<PRE><B>Question 148867</B>
# 1




Looking at the expression {{{6b^2-7b-5}}}, we can see that the first coefficient is {{{6}}}, the second coefficient is {{{-7}}}, and the last term is {{{-5}}}.



Now multiply the first coefficient {{{6}}} by the last term {{{-5}}} to get {{{(6)(-5)=-30}}}.



Now the question is: what two whole numbers multiply to {{{-30}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-7}}}?



To find these two numbers, we need to list &lt;font size=4&gt;&lt;b&gt;all&lt;/b&gt;&lt;/font&gt; of the factors of {{{-30}}} (the previous product).



Factors of {{{-30}}}:

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-30}}}. For instance, {{{1*30=-30}}}, {{{2*15=-30}}}, etc.



Since {{{-30}}} is negative, this means that one factor is positive and one is negative.



Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-7}}}:



&lt;table border=&quot;1&quot;&gt;&lt;th&gt;First Number&lt;/th&gt;&lt;th&gt;Second Number&lt;/th&gt;&lt;th&gt;Sum&lt;/th&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;1&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-30&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;1+(-30)=-29&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;2&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-15&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;2+(-15)=-13&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;3&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-10&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;3+(-10)=-7&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;5&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-6&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;5+(-6)=-1&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-1&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;30&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-1+30=29&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-2&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;15&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-2+15=13&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-3&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;10&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-3+10=7&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-5&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;6&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-5+6=1&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{3}}} and {{{-10}}} add to {{{-7}}} (the middle coefficient).



So the two numbers {{{3}}} and {{{-10}}} both multiply to {{{-30}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-7}}}



Now replace the middle term {{{-7b}}} with {{{3b-10b}}}. Remember, {{{3}}} and {{{-10}}} add to {{{-7}}}. So this shows us that {{{3b-10b=-7b}}}.



{{{6b^2+highlight(3b-10b)-5}}} Replace the second term {{{-7b}}} with {{{3b-10b}}}.



{{{(6b^2+3b)+(-10b-5)}}} Group the terms into two pairs.



{{{3b(2b+1)+(-10b-5)}}} Factor out the GCF {{{3b}}} from the first group.



{{{3b(2b+1)-5(2b+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.



{{{(3b-5)(2b+1)}}} Combine like terms. Or factor out the common term {{{2b+1}}}


---------------------------------------------



Answer:



So {{{6b^2-7b-5}}} factors to {{{(3b-5)(2b+1)}}}.



Note: you can check the answer by FOILing {{{(3b-5)(2b+1)}}} to get {{{6b^2-7b-5}}} or by graphing the original expression and the answer (the two graphs should be identical).





&lt;hr&gt;



# 2




Looking at {{{1m^2+4mn+4n^2}}} we can see that the first term is {{{1m^2}}} and the last term is {{{4n^2}}} where the coefficients are 1 and 4 respectively.


Now multiply the first coefficient 1 and the last coefficient 4 to get 4. Now what two numbers multiply to 4 and add to the  middle coefficient 4? Let&#39;s list all of the factors of 4:




Factors of 4:

1,2


-1,-2 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 4

1*4

2*2

(-1)*(-4)

(-2)*(-2)


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


&lt;table border=&quot;1&quot;&gt;&lt;th&gt;First Number&lt;/th&gt;&lt;th&gt;Second Number&lt;/th&gt;&lt;th&gt;Sum&lt;/th&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;1&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;4&lt;/td&gt;&lt;td&gt;1+4=5&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;2&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;2&lt;/td&gt;&lt;td&gt;2+2=4&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;-1&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;-4&lt;/td&gt;&lt;td&gt;-1+(-4)=-5&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;-2&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;-2&lt;/td&gt;&lt;td&gt;-2+(-2)=-4&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From this list we can see that 2 and 2 add up to 4 and multiply to 4



Now looking at the expression {{{1m^2+4mn+4n^2}}}, replace {{{4mn}}} with {{{2mn+2mn}}} (notice {{{2mn+2mn}}} adds up to {{{4mn}}}. So it is equivalent to {{{4mn}}})


{{{1m^2+highlight(2mn+2mn)+4n^2}}}



Now let&#39;s factor {{{1m^2+2mn+2mn+4n^2}}} by grouping:



