SOLUTION: 1). 3y^2 +14y+4 I think it is unfactorable because I tried different pairings. 2). 9p^2-q^2+3p I think this one is also unfactorable. Thank you so much!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 1). 3y^2 +14y+4 I think it is unfactorable because I tried different pairings. 2). 9p^2-q^2+3p I think this one is also unfactorable. Thank you so much!       Log On


   

Question 879815: 1). 3y^2 +14y+4
I think it is unfactorable because I tried different pairings.
2). 9p^2-q^2+3p
I think this one is also unfactorable.
Thank you so much!
Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
Rather use discriminant to avoid testing different combinations.
ax%5E2%2Bbx%2Bc has a discriminant number, b%5E2-4ac. If discriminant is a square integer, then the quadratic trinomial is factorable.

3y^2+14y+4,
check 14%5E2-4%2A3%2A4=196-48=148.
What happens if you form sqrt(148)?
2%2Asqrt%2837%29, irrational; so the roots of the polynomial are also irrational, so 3y^2+14y+4 is not factorable (unless you want irrational constants in your binomials).

9p^2-p+3p, assuming you made a misprint when you showed -q;
Check yourself... discriminant, %28-1%29%5E2-4%2A9%2A3....

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 3y%5E2%2B14y%2B4, we can see that the first coefficient is 3, the second coefficient is 14, and the last term is 4.



Now multiply the first coefficient 3 by the last term 4 to get %283%29%284%29=12.



Now the question is: what two whole numbers multiply to 12 (the previous product) and add to the second coefficient 14?



To find these two numbers, we need to list all of the factors of 12 (the previous product).



Factors of 12:

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12



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



These factors pair up and multiply to 12.

1*12 = 12
2*6 = 12
3*4 = 12
(-1)*(-12) = 12
(-2)*(-6) = 12
(-3)*(-4) = 12


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 14:



First NumberSecond NumberSum
1121+12=13
262+6=8
343+4=7
-1-12-1+(-12)=-13
-2-6-2+(-6)=-8
-3-4-3+(-4)=-7




From the table, we can see that there are no pairs of numbers which add to 14. So 3y%5E2%2B14y%2B4 cannot be factored.



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Answer:



So 3%2Ay%5E2%2B14%2Ay%2B4 doesn't factor at all (over the rational numbers).



So 3%2Ay%5E2%2B14%2Ay%2B4 is prime.


Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 9p%5E2%2Bp%2B3, we can see that the first coefficient is 9, the second coefficient is 1, and the last term is 3.



Now multiply the first coefficient 9 by the last term 3 to get %289%29%283%29=27.



Now the question is: what two whole numbers multiply to 27 (the previous product) and add to the second coefficient 1?



To find these two numbers, we need to list all of the factors of 27 (the previous product).



Factors of 27:

1,3,9,27

-1,-3,-9,-27



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



These factors pair up and multiply to 27.

1*27 = 27
3*9 = 27
(-1)*(-27) = 27
(-3)*(-9) = 27


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 1:



First NumberSecond NumberSum
1271+27=28
393+9=12
-1-27-1+(-27)=-28
-3-9-3+(-9)=-12




From the table, we can see that there are no pairs of numbers which add to 1. So 9p%5E2%2Bp%2B3 cannot be factored.



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Answer:



So 9%2Ap%5E2%2Bp%2B3 doesn't factor at all (over the rational numbers).



So 9%2Ap%5E2%2Bp%2B3 is prime.