Question 166139
{{{(1/9)x^2+(2/3)x-63}}} Start with the given expression



{{{(1/9)(x^2+6x-567)}}} Factor out the GCF {{{1/9}}}. This will make every term on the inside of the parenthesis have a whole coefficient.



Now let's factor {{{x^2+6x-567}}}


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Looking at the expression {{{x^2+6x-567}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{6}}}, and the last term is {{{-567}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{-567}}} to get {{{(1)(-567)=-567}}}.



Now the question is: what two whole numbers multiply to {{{-567}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{6}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-567}}} (the previous product).



Factors of {{{-567}}}:

1,3,7,9,21,27,63,81,189,567

-1,-3,-7,-9,-21,-27,-63,-81,-189,-567



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-567}}}.

1*(-567)
3*(-189)
7*(-81)
9*(-63)
21*(-27)
(-1)*(567)
(-3)*(189)
(-7)*(81)
(-9)*(63)
(-21)*(27)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{6}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>-567</font></td><td  align="center"><font color=black>1+(-567)=-566</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>-189</font></td><td  align="center"><font color=black>3+(-189)=-186</font></td></tr><tr><td  align="center"><font color=black>7</font></td><td  align="center"><font color=black>-81</font></td><td  align="center"><font color=black>7+(-81)=-74</font></td></tr><tr><td  align="center"><font color=black>9</font></td><td  align="center"><font color=black>-63</font></td><td  align="center"><font color=black>9+(-63)=-54</font></td></tr><tr><td  align="center"><font color=black>21</font></td><td  align="center"><font color=black>-27</font></td><td  align="center"><font color=black>21+(-27)=-6</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>567</font></td><td  align="center"><font color=black>-1+567=566</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>189</font></td><td  align="center"><font color=black>-3+189=186</font></td></tr><tr><td  align="center"><font color=black>-7</font></td><td  align="center"><font color=black>81</font></td><td  align="center"><font color=black>-7+81=74</font></td></tr><tr><td  align="center"><font color=black>-9</font></td><td  align="center"><font color=black>63</font></td><td  align="center"><font color=black>-9+63=54</font></td></tr><tr><td  align="center"><font color=red>-21</font></td><td  align="center"><font color=red>27</font></td><td  align="center"><font color=red>-21+27=6</font></td></tr></table>



From the table, we can see that the two numbers {{{-21}}} and {{{27}}} add to {{{6}}} (the middle coefficient).



So the two numbers {{{-21}}} and {{{27}}} both multiply to {{{-567}}} <font size=4><b>and</b></font> add to {{{6}}}



Now replace the middle term {{{6x}}} with {{{-21x+27x}}}. Remember, {{{-21}}} and {{{27}}} add to {{{6}}}. So this shows us that {{{-21x+27x=6x}}}.



{{{x^2+highlight(-21x+27x)-567}}} Replace the second term {{{6x}}} with {{{-21x+27x}}}.



{{{(x^2-21x)+(27x-567)}}} Group the terms into two pairs.



{{{x(x-21)+(27x-567)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x-21)+27(x-21)}}} Factor out {{{27}}} 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+27)(x-21)}}} Combine like terms. Or factor out the common term {{{x-21}}}






So {{{x^2+6x-567}}} factors to {{{(x+27)(x-21)}}}.

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So {{{(1/9)(x^2+6x-567)}}} becomes {{{(1/9)(x+27)(x-21)}}}




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Answer:


So {{{(1/9)x^2+(2/3)x-63}}} completely factors to {{{(1/9)(x+27)(x-21)}}}