25</td></tr><tr><td align="center">2</td><td align="center">12</td><td>2+12=14</td></tr><tr><td align="center">3</td><td align="center">8</td><td>3+8=11</td></tr><tr><td align="center">4</td><td align="center">6</td><td>4+6=10</td></tr><tr><td align="center">-1</td><td align="center">-24</td><td>-1+(-24)=-25</td></tr><tr><td align="center">-2</td><td align="center">-12</td><td>-2+(-12)=-14</td></tr><tr><td align="center">-3</td><td align="center">-8</td><td>-3+(-8)=-11</td></tr><tr><td align="center">-4</td><td align="center">-6</td><td>-4+(-6)=-10</td></tr></table>



From this list we can see that -3 and -8 add up to -11 and multiply to 24



Now looking at the expression {{{12y^2-11y+2}}}, replace {{{-11y}}} with {{{-3y+-8y}}} (notice {{{-3y+-8y}}} adds up to {{{-11y}}}. So it is equivalent to {{{-11y}}})


{{{12y^2+highlight(-3y+-8y)+2}}}



Now let's factor {{{12y^2-3y-8y+2}}} by grouping:



{{{(12y^2-3y)+(-8y+2)}}} Group like terms



{{{3y(4y-1)-2(4y-1)}}} Factor out the GCF of {{{3y}}} out of the first group. Factor out the GCF of {{{-2}}} out of the second group



{{{(3y-2)(4y-1)}}} Since we have a common term of {{{4y-1}}}, we can combine like terms


So {{{12y^2-3y-8y+2}}} factors to {{{(3y-2)(4y-1)}}}



So this also means that {{{12y^2-11y+2}}} factors to {{{(3y-2)(4y-1)}}} (since {{{12y^2-11y+2}}} is equivalent to {{{12y^2-3y-8y+2}}})




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     Answer:

So {{{12y^2-11y+2}}} factors to {{{(3y-2)(4y-1)}}}