Question 880974
Let's factor {{{25y^2-10y+1}}}



Looking at the expression {{{25y^2-10y+1}}}, we can see that the first coefficient is {{{25}}}, the second coefficient is {{{-10}}}, and the last term is {{{1}}}.



Now multiply the first coefficient {{{25}}} by the last term {{{1}}} to get {{{(25)(1)=25}}}.



Now the question is: what two whole numbers multiply to {{{25}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-10}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{25}}} (the previous product).



Factors of {{{25}}}:

1,5,25

-1,-5,-25



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



These factors pair up and multiply to {{{25}}}.

1*25 = 25
5*5 = 25
(-1)*(-25) = 25
(-5)*(-5) = 25


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-10}}}:



<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>25</font></td><td  align="center"><font color=black>1+25=26</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>5+5=10</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-25</font></td><td  align="center"><font color=black>-1+(-25)=-26</font></td></tr><tr><td  align="center"><font color=red>-5</font></td><td  align="center"><font color=red>-5</font></td><td  align="center"><font color=red>-5+(-5)=-10</font></td></tr></table>



From the table, we can see that the two numbers {{{-5}}} and {{{-5}}} add to {{{-10}}} (the middle coefficient).



So the two numbers {{{-5}}} and {{{-5}}} both multiply to {{{25}}} <font size=4><b>and</b></font> add to {{{-10}}}



Now replace the middle term {{{-10y}}} with {{{-5y-5y}}}. Remember, {{{-5}}} and {{{-5}}} add to {{{-10}}}. So this shows us that {{{-5y-5y=-10y}}}.



{{{25y^2+highlight(-5y-5y)+1}}} Replace the second term {{{-10y}}} with {{{-5y-5y}}}.



{{{(25y^2-5y)+(-5y+1)}}} Group the terms into two pairs.



{{{5y(5y-1)+(-5y+1)}}} Factor out the GCF {{{5y}}} from the first group.



{{{5y(5y-1)-1(5y-1)}}} Factor out {{{1}}} 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.



{{{(5y-1)(5y-1)}}} Combine like terms. Or factor out the common term {{{5y-1}}}



{{{(5y-1)^2}}} Condense the terms.



So {{{25y^2-10y+1}}} factors to {{{(5y-1)^2}}}.



In other words, {{{25y^2-10y+1=(5y-1)^2}}}.



Note: you can check the answer by expanding {{{(5y-1)^2}}} to get {{{25y^2-10y+1}}} or by graphing the original expression and the answer (the two graphs should be identical).



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So {{{y(25y^2-10y+1) }}} turns into {{{y(5y-1)^2 }}}


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<font color="red">Final Answer:</font> {{{25y^3-10y^2+y }}} completely factors to {{{y(5y-1)^2}}}