<PRE><B>Question 244032</B>
Are you sure that the exponents aren&#39;t twos? The cubics that you&#39;ve written out aren&#39;t factorable.



If the first one is {{{7x^2-15x+2}}}, then 



Looking at the expression {{{7x^2-15x+2}}}, we can see that the first coefficient is {{{7}}}, the second coefficient is {{{-15}}}, and the last term is {{{2}}}.



Now multiply the first coefficient {{{7}}} by the last term {{{2}}} to get {{{(7)(2)=14}}}.



Now the question is: what two whole numbers multiply to {{{14}}} (the previous product) &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to the second coefficient {{{-15}}}?



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 {{{14}}} (the previous product).



Factors of {{{14}}}:

1,2,7,14

-1,-2,-7,-14



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{14}}}.

1*14
2*7
(-1)*(-14)
(-2)*(-7)


Now let&#39;s add up each pair of factors to see if one pair adds to the middle coefficient {{{-15}}}:



&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;14&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;1+14=15&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;7&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;2+7=9&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-1&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-14&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=red&gt;-1+(-14)=-15&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;-7&lt;/font&gt;&lt;/td&gt;&lt;td  align=&quot;center&quot;&gt;&lt;font color=black&gt;-2+(-7)=-9&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;



From the table, we can see that the two numbers {{{-1}}} and {{{-14}}} add to {{{-15}}} (the middle coefficient).



So the two numbers {{{-1}}} and {{{-14}}} both multiply to {{{14}}} &lt;font size=4&gt;&lt;b&gt;and&lt;/b&gt;&lt;/font&gt; add to {{{-15}}}



Now replace the middle term {{{-15x}}} with {{{-x-14x}}}. Remember, {{{-1}}} and {{{-14}}} add to {{{-15}}}. So this shows us that {{{-x-14x=-15x}}}.



{{{7x^2+highlight(-x-14x)+2}}} Replace the second term {{{-15x}}} with {{{-x-14x}}}.



{{{(7x^2-x)+(-14x+2)}}} Group the terms into two pairs.



{{{x(7x-1)+(-14x+2)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(7x-1)-2(7x-1)}}} Factor out {{{2}}} 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-2)(7x-1)}}} Combine like terms. Or factor out the common term {{{7x-1}}}


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Answer:



So {{{7x^2-15x+2}}} factors to {{{(x-2)(7x-1)}}}.



Note: you can check the answer by FOILing {{{(x-2)(7x-1)}}} to get {{{7x^2-15x+2}}} or by graphing the original expression and the answer (the two graphs should be identical).




As for the other one, let me know if you still need help.</PRE>