Question 289215
I'm assuming that you want to factor.




Looking at the expression {{{9t^2+30t+25}}}, we can see that the first coefficient is {{{9}}}, the second coefficient is {{{30}}}, and the last term is {{{25}}}.



Now multiply the first coefficient {{{9}}} by the last term {{{25}}} to get {{{(9)(25)=225}}}.



Now the question is: what two whole numbers multiply to {{{225}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{30}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{225}}} (the previous product).



Factors of {{{225}}}:

1,3,5,9,15,25,45,75,225

-1,-3,-5,-9,-15,-25,-45,-75,-225



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



These factors pair up and multiply to {{{225}}}.

1*225 = 225
3*75 = 225
5*45 = 225
9*25 = 225
15*15 = 225
(-1)*(-225) = 225
(-3)*(-75) = 225
(-5)*(-45) = 225
(-9)*(-25) = 225
(-15)*(-15) = 225


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{30}}}:



<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>225</font></td><td  align="center"><font color=black>1+225=226</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>75</font></td><td  align="center"><font color=black>3+75=78</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>45</font></td><td  align="center"><font color=black>5+45=50</font></td></tr><tr><td  align="center"><font color=black>9</font></td><td  align="center"><font color=black>25</font></td><td  align="center"><font color=black>9+25=34</font></td></tr><tr><td  align="center"><font color=red>15</font></td><td  align="center"><font color=red>15</font></td><td  align="center"><font color=red>15+15=30</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-225</font></td><td  align="center"><font color=black>-1+(-225)=-226</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-75</font></td><td  align="center"><font color=black>-3+(-75)=-78</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>-45</font></td><td  align="center"><font color=black>-5+(-45)=-50</font></td></tr><tr><td  align="center"><font color=black>-9</font></td><td  align="center"><font color=black>-25</font></td><td  align="center"><font color=black>-9+(-25)=-34</font></td></tr><tr><td  align="center"><font color=black>-15</font></td><td  align="center"><font color=black>-15</font></td><td  align="center"><font color=black>-15+(-15)=-30</font></td></tr></table>



From the table, we can see that the two numbers {{{15}}} and {{{15}}} add to {{{30}}} (the middle coefficient).



So the two numbers {{{15}}} and {{{15}}} both multiply to {{{225}}} <font size=4><b>and</b></font> add to {{{30}}}



Now replace the middle term {{{30t}}} with {{{15t+15t}}}. Remember, {{{15}}} and {{{15}}} add to {{{30}}}. So this shows us that {{{15t+15t=30t}}}.



{{{9t^2+highlight(15t+15t)+25}}} Replace the second term {{{30t}}} with {{{15t+15t}}}.



{{{(9t^2+15t)+(15t+25)}}} Group the terms into two pairs.



{{{3t(3t+5)+(15t+25)}}} Factor out the GCF {{{3t}}} from the first group.



{{{3t(3t+5)+5(3t+5)}}} 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.



{{{(3t+5)(3t+5)}}} Combine like terms. Or factor out the common term {{{3t+5}}}



{{{(3t+5)^2}}} Condense the terms.



===============================================================



Answer:



So {{{9t^2+30t+25}}} factors to {{{(3t+5)^2}}}.



In other words, {{{9t^2+30t+25=(3t+5)^2}}}.



Note: you can check the answer by expanding {{{(3t+5)^2}}} to get {{{9t^2+30t+25}}} or by graphing the original expression and the answer (the two graphs should be identical).