<PRE><B>Question 299476</B>


Looking at the expression {{{x^2-7x+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 {{{-7x}}} with {{{-2x-5x}}}. Remember, {{{-2}}} and {{{-5}}} add to {{{-7}}}. So this shows us that {{{-2x-5x=-7x}}}.



{{{x^2+highlight(-2x-5x)+10}}} Replace the second term {{{-7x}}} with {{{-2x-5x}}}.



{{{(x^2-2x)+(-5x+10)}}} Group the terms into two pairs.



{{{x(x-2)+(-5x+10)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x-2)-5(x-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.



{{{(x-5)(x-2)}}} Combine like terms. Or factor out the common term {{{x-2}}}



===============================================================



Answer:



So {{{x^2-7x+10}}} factors to {{{(x-5)(x-2)}}}.



In other words, {{{x^2-7x+10=(x-5)(x-2)}}}.



Note: you can check the answer by expanding {{{(x-5)(x-2)}}} to get {{{x^2-7x+10}}} or by graphing the original expression and the answer (the two graphs should be identical).
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