SOLUTION: Hello, May I ask for help with the following questions? Factor the trinomials- 6x^2+31x+5 I think that this cannot be factored. Also, X^2-8xy+12y^2 I can't seem to m

Question 725215: Hello,
May I ask for help with the following questions?
Factor the trinomials-
6x^2+31x+5
I think that this cannot be factored.
Also,
X^2-8xy+12y^2
I can't seem to move on from these two questions. Thank you for your help.
Found 3 solutions by stanbon, lynnlo, solver91311:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
6x^2+31x+5
Since 5*6 = 30*1 and 30+1 = 31, you can use reverse-FOIL:
6x^2 + 30x + x+ 5
----
6x(x+5) + (x+5)
-----
= (x=5)(6x+10
=======================
X^2-8xy+12y^2
Since -2*-6 = 12 and -2+-6 = -8, you can use reverse-FOIL
---
x^2 -6xy-2xy + 12y^2
-----
x(x-6y)-2y(x-6y)
------
= (x-6y)(x-2y)
==================
Cheers,
Stan H.

Answer by lynnlo(4176)   (Show Source): You can put this solution on YOUR website!
(1)6x^2+31x+5
5+31x+6x^2
(5+x)(1+6x)
(2)x^2-8xy+12y^2
(x-6y)(x-2y)
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

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:

Then group the two pairs of factors:

Factor the GCF out of each group:

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

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it

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