SOLUTION: Can you help me factor the polynomial 3a^2+a-14

Question 259024: Can you help me factor the polynomial 3a^2+a-14
Found 2 solutions by richwmiller, drk:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
(a-2)*(3a+7)

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 :

1,2,3,6,7,14,21,42

-1,-2,-3,-6,-7,-14,-21,-42

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

These factors pair up and multiply to .

1*(-42) = -42
2*(-21) = -42
3*(-14) = -42
6*(-7) = -42
(-1)*(42) = -42
(-2)*(21) = -42
(-3)*(14) = -42
(-6)*(7) = -42

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 -42 1+(-42)=-41
2 -21 2+(-21)=-19
3 -14 3+(-14)=-11
6 -7 6+(-7)=-1
-1 42 -1+42=41
-2 21 -2+21=19
-3 14 -3+14=11
-6 7 -6+7=1

From the table, we can see that the two numbers and add to (the middle coefficient).

So the two numbers and both multiply to and add to

Now replace the middle term with . Remember, and add to . So this shows us that .

Replace the second term with .

Group the terms into two pairs.

Factor out the GCF from the first group.

Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

Combine like terms. Or factor out the common term

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

Answer:

So factors to .

In other words, .

Note: you can check the answer by expanding to get or by graphing the original expression and the answer (the two graphs should be identical).

Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Here is the original problem

there are several ways to factor. The basic idea is to get (ax+b)(cx+d)
--
step 1 - multiply 3 and 14 to get 42.
we want 2 factors of 42 that add to 1, but are different signs.
we get 7 + (-6)
--
step 2 - replace 1a with 7a - 6a like this

step 3 - group the first pair and last pair and find the GCF of each like this

finding GCF's we get

notice the parenthesis are the same; this is what we want to happen.
step 4 - find another GCF as

and you have factored.
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