Question 181957
I'll do the first three to get you started:



# 46



{{{-3t^3+3t^2-6t}}} Start with the given expression



{{{-3t(t^2-t+2)}}} Factor out the GCF {{{-3t}}}



So {{{-3t^3+3t^2-6t}}} factors to {{{-3t(t^2-t+2)}}}


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# 60



Looking at {{{10a^2+ab-2b^2}}} we can see that the first term is {{{10a^2}}} and the last term is {{{-2b^2}}} where the coefficients are 10 and -2 respectively.


Now multiply the first coefficient 10 and the last coefficient -2 to get -20. Now what two numbers multiply to -20 and add to the  middle coefficient 1? Let's list all of the factors of -20:




Factors of -20:

1,2,4,5,10,20


-1,-2,-4,-5,-10,-20 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -20

(1)*(-20)

(2)*(-10)

(4)*(-5)

(-1)*(20)

(-2)*(10)

(-4)*(5)


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



Now which of these pairs add to 1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 1


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-20</td><td>1+(-20)=-19</td></tr><tr><td align="center">2</td><td align="center">-10</td><td>2+(-10)=-8</td></tr><tr><td align="center">4</td><td align="center">-5</td><td>4+(-5)=-1</td></tr><tr><td align="center">-1</td><td align="center">20</td><td>-1+20=19</td></tr><tr><td align="center">-2</td><td align="center">10</td><td>-2+10=8</td></tr><tr><td align="center">-4</td><td align="center">5</td><td>-4+5=1</td></tr></table>



From this list we can see that -4 and 5 add up to 1 and multiply to -20



Now looking at the expression {{{10a^2+ab-2b^2}}}, replace {{{ab}}} with {{{-4ab+5ab}}} (notice {{{-4ab+5ab}}} adds up to {{{ab}}}. So it is equivalent to {{{ab}}})



{{{10a^2+highlight(-4ab+5ab)+-2b^2}}}



Now let's factor {{{10a^2-4ab+5ab-2b^2}}} by grouping:



{{{(10a^2-4ab)+(5ab-2b^2)}}} Group like terms



{{{2a(5a-2b)+b(5a-2b)}}} Factor out the GCF of {{{2a}}} out of the first group. Factor out the GCF of {{{b}}} out of the second group



{{{(2a+b)(5a-2b)}}} Since we have a common term of {{{5a-2b}}}, we can combine like terms



So {{{10a^2-4ab+5ab-2b^2}}} factors to {{{(2a+b)(5a-2b)}}}



So this also means that {{{10a^2+ab-2b^2}}} factors to {{{(2a+b)(5a-2b)}}} (since {{{10a^2+ab-2b^2}}} is equivalent to {{{10a^2-4ab+5ab-2b^2}}})




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     Answer:

So {{{10a^2+ab-2b^2}}} factors to {{{(2a+b)(5a-2b)}}}



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# 80




Looking at {{{4m^2+20m+25}}} we can see that the first term is {{{4m^2}}} and the last term is {{{25}}} where the coefficients are 4 and 25 respectively.


Now multiply the first coefficient 4 and the last coefficient 25 to get 100. Now what two numbers multiply to 100 and add to the  middle coefficient 20? Let's list all of the factors of 100:




Factors of 100:

1,2,4,5,10,20,25,50


-1,-2,-4,-5,-10,-20,-25,-50 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 100

1*100

2*50

4*25

5*20

10*10

(-1)*(-100)

(-2)*(-50)

(-4)*(-25)

(-5)*(-20)

(-10)*(-10)


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



Now which of these pairs add to 20? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 20


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">100</td><td>1+100=101</td></tr><tr><td align="center">2</td><td align="center">50</td><td>2+50=52</td></tr><tr><td align="center">4</td><td align="center">25</td><td>4+25=29</td></tr><tr><td align="center">5</td><td align="center">20</td><td>5+20=25</td></tr><tr><td align="center">10</td><td align="center">10</td><td>10+10=20</td></tr><tr><td align="center">-1</td><td align="center">-100</td><td>-1+(-100)=-101</td></tr><tr><td align="center">-2</td><td align="center">-50</td><td>-2+(-50)=-52</td></tr><tr><td align="center">-4</td><td align="center">-25</td><td>-4+(-25)=-29</td></tr><tr><td align="center">-5</td><td align="center">-20</td><td>-5+(-20)=-25</td></tr><tr><td align="center">-10</td><td align="center">-10</td><td>-10+(-10)=-20</td></tr></table>



From this list we can see that 10 and 10 add up to 20 and multiply to 100



Now looking at the expression {{{4m^2+20m+25}}}, replace {{{20m}}} with {{{10m+10m}}} (notice {{{10m+10m}}} adds up to {{{20m}}}. So it is equivalent to {{{20m}}})


{{{4m^2+highlight(10m+10m)+25}}}



Now let's factor {{{4m^2+10m+10m+25}}} by grouping:



{{{(4m^2+10m)+(10m+25)}}} Group like terms



{{{2m(2m+5)+5(2m+5)}}} Factor out the GCF of {{{2m}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(2m+5)(2m+5)}}} Since we have a common term of {{{2m+5}}}, we can combine like terms


So {{{4m^2+10m+10m+25}}} factors to {{{(2m+5)(2m+5)}}}



So this also means that {{{4m^2+20m+25}}} factors to {{{(2m+5)(2m+5)}}} (since {{{4m^2+20m+25}}} is equivalent to {{{4m^2+10m+10m+25}}})



note:  {{{(2m+5)(2m+5)}}} is equivalent to  {{{(2m+5)^2}}} since the term {{{2m+5}}} occurs twice. So {{{4m^2+20m+25}}} also factors to {{{(2m+5)^2}}}




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     Answer:

So {{{4m^2+20m+25}}} factors to {{{(2m+5)^2}}}