Question 317820
# 1




{{{36x^2+96xy+64y^2}}} Start with the given expression



{{{4(9x^2+24xy+16y^2)}}} Factor out the GCF {{{4}}}



Now let's focus on the inner expression {{{9x^2+24xy+16y^2}}}





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Looking at {{{9x^2+24xy+16y^2}}} we can see that the first term is {{{9x^2}}} and the last term is {{{16y^2}}} where the coefficients are 9 and 16 respectively.


Now multiply the first coefficient 9 and the last coefficient 16 to get 144. Now what two numbers multiply to 144 and add to the  middle coefficient 24? Let's list all of the factors of 144:




Factors of 144:

1,2,3,4,6,8,9,12,16,18,24,36,48,72


-1,-2,-3,-4,-6,-8,-9,-12,-16,-18,-24,-36,-48,-72 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 144

1*144

2*72

3*48

4*36

6*24

8*18

9*16

12*12

(-1)*(-144)

(-2)*(-72)

(-3)*(-48)

(-4)*(-36)

(-6)*(-24)

(-8)*(-18)

(-9)*(-16)

(-12)*(-12)


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



Now which of these pairs add to 24? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 24


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">144</td><td>1+144=145</td></tr><tr><td align="center">2</td><td align="center">72</td><td>2+72=74</td></tr><tr><td align="center">3</td><td align="center">48</td><td>3+48=51</td></tr><tr><td align="center">4</td><td align="center">36</td><td>4+36=40</td></tr><tr><td align="center">6</td><td align="center">24</td><td>6+24=30</td></tr><tr><td align="center">8</td><td align="center">18</td><td>8+18=26</td></tr><tr><td align="center">9</td><td align="center">16</td><td>9+16=25</td></tr><tr><td align="center">12</td><td align="center">12</td><td>12+12=24</td></tr><tr><td align="center">-1</td><td align="center">-144</td><td>-1+(-144)=-145</td></tr><tr><td align="center">-2</td><td align="center">-72</td><td>-2+(-72)=-74</td></tr><tr><td align="center">-3</td><td align="center">-48</td><td>-3+(-48)=-51</td></tr><tr><td align="center">-4</td><td align="center">-36</td><td>-4+(-36)=-40</td></tr><tr><td align="center">-6</td><td align="center">-24</td><td>-6+(-24)=-30</td></tr><tr><td align="center">-8</td><td align="center">-18</td><td>-8+(-18)=-26</td></tr><tr><td align="center">-9</td><td align="center">-16</td><td>-9+(-16)=-25</td></tr><tr><td align="center">-12</td><td align="center">-12</td><td>-12+(-12)=-24</td></tr></table>



From this list we can see that 12 and 12 add up to 24 and multiply to 144



Now looking at the expression {{{9x^2+24xy+16y^2}}}, replace {{{24xy}}} with {{{12xy+12xy}}} (notice {{{12xy+12xy}}} adds up to {{{24xy}}}. So it is equivalent to {{{24xy}}})


{{{9x^2+highlight(12xy+12xy)+16y^2}}}



Now let's factor {{{9x^2+12xy+12xy+16y^2}}} by grouping:



{{{(9x^2+12xy)+(12xy+16y^2)}}} Group like terms



{{{3x(3x+4y)+4y(3x+4y)}}} Factor out the GCF of {{{3x}}} out of the first group. Factor out the GCF of {{{4y}}} out of the second group



{{{(3x+4y)(3x+4y)}}} Since we have a common term of {{{3x+4y}}}, we can combine like terms


So {{{9x^2+12xy+12xy+16y^2}}} factors to {{{(3x+4y)(3x+4y)}}}



So this also means that {{{9x^2+24xy+16y^2}}} factors to {{{(3x+4y)(3x+4y)}}} (since {{{9x^2+24xy+16y^2}}} is equivalent to {{{9x^2+12xy+12xy+16y^2}}})



note:  {{{(3x+4y)(3x+4y)}}} is equivalent to  {{{(3x+4y)^2}}} since the term {{{3x+4y}}} occurs twice. So {{{9x^2+24xy+16y^2}}} also factors to {{{(3x+4y)^2}}}




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So our expression goes from {{{4(9x^2+24xy+16y^2)}}} and factors further to {{{4(3x+4y)^2}}}



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Answer:


So {{{36x^2+96xy+64y^2}}} factors to {{{4(3x+4y)^2}}}

    

In other words {{{36x^2+96xy+64y^2=4(3x+4y)^2}}}


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# 2




{{{8x^3+27y^3}}} Start with the given expression.



{{{(2x)^3+(3y)^3}}} Rewrite {{{8x^3}}} as {{{(2x)^3}}}. Rewrite {{{27y^3}}} as {{{(3y)^3}}}.



{{{(2x+3y)((2x)^2-(2x)(3y)+(3y)^2)}}} Now factor by using the sum of cubes formula. Remember the <a href="http://www.purplemath.com/modules/specfact2.htm">sum of cubes formula</a> is {{{A^3+B^3=(A+B)(A^2-AB+B^2)}}}



{{{(2x+3y)(4x^2-6xy+9y^2)}}} Multiply


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Answer:


So {{{8x^3+27y^3}}} factors to {{{(2x+3y)(4x^2-6xy+9y^2)}}}.


In other words, {{{8x^3+27y^3=(2x+3y)(4x^2-6xy+9y^2)}}}