Question 114003
#1


{{{18xy^3+3xy^2-10xy}}} Start with the given expression



{{{xy(18y^2+3y-10)}}} Factor out the GCF {{{xy}}}



Now let's focus on the inner expression {{{18y^2+3y-10}}}



Looking at {{{18y^2+3y-10}}} we can see that the first term is {{{18y^2}}} and the last term is {{{-10}}} where the coefficients are 18 and -10 respectively.


Now multiply the first coefficient 18 and the last coefficient -10 to get -180. Now what two numbers multiply to -180 and add to the  middle coefficient 3? Let's list all of the factors of -180:




Factors of -180:

1,2,3,4,5,6,9,10,12,15,18,20,30,36,45,60,90,180


-1,-2,-3,-4,-5,-6,-9,-10,-12,-15,-18,-20,-30,-36,-45,-60,-90,-180 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -180

(1)*(-180)

(2)*(-90)

(3)*(-60)

(4)*(-45)

(5)*(-36)

(6)*(-30)

(9)*(-20)

(10)*(-18)

(12)*(-15)

(-1)*(180)

(-2)*(90)

(-3)*(60)

(-4)*(45)

(-5)*(36)

(-6)*(30)

(-9)*(20)

(-10)*(18)

(-12)*(15)


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



Now which of these pairs add to 3? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 3


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-180</td><td>1+(-180)=-179</td></tr><tr><td align="center">2</td><td align="center">-90</td><td>2+(-90)=-88</td></tr><tr><td align="center">3</td><td align="center">-60</td><td>3+(-60)=-57</td></tr><tr><td align="center">4</td><td align="center">-45</td><td>4+(-45)=-41</td></tr><tr><td align="center">5</td><td align="center">-36</td><td>5+(-36)=-31</td></tr><tr><td align="center">6</td><td align="center">-30</td><td>6+(-30)=-24</td></tr><tr><td align="center">9</td><td align="center">-20</td><td>9+(-20)=-11</td></tr><tr><td align="center">10</td><td align="center">-18</td><td>10+(-18)=-8</td></tr><tr><td align="center">12</td><td align="center">-15</td><td>12+(-15)=-3</td></tr><tr><td align="center">-1</td><td align="center">180</td><td>-1+180=179</td></tr><tr><td align="center">-2</td><td align="center">90</td><td>-2+90=88</td></tr><tr><td align="center">-3</td><td align="center">60</td><td>-3+60=57</td></tr><tr><td align="center">-4</td><td align="center">45</td><td>-4+45=41</td></tr><tr><td align="center">-5</td><td align="center">36</td><td>-5+36=31</td></tr><tr><td align="center">-6</td><td align="center">30</td><td>-6+30=24</td></tr><tr><td align="center">-9</td><td align="center">20</td><td>-9+20=11</td></tr><tr><td align="center">-10</td><td align="center">18</td><td>-10+18=8</td></tr><tr><td align="center">-12</td><td align="center">15</td><td>-12+15=3</td></tr></table>



From this list we can see that -12 and 15 add up to 3 and multiply to -180



Now looking at the expression {{{18y^2+3y-10}}}, replace {{{3y}}} with {{{-12y+15y}}} (notice {{{-12y+15y}}} adds up to {{{3y}}}. So it is equivalent to {{{3y}}})


{{{18y^2+highlight(-12y+15y)+-10}}}



Now let's factor {{{18y^2-12y+15y-10}}} by grouping:



{{{(18y^2-12y)+(15y-10)}}} Group like terms



{{{6y(3y-2)+5(3y-2)}}} Factor out the GCF of {{{6y}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(6y+5)(3y-2)}}} Since we have a common term of {{{3y-2}}}, we can combine like terms


So {{{18y^2+3y-10}}} factors to {{{(6y+5)(3y-2)}}}



{{{xy(6y+5)(3y-2)}}} Now reintroduce the GCF




--------------------
Answer:


So {{{18xy^3+3xy^2-10xy}}} factors to {{{xy(6y+5)(3y-2)}}}




<hr>


#2




Looking at {{{15x^2+7x-2}}} we can see that the first term is {{{15x^2}}} and the last term is {{{-2}}} where the coefficients are 15 and -2 respectively.


Now multiply the first coefficient 15 and the last coefficient -2 to get -30. Now what two numbers multiply to -30 and add to the  middle coefficient 7? Let's list all of the factors of -30:




Factors of -30:

1,2,3,5,6,10,15,30


-1,-2,-3,-5,-6,-10,-15,-30 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -30

(1)*(-30)

(2)*(-15)

(3)*(-10)

(5)*(-6)

(-1)*(30)

(-2)*(15)

(-3)*(10)

(-5)*(6)


note: remember, the product of a negative and a positive number is a negative 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">-30</td><td>1+(-30)=-29</td></tr><tr><td align="center">2</td><td align="center">-15</td><td>2+(-15)=-13</td></tr><tr><td align="center">3</td><td align="center">-10</td><td>3+(-10)=-7</td></tr><tr><td align="center">5</td><td align="center">-6</td><td>5+(-6)=-1</td></tr><tr><td align="center">-1</td><td align="center">30</td><td>-1+30=29</td></tr><tr><td align="center">-2</td><td align="center">15</td><td>-2+15=13</td></tr><tr><td align="center">-3</td><td align="center">10</td><td>-3+10=7</td></tr><tr><td align="center">-5</td><td align="center">6</td><td>-5+6=1</td></tr></table>



From this list we can see that -3 and 10 add up to 7 and multiply to -30



Now looking at the expression {{{15x^2+7x-2}}}, replace {{{7x}}} with {{{-3x+10x}}} (notice {{{-3x+10x}}} adds up to {{{7x}}}. So it is equivalent to {{{7x}}})


{{{15x^2+highlight(-3x+10x)+-2}}}



Now let's factor {{{15x^2-3x+10x-2}}} by grouping:



{{{(15x^2-3x)+(10x-2)}}} Group like terms



{{{3x(5x-1)+2(5x-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)(5x-1)}}} Since we have a common term of {{{5x-1}}}, we can combine like terms


So {{{15x^2-3x+10x-2}}} factors to {{{(3x+2)(5x-1)}}}



So this also means that {{{15x^2+7x-2}}} factors to {{{(3x+2)(5x-1)}}} (since {{{15x^2+7x-2}}} is equivalent to {{{15x^2-3x+10x-2}}})



<hr>


#3




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


Now multiply the first coefficient 25 and the last coefficient 4 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 {{{25x^2+20x+4}}}, replace {{{20x}}} with {{{10x+10x}}} (notice {{{10x+10x}}} adds up to {{{20x}}}. So it is equivalent to {{{20x}}})


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



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



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



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



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


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



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