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!

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(39625) (Show Source): You can put this solution on YOUR website!
Rather use discriminant to avoid testing different combinations.
has a discriminant number, . If discriminant is a square integer, then the quadratic trinomial is factorable.

3y^2+14y+4,
check .
What happens if you form sqrt(148)?
, 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, ....
Answer by richwmiller(17219) (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 , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

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

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

Factors of :

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 .

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 :

First Number Second Number Sum
1 12 1+12=13
2 6 2+6=8
3 4 3+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 . So cannot be factored.

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

Answer:

So doesn't factor at all (over the rational numbers).

So is prime.

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

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

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

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

Factors of :

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 .

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 :

First Number Second Number Sum
1 27 1+27=28
3 9 3+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 . So cannot be factored.

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

Answer:

So doesn't factor at all (over the rational numbers).

So is prime.

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