<PRE><B>Question 540990</B>
I&#39;m assuming you&#39;re wondering how to factor {{{b^2-7b+10}}} to get {{{(b-5)(b-2)}}}




Looking at the expression {{{b^2-7b+10}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-7}}}, and the last term is {{{10}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{10}}} to get {{{(1)(10)=10}}}.



Now the question is: what two whole numbers multiply to {{{10}}} (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 {{{10}}} (the previous product).



Factors of {{{10}}}:

1,2,5,10

-1,-2,-5,-10



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{10}}}.

1*10 = 10
2*5 = 10
(-1)*(-10) = 10
(-2)*(-5) = 10


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;&lt;font color=black&gt;1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+10=11&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+5=7&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-10&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-1+(-10)=-11&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-2&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-5&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-2+(-5)=-7&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-2}}} and {{{-5}}} add to {{{-7}}} (the middle coefficient).



So the two numbers {{{-2}}} and {{{-5}}} both multiply to {{{10}}} &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 {{{-2b-5b}}}. Remember, {{{-2}}} and {{{-5}}} add to {{{-7}}}. So this shows us that {{{-2b-5b=-7b}}}.



{{{b^2+highlight(-2b-5b)+10}}} Replace the second term {{{-7b}}} with {{{-2b-5b}}}.



{{{(b^2-2b)+(-5b+10)}}} Group the terms into two pairs.



{{{b(b-2)+(-5b+10)}}} Factor out the GCF {{{b}}} from the first group.



{{{b(b-2)-5(b-2)}}} 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.



{{{(b-5)(b-2)}}} Combine like terms. Or factor out the common term {{{b-2}}}



===============================================================



Answer:



So {{{b^2-7b+10}}} factors to {{{(b-5)(b-2)}}}.



In other words, {{{b^2-7b+10=(b-5)(b-2)}}}.



Note: you can check the answer by expanding {{{(b-5)(b-2)}}} to get {{{b^2-7b+10}}} or by graphing the original expression and the answer (the two graphs should be identical).



If you need more help, email me at &lt;a href=&quot;mailto:jim_thompson5910@hotmail.com&quot;&gt;jim_thompson5910@hotmail.com&lt;/a&gt;


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Jim
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