<PRE><B>Question 148860</B>




1)



{{{5x^2-10x}}} Start with the given expression



{{{5x(x-2)}}} Factor out the GCF {{{5x}}}.


&lt;hr&gt;


2) 

Correct. You can check the answer by distributing



&lt;hr&gt;


3) 



Good so far, but can factor {{{a^2-b^2}}} into {{{(a+b)(a-b)}}}


So {{{3a^3b-3ab^3}}} completely factors to {{{3ab(a+b)(a-b)}}}



&lt;hr&gt;


4)




Looking at the expression {{{a^2+2a-24}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{2}}}, and the last term is {{{-24}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{-24}}} to get {{{(1)(-24)=-24}}}.



Now the question is: what two whole numbers multiply to {{{-24}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{2}}}?



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 {{{-24}}} (the previous product).



Factors of {{{-24}}}:

1,2,3,4,6,8,12,24

-1,-2,-3,-4,-6,-8,-12,-24



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-24}}}. For instance, {{{1*24=-24}}}, {{{2*12=-24}}}, etc.



Since {{{-24}}} 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 {{{2}}}:



&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;-24&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;1+(-24)=-23&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;-12&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;2+(-12)=-10&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;-8&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;3+(-8)=-5&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;4&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-6&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;4+(-6)=-2&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;24&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-1+24=23&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;12&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-2+12=10&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;8&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-3+8=5&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-4&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;6&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-4+6=2&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-4}}} and {{{6}}} add to {{{2}}} (the middle coefficient).



So the two numbers {{{-4}}} and {{{6}}} both multiply to {{{-24}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{2}}}



Now replace the middle term {{{2a}}} with {{{-4a+6a}}}. Remember, {{{-4}}} and {{{6}}} add to {{{2}}}. So this shows us that {{{-4a+6a=2a}}}.



{{{a^2+highlight(-4a+6a)-24}}} Replace the second term {{{2a}}} with {{{-4a+6a}}}.



{{{(a^2-4a)+(6a-24)}}} Group the terms into two pairs.



{{{a(a-4)+(6a-24)}}} Factor out the GCF {{{a}}} from the first group.



{{{a(a-4)+6(a-4)}}} Factor out {{{6}}} 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+6)(a-4)}}} Combine like terms. Or factor out the common term {{{a-4}}}


---------------------------------------------



Answer:



So {{{a^2+2a-24}}} factors to {{{(a+6)(a-4)}}}.



Note: you can check the answer by FOILing {{{(a+6)(a-4)}}} to get {{{a^2+2a-24}}} or by graphing the original expression and the answer (the two graphs should be identical).



&lt;hr&gt;


5)




Looking at the expression {{{4b^2-28b+49}}}, we can see that the first coefficient is {{{4}}}, the second coefficient is {{{-28}}}, and the last term is {{{49}}}.



Now multiply the first coefficient {{{4}}} by the last term {{{49}}} to get {{{(4)(49)=196}}}.



Now the question is: what two whole numbers multiply to {{{196}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-28}}}?



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 {{{196}}} (the previous product).



Factors of {{{196}}}:

1,2,4,7,14,28,49,98,196

-1,-2,-4,-7,-14,-28,-49,-98,-196



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{196}}}. For instance, {{{1*196=196}}}, {{{2*98=196}}}, etc.



Since {{{196}}} 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 {{{-28}}}:



&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;196&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;1+196=197&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;98&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;2+98=100&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;4&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;49&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;4+49=53&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;7&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;28&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;7+28=35&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;14&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;14&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;14+14=28&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;-196&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-1+(-196)=-197&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;-98&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-2+(-98)=-100&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-4&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-49&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-4+(-49)=-53&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-7&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-28&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-7+(-28)=-35&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;-14&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-14&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;-14+(-14)=-28&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-14}}} and {{{-14}}} add to {{{-28}}} (the middle coefficient).



So the two numbers {{{-14}}} and {{{-14}}} both multiply to {{{196}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-28}}}



Now replace the middle term {{{-28b}}} with {{{-14b-14b}}}. Remember, {{{-14}}} and {{{-14}}} add to {{{-28}}}. So this shows us that {{{-14b-14b=-28b}}}.



{{{4b^2+highlight(-14b-14b)+49}}} Replace the second term {{{-28b}}} with {{{-14b-14b}}}.



{{{(4b^2-14b)+(-14b+49)}}} Group the terms into two pairs.



{{{2b(2b-7)+(-14b+49)}}} Factor out the GCF {{{2b}}} from the first group.



{{{2b(2b-7)-7(2b-7)}}} Factor out {{{7}}} 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.



{{{(2b-7)(2b-7)}}} Combine like terms. Or factor out the common term {{{2b-7}}}



{{{(2b-7)^2}}} Collect and condense the terms.


---------------------------------------------



Answer:



So {{{4b^2-28b+49}}} factors to {{{(2b-7)^2}}}.



Note: you can check the answer by FOILing {{{(2b-7)^2}}} to get {{{4b^2-28b+49}}} or by graphing the original expression and the answer (the two graphs should be identical).



&lt;hr&gt;



6) Correct. You can check the answer by distributing
</PRE>