Question 211198
I'll do the first two to get you going


# 1




{{{x^3+64y^3}}} Start with the given expression.



{{{(x)^3+(4y)^3}}} Rewrite {{{x^3}}} as {{{(x)^3}}}. Rewrite {{{64y^3}}} as {{{(4y)^3}}}.



{{{(x+4y)((x)^2-(x)(4y)+(4y)^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)}}}



{{{(x+4y)(x^2-4xy+16y^2)}}} Multiply


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

Answer:


So {{{x^3+64y^3}}} factors to {{{(x+4y)(x^2-4xy+16y^2)}}}.


In other words, {{{x^3+64y^3=(x+4y)(x^2-4xy+16y^2)}}}




<hr>


# 2



Looking at the expression {{{4r^2+21r+5}}}, we can see that the first coefficient is {{{4}}}, the second coefficient is {{{21}}}, and the last term is {{{5}}}.



Now multiply the first coefficient {{{4}}} by the last term {{{5}}} to get {{{(4)(5)=20}}}.



Now the question is: what two whole numbers multiply to {{{20}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{21}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{20}}} (the previous product).



Factors of {{{20}}}:

1,2,4,5,10,20

-1,-2,-4,-5,-10,-20



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{20}}}.

1*20 = 20
2*10 = 20
4*5 = 20
(-1)*(-20) = 20
(-2)*(-10) = 20
(-4)*(-5) = 20


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{21}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=red>1</font></td><td  align="center"><font color=red>20</font></td><td  align="center"><font color=red>1+20=21</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>2+10=12</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>4+5=9</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-20</font></td><td  align="center"><font color=black>-1+(-20)=-21</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>-2+(-10)=-12</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>-4+(-5)=-9</font></td></tr></table>



From the table, we can see that the two numbers {{{1}}} and {{{20}}} add to {{{21}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{20}}} both multiply to {{{20}}} <font size=4><b>and</b></font> add to {{{21}}}



Now replace the middle term {{{21r}}} with {{{r+20r}}}. Remember, {{{1}}} and {{{20}}} add to {{{21}}}. So this shows us that {{{r+20r=21r}}}.



{{{4r^2+highlight(r+20r)+5}}} Replace the second term {{{21r}}} with {{{r+20r}}}.



{{{(4r^2+r)+(20r+5)}}} Group the terms into two pairs.



{{{r(4r+1)+(20r+5)}}} Factor out the GCF {{{r}}} from the first group.



{{{r(4r+1)+5(4r+1)}}} Factor out {{{5}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(r+5)(4r+1)}}} Combine like terms. Or factor out the common term {{{4r+1}}}



===============================================================


<a name="ans">


Answer:



So {{{4r^2+21r+5}}} factors to {{{(r+5)(4r+1)}}}.



In other words, {{{4r^2+21r+5=(r+5)(4r+1)}}}.



Note: you can check the answer by expanding {{{(r+5)(4r+1)}}} to get {{{4r^2+21r+5}}} or by graphing the original expression and the answer (the two graphs should be identical).