Question 267280
<font face="Garamond" size="+2">


First let's check to see if your factorization is correct.  We know that if we say that *[tex \Large a] and *[tex \Large b] are the factors of *[tex \Large c], then we must be able to multiply *[tex \Large a] times *[tex \Large b] and get *[tex \Large c] as a result.  Let's check your factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(5x\ +\ 2\right)\left(5x\right)\ =\ 25x^2\ +\ 10x]


But you started with *[tex \LARGE 25x^2\ +\ 10x\ +\ 1] and we can say for certain that *[tex \LARGE 25x^2\ +\ 10x\ +\ 1\ \neq\ 25x^2\ +\ 10x] (At least until someone invents a mathematics where 1 equals 0).


No wonder you are lost.


Let's see:  5 times 5 is 25, 1 times 1 is 1, and 5 plus 5 is 10.  So how about:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(5x\ +\ 1\right)\left(5x\ +\ 1\right)]


Let's check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\,\cdot\,5x\ =\ 25x^2\ \ \ ] First


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\,\cdot\,1\ =\ 5x\ \ \ ] Outside


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\,\cdot\,5x\ =\ 5x\ \ \ ] Inside


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\,\cdot\,1\ =\ 1\ \ \ ] Last


Putting it all together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25x^2\ +\ 5x\ +\ 5x\ +\ 1]


Collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25x^2\ +\ 10x\ +\ 1]


Yep, that checks.  Your factors are indeed:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(5x\ +\ 1\right)\left(5x\ +\ 1\right)]


Which can also be expressed:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(5x\ +\ 1\right)^2]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>