Question 725215
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Of course it can be factored.  Here is how to tell quickly.  Square the coefficient on the first degree term:  31 times 31 is 961.  Multiply 4 times the lead coefficient times the constant term.  4 times 6 times 5 is 120.  Subtract the second result from the first:  961 minus 120 is 841.  Then take the square root of that result.  If it is a perfect square, and in this case square root of 841 is exactly 29, then the quadratic trinomial is factorable over the rational numbers.


Here is a process that you might try.  Multiply the lead coefficient times the constant term.  List the two number factorizations of this product.  Find a pair of factors that when summed equal the first degree term coefficient.  6 times 5 is 30.  30 times 1 is 30, and 30 plus 1 is 31 which is your middle term coefficient.  Re-write your trinomial into a four term polynomial splitting the middle term as suggested by the above factorization:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x^2\ +\ 30x\ +\ x\ +\ 5]


Then group the two pairs of factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(6x^2\ +\ 30x\right)\ +\ \left(x\ +\ 5\right)]


Factor the GCF out of each group:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x\left(x\ +\ 5)\ +\ 1\left(x\ +\ 5\right)]


Properly performed, the two binomial factors should be identical for a factorable quadratic, so factor out the binomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(6x\ +\ 1\right)\left(x\ +\ 5\right)\ =\ 6x^2\ +\ 31x\ +\ 5]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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