{{{(1m^2+2mn)+(2mn+4n^2)}}} Group like terms



{{{m(m+2n)+2n(m+2n)}}} Factor out the GCF of {{{m}}} out of the first group. Factor out the GCF of {{{2n}}} out of the second group



{{{(m+2n)(m+2n)}}} Since we have a common term of {{{m+2n}}}, we can combine like terms


So {{{1m^2+2mn+2mn+4n^2}}} factors to {{{(m+2n)(m+2n)}}}



So this also means that {{{1m^2+4mn+4n^2}}} factors to {{{(m+2n)(m+2n)}}} (since {{{1m^2+4mn+4n^2}}} is equivalent to {{{1m^2+2mn+2mn+4n^2}}})



note:  {{{(m+2n)(m+2n)}}} is equivalent to  {{{(m+2n)^2}}} since the term {{{m+2n}}} occurs twice. So {{{1m^2+4mn+4n^2}}} also factors to {{{(m+2n)^2}}}




------------------------------------------------------------




     Answer:

So {{{m^2+4mn+4n^2}}} factors to {{{(m+2n)^2}}}




&lt;hr&gt;


# 3




Looking at the expression {{{2a^2-13a+15}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{-13}}}, and the last term is {{{15}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{15}}} to get {{{(2)(15)=30}}}.



Now the question is: what two whole numbers multiply to {{{30}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-13}}}?



To find these two numbers, we need to list &lt;font size=4&gt;&lt;b&gt;all&lt;/b&gt;&lt;/font&gt; of the factors of {{{30}}} (the previous product).



Factors of {{{30}}}:

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{30}}}. For instance, {{{1*30=30}}}, {{{2*15=30}}}, etc.



Since {{{30}}} is positive, this means that either

a) both factors are positive, or...
b) both factors are negative.



Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-13}}}:



&lt;table border=&quot;1&quot;&gt;&lt;th&gt;First Number&lt;/th&gt;&lt;th&gt;Second Number&lt;/th&gt;&lt;th&gt;Sum&lt;/th&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;1&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;30&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;1+30=31&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;2&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;15&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;2+15=17&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;3&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;10&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;3+10=13&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;5&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;6&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;5+6=11&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-1&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-30&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-1+(-30)=-31&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-2&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-15&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-2+(-15)=-17&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-3&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-10&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-3+(-10)=-13&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-5&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-6&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-5+(-6)=-11&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-3}}} and {{{-10}}} add to {{{-13}}} (the middle coefficient).



So the two numbers {{{-3}}} and {{{-10}}} both multiply to {{{30}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-13}}}



Now replace the middle term {{{-13a}}} with {{{-3a-10a}}}. Remember, {{{-3}}} and {{{-10}}} add to {{{-13}}}. So this shows us that {{{-3a-10a=-13a}}}.



{{{2a^2+highlight(-3a-10a)+15}}} Replace the second term {{{-13a}}} with {{{-3a-10a}}}.



{{{(2a^2-3a)+(-10a+15)}}} Group the terms into two pairs.



{{{a(2a-3)+(-10a+15)}}} Factor out the GCF {{{a}}} from the first group.



{{{a(2a-3)-5(2a-3)}}} 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.



{{{(a-5)(2a-3)}}} Combine like terms. Or factor out the common term {{{2a-3}}}


---------------------------------------------



Answer:



So {{{2a^2-13a+15}}} factors to {{{(a-5)(2a-3)}}}.



Note: you can check the answer by FOILing {{{(a-5)(2a-3)}}} to get {{{2a^2-13a+15}}} or by graphing the original expression and the answer (the two graphs should be identical).




&lt;hr&gt;



# 4




{{{z^3+9z^2+18z}}} Start with the given expression



{{{z(z^2+9z+18)}}} Factor out the GCF {{{z}}}



Now let&#39;s focus on the inner expression {{{z^2+9z+18}}}





------------------------------------------------------------




Looking at {{{1z^2+9z+18}}} we can see that the first term is {{{1z^2}}} and the last term is {{{18}}} where the coefficients are 1 and 18 respectively.


