Question 191479


{{{35y^3-60y^2-20y}}} Start with the given expression



{{{5y(7y^2-12y-4)}}} Factor out the GCF {{{5y}}}



Now let's focus on the inner expression {{{7y^2-12y-4}}}





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Looking at {{{7y^2-12y-4}}} we can see that the first term is {{{7y^2}}} and the last term is {{{-4}}} where the coefficients are 7 and -4 respectively.


Now multiply the first coefficient 7 and the last coefficient -4 to get -28. Now what two numbers multiply to -28 and add to the  middle coefficient -12? Let's list all of the factors of -28:




Factors of -28:

1,2,4,7,14,28


-1,-2,-4,-7,-14,-28 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -28

(1)*(-28)

(2)*(-14)

(4)*(-7)

(-1)*(28)

(-2)*(14)

(-4)*(7)


note: remember, the product of a negative and a positive number is a negative number



Now which of these pairs add to -12? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -12


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-28</td><td>1+(-28)=-27</td></tr><tr><td align="center">2</td><td align="center">-14</td><td>2+(-14)=-12</td></tr><tr><td align="center">4</td><td align="center">-7</td><td>4+(-7)=-3</td></tr><tr><td align="center">-1</td><td align="center">28</td><td>-1+28=27</td></tr><tr><td align="center">-2</td><td align="center">14</td><td>-2+14=12</td></tr><tr><td align="center">-4</td><td align="center">7</td><td>-4+7=3</td></tr></table>



From this list we can see that 2 and -14 add up to -12 and multiply to -28



Now looking at the expression {{{7y^2-12y-4}}}, replace {{{-12y}}} with {{{2y-14y}}} (notice {{{2y-14y}}} adds up to {{{-12y}}}. So it is equivalent to {{{-12y}}})


{{{7y^2+highlight(2y-14y)-4}}}



Now let's factor {{{7y^2+2y-14y-4}}} by grouping:



{{{(7y^2+2y)+(-14y-4)}}} Group like terms



{{{y(7y+2)-2(7y+2)}}} Factor out the GCF of {{{y}}} out of the first group. Factor out the GCF of {{{-2}}} out of the second group



{{{(y-2)(7y+2)}}} Since we have a common term of {{{7y+2}}}, we can combine like terms


So {{{7y^2+2y-14y-4}}} factors to {{{(y-2)(7y+2)}}}



So this also means that {{{7y^2-12y-4}}} factors to {{{(y-2)(7y+2)}}} (since {{{7y^2-12y-4}}} is equivalent to {{{7y^2+2y-14y-4}}})




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So our expression goes from {{{5y(7y^2-12y-4)}}} and factors further to {{{5y(y-2)(7y+2)}}}



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Answer:


So {{{35y^3-60y^2-20y}}} completely factors to {{{5y(y-2)(7y+2)}}}