Question 140851
# 1



{{{4x^5y^3+16x^3y^2+32xy}}} Start with the given expression



{{{4xy(x^4y^2+4x^2y+8)}}} Factor out the GCF {{{4xy}}}



Now let's focus on the inner expression {{{x^4y^2+4x^2y+8}}}





------------------------------------------------------------




Looking at {{{1x^4y^2+4x^2y+8}}} we can see that the first term is {{{1x^4y^2}}} and the last term is {{{8}}} where the coefficients are 1 and 8 respectively.


Now multiply the first coefficient 1 and the last coefficient 8 to get 8. Now what two numbers multiply to 8 and add to the  middle coefficient 4? Let's list all of the factors of 8:




Factors of 8:

1,2,4,8


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


These factors pair up and multiply to 8

1*8

2*4

(-1)*(-8)

(-2)*(-4)


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



Now which of these pairs add to 4? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 4


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">8</td><td>1+8=9</td></tr><tr><td align="center">2</td><td align="center">4</td><td>2+4=6</td></tr><tr><td align="center">-1</td><td align="center">-8</td><td>-1+(-8)=-9</td></tr><tr><td align="center">-2</td><td align="center">-4</td><td>-2+(-4)=-6</td></tr></table>

None of these pairs of factors add to 4. So the expression {{{1x^4y^2+4x^2y+8}}} cannot be factored


------------------------------------------------------------






Answer:

So {{{4x^5y^3+16x^3y^2+32xy}}} factors to {{{4xy(x^4y^2+4x^2y+8)}}}

    



<hr>



# 2




{{{9x^2=72x}}} Start with the given equation



{{{9x^2-72x=0}}}  Subtract 72x from both sides. 



{{{9x(x-8)=0}}} Factor out the GCF 9x




Now set each factor equal to zero:


{{{9x=0}}} or  {{{x-8=0}}} 



{{{x=0}}} or  {{{x=8}}}    Now solve for x in each case



So our answers are
 

 {{{x=0}}} or  {{{x=8}}} 






<hr>


# 3

{{{21x^3-15x^2+28x-20}}} Start with the given expression


{{{(21x^3-15x^2)+(28x-20)}}} Group like terms



{{{3x^2(7x-5)+4(7x-5)}}} Factor out the GCF {{{3x^2}}} out of the first group. Factor out the GCF {{{4}}} out of the second group



{{{(3x^2+4)(7x-5)}}} Since we have the common term {{{7x-5}}}, we can combine like terms


So {{{21x^3-15x^2+28x-20}}} factors to {{{(3x^2+4)(7x-5)}}}