Now multiply the first coefficient 1 and the last coefficient 18 to get 18. Now what two numbers multiply to 18 and add to the  middle coefficient 9? Let&#39;s list all of the factors of 18:




Factors of 18:

1,2,3,6,9,18


-1,-2,-3,-6,-9,-18 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 18

1*18

2*9

3*6

(-1)*(-18)

(-2)*(-9)

(-3)*(-6)


note: remember two negative numbers multiplied together make a positive number



Now which of these pairs add to 9? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 9


&lt;table border=&quot;1&quot;&gt;&lt;th&gt;First Number&lt;/th&gt;&lt;th&gt;Second Number&lt;/th&gt;&lt;th&gt;Sum&lt;/th&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;1&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;18&lt;/td&gt;&lt;td&gt;1+18=19&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;2&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;9&lt;/td&gt;&lt;td&gt;2+9=11&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;3&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;6&lt;/td&gt;&lt;td&gt;3+6=9&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;-1&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;-18&lt;/td&gt;&lt;td&gt;-1+(-18)=-19&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;-2&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;-9&lt;/td&gt;&lt;td&gt;-2+(-9)=-11&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td align=&quot;center&quot;&gt;-3&lt;/td&gt;&lt;td align=&quot;center&quot;&gt;-6&lt;/td&gt;&lt;td&gt;-3+(-6)=-9&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From this list we can see that 3 and 6 add up to 9 and multiply to 18



Now looking at the expression {{{1z^2+9z+18}}}, replace {{{9z}}} with {{{3z+6z}}} (notice {{{3z+6z}}} adds up to {{{9z}}}. So it is equivalent to {{{9z}}})


{{{1z^2+highlight(3z+6z)+18}}}



Now let&#39;s factor {{{1z^2+3z+6z+18}}} by grouping:



{{{(1z^2+3z)+(6z+18)}}} Group like terms



{{{z(z+3)+6(z+3)}}} Factor out the GCF of {{{z}}} out of the first group. Factor out the GCF of {{{6}}} out of the second group



{{{(z+6)(z+3)}}} Since we have a common term of {{{z+3}}}, we can combine like terms


So {{{1z^2+3z+6z+18}}} factors to {{{(z+6)(z+3)}}}



So this also means that {{{1z^2+9z+18}}} factors to {{{(z+6)(z+3)}}} (since {{{1z^2+9z+18}}} is equivalent to {{{1z^2+3z+6z+18}}})




------------------------------------------------------------





So our expression goes from {{{z(z^2+9z+18)}}} and factors further to {{{z(z+6)(z+3)}}}



------------------

Answer:


So {{{z^3+9z^2+18z}}} factors to {{{z(z+6)(z+3)}}}

    


&lt;hr&gt;



# 5




{{{x^3+125}}} Start with the given expression.



{{{(x)^3+(5)^3}}} Rewrite {{{x^3}}} as {{{(x)^3}}}. Rewrite {{{125}}} as {{{(5)^3}}}.



{{{(x+5)((x)^2-(x)(5)+(5)^2)}}} Now factor by using the sum of cubes formula. Remember the &lt;a href=&quot;http://www.purplemath.com/modules/specfact2.htm&quot;&gt;sum of cubes formula&lt;/a&gt; is {{{A^3+B^3=(A+B)(A^2-AB+B^2)}}}



{{{(x+5)(x^2-5x+25)}}} Multiply


-----------------------------------

Answer:


So {{{x^3+125}}} factors to {{{(x+5)(x^2-5x+25)}}}.


In other words, {{{x^3+125=(x+5)(x^2-5x+25)}}}




&lt;hr&gt;


# 6



{{{a^4-ab^3}}} Start with the given expression



{{{a(a^3-b^3)}}} Factor out the GCF {{{a}}}



Now let&#39;s focus on the inner expression {{{a^3-b^3}}}


------------------------------------------------------------



{{{(a)^3-(b)^3}}} Rewrite {{{a^3}}} as {{{(a)^3}}}. Rewrite {{{b^3}}} as {{{(b)^3}}}.



{{{(a-b)((a)^2+(a)(b)+(b)^2)}}} Now factor by using the difference of cubes formula. Remember the &lt;a href=&quot;http://www.purplemath.com/modules/specfact2.htm&quot;&gt;difference of cubes formula&lt;/a&gt; is {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}}



{{{(a-b)(a^2+ab+b^2)}}} Multiply




So {{{a^3-b^3}}} factors to {{{(a-b)(a^2+ab+b^2)}}}.




-----------------------------------

Answer:


So {{{a^4-b^3}}} factors to {{{a(a-b)(a^2+ab+b^2)}}}.
</PRE>