Question 207241
{{{3x^10-9x^9+6x^8}}} Start with the given expression.



{{{3x^8(x^2-3x+2)}}} Factor out the GCF {{{3x^8}}}.



Now let's try to factor the inner expression {{{x^2-3x+2}}}



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Looking at the expression {{{x^2-3x+2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-3}}}, and the last term is {{{2}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{2}}} to get {{{(1)(2)=2}}}.



Now the question is: what two whole numbers multiply to {{{2}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-3}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{2}}} (the previous product).



Factors of {{{2}}}:

1,2

-1,-2



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



These factors pair up and multiply to {{{2}}}.

1*2 = 2
(-1)*(-2) = 2


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-3}}}:



<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>2</font></td><td  align="center"><font color=black>1+2=3</font></td></tr><tr><td  align="center"><font color=red>-1</font></td><td  align="center"><font color=red>-2</font></td><td  align="center"><font color=red>-1+(-2)=-3</font></td></tr></table>



From the table, we can see that the two numbers {{{-1}}} and {{{-2}}} add to {{{-3}}} (the middle coefficient).



So the two numbers {{{-1}}} and {{{-2}}} both multiply to {{{2}}} <font size=4><b>and</b></font> add to {{{-3}}}



Now replace the middle term {{{-3x}}} with {{{-x-2x}}}. Remember, {{{-1}}} and {{{-2}}} add to {{{-3}}}. So this shows us that {{{-x-2x=-3x}}}.



{{{x^2+highlight(-x-2x)+2}}} Replace the second term {{{-3x}}} with {{{-x-2x}}}.



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



{{{x(x-1)+(-2x+2)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x-1)-2(x-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)(x-1)}}} Combine like terms. Or factor out the common term {{{x-1}}}



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So {{{3x^8(x^2-3x+2)}}} then factors further to {{{3x^8(x-2)(x-1)}}}



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Answer:



So {{{3x^10-9x^9+6x^8}}} completely factors to {{{3x^8(x-2)(x-1)}}}.



In other words, {{{3x^10-9x^9+6x^8=3x^8(x-2)(x-1)}}}.